11. Lyapunov Certificates, Safe Model-Based RL & Learning-Based MPC

Regions of attraction under uncertainty, Lyapunov-based safe RL, neural certificates, learning-based and certified approximate MPC, statistical robust synthesis

Before you start

This module assumes:

A route through this module

First build the core calculation: Lyapunov decrease and sublevel sets, uncertain decrease checks, the sample-to-region argument, recursive feasibility. Try the Easy practice as you go, then the Medium applications. Then study expected-budget Lyapunov functions, learned certificates and approximate MPC. The final statistical section explains the probability attached to a robust-control certificate.

The Hard practice checks proofs and assumptions. The original longer exercises remain for a deeper pass. If a calculation stalls, follow its prerequisite review link and return after practising that skill.

Background — check the skills used next

Check quadratic decrease, ellipsoid membership and the distinction between a mean estimate and an uncertainty bound.

Show hint

Evaluate $V$ at the successor before subtracting its current value.

Show worked check

For $x^+=x/2$ and $V=x^2$, $\Delta V=x^2/4-x^2=-3x^2/4$. The set $V\le1$ is $[-1,1]$ and maps into $[-1/2,1/2]$. If a model predicts only a mean successor, uncertainty must still be included before declaring this decrease valid. Review Lyapunov functions, quadratic forms, and confidence bounds.

Book contents · Apply this chapter to a real decision · Glossary

Contents
1. Lyapunov Stability and Regions of Attraction 2. Safe Model-Based RL with Stability Guarantees (Berkenkamp et al.) 3. Lyapunov-Based Safe RL (Chow et al.) 4. Neural Lyapunov, Barrier and Contraction Certificates 5. Learning-Based and Robust MPC 6. Certified Approximate MPC with Neural Networks 7. Statistical Guarantees Meet Robust Control (Fiedler, Scherer, Trimpe) Walkthrough: Certifying a Region of Attraction from Samples Interactive: Certified Invariant Set Under Model Uncertainty Application lab & chapter review Exercises Graded practice: Easy → Medium → Hard Key Papers Flashcards

Module 10 certified that a set is never left; this module also asks that the state is brought back. A Lyapunov function certifies both, and "safe" becomes "inside a certified region of attraction": the pendulum may swing but never falls. Berkenkamp et al. certify such a region from finitely many GP-modelled samples, so an RL agent can improve its policy and explore without leaving it. Chow et al. use a Lyapunov function for the constraint budget of a CMDP, turning a trajectory-level constraint into one linear constraint per state. Neural certificates learn Lyapunov and barrier functions with the controller and verify them afterwards. Model predictive control supplies the robust, recursively feasible backbone, and the Trimpe and Pauli groups show how to keep its guarantees when the MPC is replaced by a network or combined with learned uncertainty bounds.

Notation for this module (and clashes)
State $x\in X\subseteq\mathbb R^q$, input $u\in U\subseteq\mathbb R^p$, $x_{t+1}=f(x_t,u_t)$, policy $\pi$, closed loop $F_\pi(x):=f(x,\pi(x))$, $F_\pi(0)=0$ (see Primer D). Lyapunov function $V$ (Berkenkamp et al.: $v$), sublevel set $\mathcal V(c):=\{x\in X:V(x)\le c\}$, $\Delta V(x):=V(F_\pi(x))-V(x)$, region of attraction $\mathcal R_\pi$. Lipschitz constants $L_V,L_f,L_\pi,L_{\Delta V}$ (see Primer B) refer to the 1-norm (see Primer A), as in the source. GP notation of Module 3: $\mu_n$, $\sigma_n$, band $\mu_n\pm\beta_n\sigma_n$ (Chowdhury–Gopalan convention), information gain $\gamma_n$.
Reference — notation changes across papers
Clashes. Berkenkamp et al. write $f=h+g$ with known prior $h$ and unknown error $g$: not the barrier $h$ of Module 10, not the SafeOpt threshold, not the input matrix $g$ of $\dot x=f(x)+g(x)u$. $\gamma$ is a discount in Sections 2–3 but the information gain inside $\beta_n$; $\sigma$ without index is a noise scale. Section 3 translates Chow et al. into Module 8's CMDP notation ($s$, $r$, constraint cost $c$, budget $d$). In Sections 5–7, $N$ is an MPC horizon or a sample size and $\beta$ the scenario-approach confidence, unrelated to $\beta_n$. $\lambda$ is a Lagrange multiplier in Sections 2–3, a contraction rate in Section 4's table, an input Lipschitz constant in Section 6 (Hertneck et al.'s notation), and the GP regulariser (likelihood noise variance) in Section 7.

1. Lyapunov Stability and Regions of Attraction

Stability is a safety notion because it encodes recoverability: a controller that stabilises the upright pendulum on a set $S$ brings any state of $S$ home, so an exploratory action that stays in $S$ can always be undone (safety notion (e) of Module 1). Everything below is discrete time, because learning algorithms see sampled data.

Definition — Stability, region of attraction, inner estimate
For $x_{t+1}=F_\pi(x_t)$ with $F_\pi(0)=0$, the origin is Lyapunov stable if for every $\varepsilon\gt0$ there is $\delta\gt0$ with $\|x_0\|\lt\delta\Rightarrow\|x_t\|\lt\varepsilon$ for all $t$, and asymptotically stable if moreover $x_t\to0$ for all $x_0$ near $0$. Its region of attraction is $\mathcal R_\pi:=\{x_0\in X:\lim_{t\to\infty}x_t=0\}$. A set $S$ is forward invariant if $x_0\in S\Rightarrow x_t\in S$ for all $t$; a forward-invariant $S\subseteq\mathcal R_\pi$ is an inner estimate of the ROA, and certifying one is what "safe" means in Section 2. A Lyapunov candidate is a continuous, positive definite $V$ ($V(0)=0$, $V(x)\gt0$ otherwise); along the loop it changes by $\Delta V(x)$, the discrete analogue of the Lie derivative $\dot V=\nabla V^\top f$ of Module 10.
Theorem — Sublevel sets with strict decrease are inner estimates of the ROA (Berkenkamp et al. 2017, Thm. 1, after Khalil)
Assumptions. (A1) $F_\pi$ is continuous with $F_\pi(0)=0$, defined on a neighbourhood of the sublevel set, with successors in $X$; $0$ is an interior point of $X$. (A2) $V$ is continuous and positive definite. (A3) $\mathcal V(c)$ is compact (closed and bounded; see Primer 0) for some $c\gt0$. (A4) $\Delta V(x)\lt0$ for all $x\in\mathcal V(c)\setminus\{0\}$.
Statement. $\mathcal V(c)$ is forward invariant, every trajectory starting in it converges to the origin, so $\mathcal V(c)\subseteq\mathcal R_\pi$, and the origin is asymptotically stable.
In words. One strict inequality on a set replaces an infinite-horizon convergence test. Berkenkamp et al., NeurIPS 2017 state it for Lipschitz $f$ and continuously differentiable $V$; the list above is what the proof uses.

Proof idea (full proof: Exercise 11.1). Invariance: $V(x_t)\le c\Rightarrow V(x_{t+1})\le V(x_t)\le c$ (strict unless $x_t=0$). Convergence: $V(x_t)$ decreases to some $v^\ast\ge0$ (a bounded monotone sequence; see Primer 0); if $v^\ast\gt0$, the trajectory stays in the compact annulus $\{v^\ast\le V\le c\}$ where the continuous $\Delta V$ has a maximum $-\eta\lt0$, so $V$ would drop by $\eta$ every step. Hence $V(x_t)\to0$ and, by positive definiteness and compactness, $x_t\to0$.

Proof — The analysis facts behind the proof idea, and Lyapunov stability

Facts used (Primer 0). In $\mathbb R^q$ a set is compact exactly when it is closed and bounded, and a continuous function on a nonempty compact set attains its minimum and maximum. The annulus $\{v^\ast\le V\le c\}$ is a closed subset of the compact $\mathcal V(c)$ and excludes the origin (because $V(0)=0\lt v^\ast$), so the continuous $\Delta V$, negative on it by (A4), has a maximum $-\eta\lt0$ there. A non-increasing real sequence that is bounded below converges.

From $V(x_t)\to0$ to $x_t\to0$. Otherwise a subsequence stays at distance at least $a\gt0$ from the origin. It lies in the compact $\mathcal V(c)$, so a further subsequence converges to some $x_*$ with $\|x_*\|\ge a$, and continuity gives $V(x_*)=\lim V(x_{t_k})=0$, contradicting positive definiteness.

Lyapunov stability. Fix $\varepsilon\gt0$. If $\{x\in\mathcal V(c):\|x\|\ge\varepsilon\}$ is nonempty, it is compact and $V\gt0$ on it, so $m_\varepsilon:=\min\{V(x):x\in\mathcal V(c),\ \|x\|\ge\varepsilon\}\gt0$; otherwise put $m_\varepsilon:=c$. Continuity of $V$ at $0$ (an interior point, by (A1)) gives $\delta\gt0$ with $\|x_0\|\lt\delta\Rightarrow V(x_0)\lt\min(c,m_\varepsilon)$. As $V(x_t)$ never increases, $V(x_t)\lt m_\varepsilon$ for all $t$, so no iterate reaches $\|x_t\|\ge\varepsilon$. With convergence, this is asymptotic stability.

Why each assumption matters. (A4) must hold on the whole set: one point with $\Delta V\ge0$ can be the exit door. (A3) excludes unbounded sublevel sets (a $V$ that is not radially unbounded; see Primer D), along which trajectories can drift to infinity while $V$ decreases. (A1) is the subtle one: with $V=x^2$ and $F(x)=x/2$ for $x\le1$, $F(x)=(1+x)/2$ for $x\gt1$, $\Delta V\lt0$ for all $x\ne0$ and all sublevel sets are compact, yet $x_0=2$ gives $x_t=1+2^{-t}\to1$: the limit point would be a fixed point of the continuous extension of $F$.

Linear systems, linearisation, and the largest certified level

For $x_{t+1}=Ax_t$ and $V=x^\top Px$ (see Primer A), $\Delta V(x)=x^\top(A^\top PA-P)x$, and the Stein equation $A^\top PA-P=-Q$, $Q\succ0$, has a unique solution $P\succ0$ exactly when $A$ is Schur stable: the discrete-time Lyapunov LMI of Module 2 (see Boyd et al., SIAM 1994). Then every sublevel set is an inner estimate and $\mathcal R_\pi=\mathbb R^q$. For a nonlinear loop $F_\pi(x)=Ax+r(x)$ (see Primer B), $A=\partial F_\pi/\partial x(0)$, the same $P$ certifies a neighbourhood of the origin.

Derivation — A local ROA estimate from the linearisation

Let $A$ be Schur stable, $A^\top PA-P=-I$, and $\|r(x)\|_2\le k\|x\|_2^2$ on $\|x\|_2\le\rho$ (true for twice continuously differentiable $F_\pi$). Expand $V(F_\pi(x))=(Ax+r)^\top P(Ax+r)$:

$$\Delta V(x)=\underbrace{x^\top(A^\top PA-P)x}_{=-\|x\|^2}+2x^\top A^\top Pr(x)+r(x)^\top Pr(x).$$

Cauchy–Schwarz, the operator norm (see Primer A) and $\|r\|\le k\|x\|^2$ bound the remainder terms:

$$\begin{aligned}\Delta V(x)&\le-\|x\|^2+2\|A^\top P\|\,k\|x\|^3+\|P\|\,k^2\|x\|^4\\&=-\|x\|^2\Big(1-2\|A^\top P\|k\|x\|-\|P\|k^2\|x\|^2\Big).\end{aligned}$$

Write $a=\|P\|k^2$ and $b=2\|A^\top P\|k$. The bracket $1-b\|x\|-a\|x\|^2$ equals $1$ at $x=0$ and decreases in $\|x\|$, so it is positive for $\|x\|\lt r_+$, the positive root of $ar^2+br=1$. Rationalising the quadratic formula, $r_+=(-b+\sqrt{b^2+4a})/(2a)=2/(b+\sqrt{b^2+4a})$; the second form also covers $a=0$, and $k=0$ (no remainder) gives $r_+=+\infty$. The remainder bound holds only for $\|x\|\le\rho$, so take $r^\ast=\min(\rho,r_+)$. Since $x^\top Px\ge\lambda_{\min}(P)\|x\|^2$, $V(x)\le c$ implies $\|x\|^2\le c/\lambda_{\min}(P)$: every $\mathcal V(c)$ with $c\lt\lambda_{\min}(P)\,r^{\ast2}$ lies in the ball $\|x\|\lt r^\ast$ and satisfies (A1)–(A4). The estimate is usually tiny, because it pays for the worst-case remainder in every direction; hence the sample-based certificates of Section 2.

Largest certified level. For a fixed $V$, ROA estimation is a line search over the scalar level $c$. With known $F_\pi$, the first failing level is the infimum (Primer 0; it need not be attained, and $\inf\varnothing=+\infty$)

$$c^\ast=\inf\big\{V(x):\ x\in X\setminus\{0\},\ \Delta V(x)\ge0\big\},$$

since every compact $\mathcal V(c)$ with $c\lt c^\ast$ satisfies (A4). For polynomial systems, sufficient certificates can be sought using sum-of-squares relaxations or suitable LMIs via the S-procedure (Module 2); local sector constraints are introduced in Module 2; for their later use with neural controllers, see Module 14. With unknown $f$, $\Delta V$ cannot be evaluated at all: the starting point of Section 2. Candidates come from physical energy or from value functions: if the stage cost $r(x,u)$ is positive away from the origin, the undiscounted cost-to-go (see Primer E) satisfies $J(x)=r(x,\pi(x))+J(F_\pi(x))$, so $\Delta J=-r\lt0$.

The undiscounted argument requires a finite, continuous cost-to-go with $J(0)=0$. With a discount $0\lt\gamma\lt1$, $J=r+\gamma J\circ F_\pi$ gives $J(F_\pi(x))=(J(x)-r(x,\pi(x)))/\gamma$, hence $\Delta J=\big((1-\gamma)J-r\big)/\gamma$, which can be positive where $r$ is small compared with $(1-\gamma)J$. A discounted value is therefore only a candidate whose decrease still has to be checked.

Background — Sum-of-squares certificates

A polynomial $p$ is a sum of squares (SOS) if $p=\sum_j q_j^2$ for polynomials $q_j$; this guarantees $p\ge0$. With a vector $m(x)$ of monomials, $p$ is SOS exactly when $p=m(x)^\top Qm(x)$ for some $Q\succeq0$ (factor $Q=L^\top L$, then $p=\|Lm(x)\|^2$), and matching coefficients makes the search for $Q$ an SDP. Example: $x^4+2x^2+1=(x^2+1)^2=m^\top Qm$ with $m=(1,x,x^2)^\top$ and $Q=vv^\top$, $v=(1,0,1)^\top$.

For a polynomial loop, fixed polynomial $V$ and fixed $c$, one sufficient decrease certificate is: $-\Delta V(x)-\varepsilon\|x\|^2-s(x)\big(c-V(x)\big)$ is SOS for some SOS multiplier $s$ and $\varepsilon\gt0$. Where $V(x)\le c$ the subtracted term $s(x)(c-V(x))$ is nonnegative, so $-\Delta V(x)\ge\varepsilon\|x\|^2\gt0$ for $x\ne0$: this is the S-procedure with a polynomial multiplier, and it proves decrease only on $\mathcal V(c)$, which is exactly what an ROA estimate needs. Searching jointly over $V$, $s$ and $c$ is bilinear, hence nonconvex, and failure of this sufficient test does not prove instability.

Connection to Module 10
A Lyapunov sublevel set is a barrier set: with $h(x):=c-V(x)$, (A4) gives $h(F_\pi(x))\gt h(x)$ on $\mathcal V(c)\setminus\{0\}$, so $\{h\ge0\}$ is forward invariant (Module 10); in the words of Dawson, Gao & Fan, T-RO 2023, every Lyapunov function implies a family of barrier functions defined by its sublevel sets. The Lyapunov certificate buys convergence on top of invariance, at the price of requiring decrease on the whole set, not only at its boundary.

2. Safe Model-Based RL with Stability Guarantees (Berkenkamp et al.)

Berkenkamp, Turchetta, Schoellig & Krause, NeurIPS 2017 turn Section 1 into a learning algorithm: which sublevel set can be certified with high probability from finitely many noisy evaluations of partly unknown dynamics, and how can the policy be improved and data be collected without ever leaving it?

Assumption — Setting of Berkenkamp et al. (2017), Sec. 2
Deterministic dynamics $x_{t+1}=f(x_t,u_t)=h(x_t,u_t)+g(x_t,u_t)$: known prior model $h$, unknown error $g$; noisy measurements of $f(x,u)$ are obtained by driving the system to $x$ and applying $u$.
Assumption 1 (continuity). $h$, $g$ are $L_h$-, $L_g$-Lipschitz in the 1-norm, so $f$ is $L_f$-Lipschitz with $L_f\le L_h+L_g$; policies lie in a class $\Pi_L$ of $L_\pi$-Lipschitz functions.
Assumption 2 (well-calibrated model). With posterior mean $\mu_n$, covariance $\Sigma_n$ and $\sigma_n:=\operatorname{trace}\big(\Sigma_n^{1/2}\big)$ (matrix square root and trace: Primer A; for independent outputs the sum of the marginal standard deviations, their Remark 1), there is $\beta_n\gt0$ such that with probability at least $1-\delta$, for all $n\ge0$, $x\in X$, $u\in U$: $\|f(x,u)-\mu_n(x,u)\|_1\le\beta_n\sigma_n(x,u)$.
Lyapunov function and initial policy. $V$ is continuously differentiable, positive definite and $L_V$-Lipschitz on the compact $X$ (the paper takes $\nabla V\ne0$ away from $0$ so that sublevel sets are connected, i.e. in one piece; on a restricted $X$ this does not follow, so assume connectedness of the relevant sublevel sets directly); $\pi_0$ renders the origin asymptotically stable on a small set $S_0^x$ (e.g. an LQR designed on $h$; LQR: Primer D).

Confidence intervals on $V(f)$. Let $L_V$ be a Lipschitz constant of $V$ on a set containing both the true successors $f(x,u)$ and the means $\mu_{n-1}(x,u)$. Since $|V(f)-V(\mu)|\le L_V\|f-\mu\|_1\le L_V\beta_n\sigma$, Assumption 2 gives, with probability at least $1-\delta$,

$$V\big(f(x,u)\big)\in Q_n(x,u):=\Big[V\big(\mu_{n-1}(x,u)\big)\pm L_V\beta_n\sigma_{n-1}(x,u)\Big].$$

As in SafeOpt (Module 4) the intervals are intersected, $C_n:=C_{n-1}\cap Q_n$, with $C_0=(-\infty,V(x)-L_{\Delta V}\tau)$ on the initial safe set $S_0=\{(x,\pi_0(x)):x\in S_0^x\}$ and $C_0=\mathbb R$ elsewhere; $u_n:=\sup C_n$, $l_n:=\inf C_n$ (supremum and infimum, because $C_0$ is open or unbounded; $\pm\infty$ allowed). The sound version used below takes $C_0=\mathbb R$ everywhere and intersects only valid confidence intervals: on the calibration event every $C_n$ contains the true $V(f(x,u))$, so an empty intersection signals a violated assumption. (The paper scales by $\beta_n$, not $\beta_n^{1/2}$: our convention.) The decrease condition cannot be checked at infinitely many states, so let $X_\tau\subset X$ be a grid with $\|x-[x]_\tau\|_1\le\tau$ for all $x\in X$, $[x]_\tau$ the nearest grid point.

Theorem — Discretised region-of-attraction certificate (Berkenkamp et al. 2017, Theorem 2; repaired form)
Assumptions. Assumptions 1 and 2; $V$ as above; $X_\tau$ as above; $L_{\Delta V}:=L_VL_f(L_\pi+1)+L_V$. Use valid intervals with $C_0=\mathbb R$, the domain assumptions of Section 1, a fixed current $\pi,V$, and an independent proof of strict decrease on $\mathcal V(c_0)\setminus\{0\}$ for $0\lt c_0\le c$.
Statement. If for some $n\ge0$ and $c\gt0$ $$u_n\big(x,\pi(x)\big)\lt V(x)-L_{\Delta V}\,\tau\quad\text{for all }x\in X_\tau\text{ with }c_0-L_V\tau\lt V(x)\le c+L_V\tau,$$ then with probability at least $1-\delta$, $V(f(x,\pi(x)))\lt V(x)$ for all $x\in\mathcal V(c)\setminus\{0\}$, and $\mathcal V(c)$ is a region of attraction of the true system under $\pi$. (The paper's level set $\mathcal V(c)$ excludes the origin by definition; ours includes it.)
In words. Demand a decrease greater than $L_{\Delta V}\tau$ from the upper confidence bound at every grid point in the expanded annulus: the confidence bound pays for the unknown dynamics, the margin for the unchecked states between grid points.
Why each assumption matters. Lipschitz continuity is the only bridge from finitely many points to the continuum; $L_{\Delta V}$ is the Lipschitz constant of $\Delta V$ given by the triangle inequality (walkthrough, Step 2). Calibration is the only statistical ingredient, so any calibrated model works; and because it holds jointly for all $n,x,u$, all grid points and iterations share one failure event of probability $\delta$.
Proof — Why the annular grid test certifies all of $\mathcal V(c)$

Work on the calibration event $E$ of Assumption 2 (probability at least $1-\delta$), on which $V(f(\bar x,\pi(\bar x)))\le u_n(\bar x,\pi(\bar x))$ at every grid point. Take $x\in\mathcal V(c)$ with $V(x)\gt c_0$ and its nearest grid point $\bar x=[x]_\tau$. Since $\|x-\bar x\|_1\le\tau$ and $V$ is $L_V$-Lipschitz, $c_0-L_V\tau\lt V(\bar x)\le c+L_V\tau$: $\bar x$ is one of the tested points. With $L_{\Delta V}$ the Lipschitz constant of $\Delta V$ (walkthrough, Step 2),

$$\Delta V(x)\le\Delta V(\bar x)+L_{\Delta V}\tau\le u_n\big(\bar x,\pi(\bar x)\big)-V(\bar x)+L_{\Delta V}\tau\lt0.$$

On $\mathcal V(c_0)\setminus\{0\}$ decrease holds by the independent certificate. Hence $\Delta V\lt0$ on all of $\mathcal V(c)\setminus\{0\}$, and the theorem of Section 1 gives invariance and convergence. If the largest passing level is not attained, any passing level below the supremum can be used.

Caveat — three things the statement glosses over
  • Near the origin the test must fail: on the good event $u_n\ge V(f)\ge0$, so no grid point with $V(x)\le L_{\Delta V}\tau$ passes. The source attempts to make its original theorem usable because $C_0$ declares $S_0$ to pass by fiat: the neighbourhood of the origin is meant to be certified by prior knowledge, not by data (the CDC 2016 version estimates $S_n=S_0\cup\mathcal V(c_n)$). Strictly, for pairs in $S_0$ with $V(x)\le L_{\Delta V}\tau$ the interval $C_0=(-\infty,V(x)-L_{\Delta V}\tau)$ cannot contain $V(f)\ge0$, so the paper's Corollary 1 ($V(f)\in C_n$ with probability $1-\delta$) fails there. Knowing that $S_0^x$ lies in the region of attraction of $\pi_0$ does not repair this: it neither makes $C_0$ a valid confidence interval nor gives $\Delta V\lt0$ for the chosen $V$, and it says nothing about a new policy. A sound version (our reading) initialises $C_0=\mathbb R$ everywhere, certifies the current policy separately on an inner sublevel set $\mathcal V(c_0)$ (decrease of the same $V$ proved from prior knowledge, e.g. a robust version of the linearisation argument of Section 1 for a policy that agrees with the LQR $\pi_0$ there), and applies the grid test with valid intervals to all grid points with $c_0-L_V\tau\lt V\le c+L_V\tau$, which covers every cell outside $\mathcal V(c_0)$ (so $c_0$ must exceed roughly $(L_{\Delta V}+L_V)\tau$).
  • The boundary of the level set: the proof applies the grid condition at $[x]_\tau$ for every $x\in\mathcal V(c)$, but near $\partial\mathcal V(c)$ that grid point can lie just outside $\mathcal V(c)$, where the hypothesis says nothing. As printed, the proof does not address this; checking the grid condition on $\mathcal V(c+L_V\tau)\cap X_\tau$ repairs it, since $V([x]_\tau)\le V(x)+L_V\tau$ (our reading; the explorer uses the repaired test).
  • Practice uses other constants: the experiments fix $\beta_n=2$, which the authors say "corresponds to a high-probability decrease condition per-state, rather than jointly over the state space", and local Lipschitz constants. The theoretical $\beta_n=B_g+4\sigma\sqrt{\gamma_n+1+\ln(1/\delta)}$ (Lemma 3) needs a known RKHS-norm bound $B_g$ (Module 5 shows what heuristics cost) and pairs a $\sigma^2$-regularised posterior with a Chowdhury–Gopalan-type bound stated for regulariser $1+\eta$; for any regulariser use the corrected bound of Fiedler et al. (Module 3).

Safe policy optimisation and safe exploration

Safety is a property of states under a fixed policy, so the policy is chosen together with the level. The following are the source equations (2–3); for use here, replace their grid constraint by the repaired annular test and current-policy inner certificate above:

$$\begin{aligned}&\mathcal D_n=\big\{(x,u)\in X_\tau\times U:\ u_n(x,u)-V(x)\lt-L_{\Delta V}\tau\big\},\\&(\pi_n,c_n)=\operatorname*{arg\,max}_{\pi\in\Pi_L,\ c\gt0}c\ \ \text{s.t.}\ \ (x,\pi(x))\in\mathcal D_n\ \ \forall x\in\mathcal V(c)\cap X_\tau .\end{aligned}$$

Theorem 3 follows from a sound version of Theorem 2 (in this page, use the repaired test): with probability at least $1-\delta$, $\mathcal V(c_n)\subseteq\mathcal R_{\pi_n}$ for all $n\gt0$. Problem (3) is intractable and ignores performance, so the practical algorithm minimises the Lagrangian (their eq. 7; Lagrangian penalties: Primer B)

$$\sum_{x\in X_\tau}\Big[r(x,\pi_\theta(x))+\gamma J_{\pi_\theta}\big(\mu_{n-1}(x,\pi_\theta(x))\big)+\lambda\big(u_n(x,\pi_\theta(x))-V(x)+L_{\Delta V}\tau\big)\Big]$$

over $\pi_\theta\in\Pi_L$, with $r\ge0$ a cost, $J_{\pi_\theta}$ the discounted cost-to-go under the mean dynamics, $\lambda=1$, and the value function as Lyapunov function, $V=J$; $c_n$ is computed afterwards for the fixed policy, and one reverts to the previous (certified) policy if the set shrinks. Two consequences: a discounted $J$ need not decrease at all (Section 1), which is exactly what the certification step checks; and since $V=J$ changes with the policy, the intervals $Q_n$ for $V(f)$ must be recomputed for the new $V$ before certifying. Since $L_{\Delta V}$ grows with $L_\pi$, the constraint term also regularises the policy's Lipschitz constant. For exploration, safety is generalised from already-safe pairs (eq. 4), the pairs whose successor provably stays in $\mathcal V(c_n)$ form $S_n$ (eq. 5), and the most uncertain one is sampled (eq. 6):

$$\begin{aligned}\mathcal D_n&=\bigcup_{(x,u)\in S_{n-1}}\big\{z':\ u_n(x,u)-V(x)+L_{\Delta V}\|z'-(x,u)\|_1\lt-L_{\Delta V}\tau\big\},\\S_n&=\bigcup_{z\in S_{n-1}}\big\{z'\in(\mathcal V(c_n)\cap X_\tau)\times U_\tau:\ u_n(z)+L_VL_f\|z-z'\|_1\le c_n\big\},\end{aligned}$$
$$(x_n,a_n)\in\operatorname*{arg\,max}_{(x,u)\in S_n}\ u_n(x,u)-l_n(x,u),$$

with $U_\tau$ a discretised action set, $z=(x,u)$ and $z'=(x',u')$ stacked state–input pairs, and $\pi_n$ as backup policy at the boundary of $\mathcal V(c_n)$ (the backup idea of Module 6). The sampled input is written $a_n$ because $u_n(x,u)$ already denotes the upper confidence bound.

Definition — The oracle baseline $\mathcal R_\epsilon(S_0)$ of Theorem 4 (Berkenkamp et al. 2017, App. A.3)

Everything lives on the finite set $Z=X_\tau\times U_\tau$ of grid pairs, with $V$ and the Lipschitz constants fixed; $S_0\subseteq Z$ is the initial safe set. The oracle knows $v_f(z):=V(f(z))$ to accuracy $\epsilon$ on the pairs it may visit. From a set $S$ of such pairs it (i) generalises the decrease condition by Lipschitz continuity (their eq. 10),

$$D_\epsilon(S)=S_0\cup\big\{z'\in Z:\ \exists z=(x,u)\in S,\ v_f(z)-V(x)+\epsilon+L_{\Delta V}\|z'-z\|_1\lt-L_{\Delta V}\tau\big\},$$

(ii) takes the largest level $c_\epsilon(S)$ for which some $\pi\in\Pi_L$ has $(x,\pi(x))\in D_\epsilon(S)$ at every grid state of $\mathcal V(c)$ (eq. 11; assumed attained, ties broken by a fixed rule), and (iii) adds every pair whose successor provably returns into that level set (eq. 12):

$$R_\epsilon(S)=S\cup\big\{z'\in(\mathcal V(c_\epsilon(S))\cap X_\tau)\times U_\tau:\ \exists z\in S,\ v_f(z)+\epsilon+L_VL_f\|z-z'\|_1\le c_\epsilon(S)\big\}.$$

Iterating gives nested sets $S_0\subseteq R_\epsilon(S_0)\subseteq R_\epsilon(R_\epsilon(S_0))\subseteq\cdots$ inside the finite $Z$, so they stop growing after finitely many steps; the final set is $\mathcal R_\epsilon(S_0)$, and $\mathcal R_0(S_0)$ is the perfect oracle's ($\epsilon=0$). $|\mathcal R_0(S_0)|$ counts pairs, not volume. Equations (4)–(5) above are the same operations with $u_n$ in place of $v_f+\epsilon$.

Theorem — Safe exploration (Berkenkamp et al. 2017, Theorem 4)
Assumptions. Those of Theorem 2 as printed (with the seed $S_0$ passing by fiat, see the caveat above); $\sigma$-sub-Gaussian noise; $\|g\|_k\le B_g$; $\beta_n=B_g+4\sigma\sqrt{\gamma_n+1+\ln(1/\delta)}$; data collected by (6) within the Lipschitz-based sets (4)–(5), every selected pair being safely reachable (see Algorithm 1). Let $n^\ast$ be the smallest integer with $$\frac{n^\ast}{\beta_{n^\ast}^2\gamma_{n^\ast}}\ge\frac{C\,q\,L_V^2\,\big(|\mathcal R_0(S_0)|+1\big)}{\epsilon^2},\qquad C=\frac{8}{\log(1+\sigma^{-2})},$$ which exists when $\beta_n^2\gamma_n/n\to0$; $\mathcal R_\epsilon(S_0)$ and $\mathcal R_0(S_0)$ are the oracle sets of the box above. (The printed theorem has $L_V^2$ in the denominator and an unsubscripted $\mathcal R(S_0)$. Its appendix bounds the interval widths by $u_n-l_n\le L_V\sqrt{C q\beta_n^2\gamma_n/(n-n_0)}$ (Corollary 2), so $L_V^2$ belongs in the numerator, and its counting argument (Lemma 10, Corollary 4) runs over the elements of $\mathcal R_0(S_0)$, the set that contains every $S_n$ (Lemma 9). The display is this appendix-consistent correction: our reading.)
Statement. For any $\epsilon\gt0$, $\delta\in(0,1)$, jointly with probability at least $1-\delta$ for all $n\gt0$: (i) $\mathcal V(c_n)\subseteq\mathcal R_{\pi_n}$; (ii) $f(x,u)\in\mathcal R_{\pi_n}$ for all $(x,u)\in S_n$; (iii) $\mathcal R_\epsilon(S_0)\subseteq S_{n^\ast}\subseteq\mathcal R_0(S_0)$.
In words. Every sample is safe, and after finitely many samples the algorithm certifies as much as an $\epsilon$-accurate oracle, never more than a perfect one. The proof counts: while $S_n$ stops growing, sampling the widest interval drives all widths on $S_n$ below $\epsilon$ (Corollary 2), which unlocks a new pair, and there are at most $|\mathcal R_0(S_0)|$ pairs to unlock.
Scope. The theorem concerns the source's idealised sets with the seed-by-fiat convention. The repaired certificate (annular test, independent inner certificate) keeps the safety parts (i)–(ii), but part (iii) and the sample bound would need a new proof for it.
Algorithm 1: Safe Lyapunov Learning (practical version of Berkenkamp et al. 2017, with the repaired certificate)
  1. Input: initial policy $\pi_0$ with a certified inner decrease region for $V$, valid $C_0=\mathbb R$, GP model $\mathcal{GP}(\mu(z),k(z,z'))$ with prior mean given by $h$
  2. for $n=1,2,\dots$ do
  3. compute $\pi_n$ by stochastic gradient descent on the Lagrangian (7)
  4. prove that the current $\pi_n$ decreases $V$ on $\mathcal V(c_0)\setminus\{0\}$, then take the largest tested $c_n\ge c_0$ with $u_n(x,\pi_n(x))-V(x)\lt-L_{\Delta V}\tau$ at every grid node with $c_0-L_V\tau\lt V(x)\le c_n+L_V\tau$; if none passes, keep the previous certified policy, $V$ and level // line search on the level
  5. $S_n=\{(x,u)\in\mathcal V(c_n)\times U_\tau:\ u_n(x,u)\le c_n\}$ // successor provably inside $\mathcal V(c_n)$
  6. select $(x_n,a_n)\in S_n$ by rule (6) among pairs whose state can be reached along a certified safe trajectory (or by an allowed reset); use $\pi_n$ as backup // $S_n$ certifies the successor of $(x,u)$, not a safe path to $x$
  7. update the GP with the measurement $f(x_n,a_n)+\epsilon_n$; if $V$ changes, recompute the intervals for the new $V$

The repaired practical version has the safety guarantee when all certification assumptions hold; a fixed penalty or a heuristic GP band alone does not establish them. The source exploration guarantee concerns its idealised sets and is not automatically inherited by this repair. On a simulated torque-limited pendulum with a wrong prior (lower mass, no friction), a policy network with two hidden layers of 32 ReLU units (see Primer E) is improved from 50 safely collected points; the certified set grows inside the true region of attraction and the pendulum never falls. The authors name the main limitation: grid verification suffers from the curse of dimensionality.

The precursor (CDC 2016). Berkenkamp, Moriconi, Schoellig & Krause, CDC 2016 treat a fixed policy in continuous time: with $\|g_\pi\|_k^2\le B_g$ (Srinivas convention, $\beta_n^{1/2}$, $\beta_n=2B_g+300\gamma_n\log^3(n/\delta)$ in their Lemma 1; their Assumption 1 writes the unsquared $\|g_\pi\|_k\le B_g$), $\dot V=\nabla V^\top(f_\pi+g_\pi)$ is affine in $g_\pi$, hence a GP (see Primer C); their Theorem 1 certifies the largest $c$ with $u_n(x)\lt-L\tau$ on $\mathcal V(c)\cap X_\tau$ ($L$ a Lipschitz constant of $\dot V$), $S_n=S_0\cup\mathcal V(c_n)$ lies in the ROA with probability at least $1-\delta$, and data are taken where $\sigma_{n-1}$ is largest. The experiment models the error as a GP sample with $\beta_n^{1/2}=2$ and $\tau=0.002$: a Bayesian, pointwise reading of the band, not the frequentist assumption of the theorem (Module 3).

Connection to Modules 4 and 6
This is SafeOpt (Module 4) with the Lyapunov decrease as safety function: intersected confidence intervals, Lipschitz generalisation, a safe set that only grows, sampling of the most uncertain safe point. New is that the safe set lives in state space and is indexed by a policy, and the backup policy of GoSafe (Module 6) appears naturally as $\pi_n$ itself.

3. Lyapunov-Based Safe RL (Chow et al.)

Chow, Nachum, Duenez-Guzman & Ghavamzadeh, NeurIPS 2018 use "Lyapunov" for a different object. Safety is the CMDP constraint of Module 8, an expected cumulative cost below a budget, and the goal is a policy update that never leaves the feasible set, unlike Lagrangian methods whose intermediate policies routinely violate it. The trick: replace one trajectory-level constraint by one local, linear constraint per state, certified by a function whose expected decrease pays for the constraint cost.

Translation table (Chow et al. → this page)
Chow et al.here (Module 8)meaning
$x$, $a$, $x_0$; $c(x,a)$ minimised$s$, $a$, $s_0$; $r=-c$ maximisedstates, actions, task objective
$d(x)\in[0,D_{\max}]$, $d_0$, $D_\pi(x)$$c(s)\in[0,c_{\max}]$, $d$, $V_c^\pi(s)$constraint cost, budget, constraint value
$\epsilon(x)$, $\tilde\epsilon(x)$$\epsilon(s)$auxiliary constraint cost
The 2018 paper uses a transient, undiscounted CMDP: transient states $S'$ plus a terminal state reached within $T$ steps under every policy ($T^\ast\le T$); the 2019 follow-up is discounted.
Definition — Bellman operators, Lyapunov functions and induced policies (Chow et al. 2018, eq. 1)
$T_{\pi,h}[W](s):=\sum_a\pi(a|s)\big[h(s,a)+\sum_{s'\in S'}P(s'|s,a)W(s')\big]$ for $s\in S'$ (the policy-evaluation operator of Primer E, restricted to transient states); the constraint value $V_c^\pi(s)=\mathbb E\big[\sum_{t=0}^{T^\ast-1}c(s_t)\,\big|\,s_0=s,\pi\big]$ is the unique fixed point of $T_{\pi,c}$. For a feasible baseline $\pi_B$ ($V_c^{\pi_B}(s_0)\le d$) the Lyapunov functions are $$\mathcal L_{\pi_B}(s_0,d):=\Big\{L\ge0:\ T_{\pi_B,c}[L]\le L\text{ on }S',\ \ L=0\text{ at termination},\ \ L(s_0)\le d\Big\},$$ and the $L$-induced policies at $s$ are $\mathcal F_L(s):=\{\pi(\cdot|s)\in\Delta:\ T_{\pi,c}[L](s)\le L(s)\}$, with $\Delta=\{p\in\mathbb R^{|A|}:p_a\ge0,\ \sum_ap_a=1\}$ the probability simplex of action distributions (Primer E). Both sets are nonempty: $V_c^{\pi_B}\in\mathcal L_{\pi_B}(s_0,d)$, since it is nonnegative ($c\ge0$), zero at termination, a fixed point of $T_{\pi_B,c}$ (the inequality holds with equality) and $V_c^{\pi_B}(s_0)\le d$ by feasibility; and for every $L\in\mathcal L_{\pi_B}(s_0,d)$, $\pi_B(\cdot|s)\in\mathcal F_L(s)$, which is the defining inequality $T_{\pi_B,c}[L]\le L$ read at $s$.
Why "Lyapunov"?
For a state cost, $T_{\pi,c}[L](s)\le L(s)$ reads $\mathbb E_\pi[L(s_{t+1})\mid s_t=s]\le L(s)-c(s)$: $L$ decreases in expectation by at least the cost just paid, so $L(s_t)+\sum_{k\lt t}c(s_k)$ is a supermartingale and $V_c^\pi(s_0)\le L(s_0)\le d$. The decrease no longer certifies convergence but that the budget is never overspent: a CMDP guarantee (expected cumulative cost), not a per-trajectory one.

Can an optimal policy be Lyapunov-induced? In principle yes: their Lemma 1 gives an auxiliary cost $\epsilon$ with $L_\epsilon(s):=\mathbb E[\sum_{t\lt T^\ast}(c(s_t)+\epsilon(s_t))\mid\pi_B,s]=V_c^{\pi^\ast}(s)$, bounded by $\epsilon^\ast(s):=2Tc_{\max}D_{TV}(\pi^\ast\|\pi_B)(s)$.

Theorem — An optimal policy is Lyapunov-induced (Chow et al. 2018, Assumption 1, Theorems 1–2)
Assumption 1. $\max_{s\in S'}\epsilon^\ast(s)\le c_{\max}\min\Big\{\dfrac{d-V_c^{\pi_B}(s_0)}{Tc_{\max}},\ \dfrac{Tc_{\max}-\bar D}{Tc_{\max}+\bar D}\Big\}$ with $\bar D=\max_{s\in S'}\max_\pi V_c^\pi(s)$: the baseline is close to optimal.
Statement. Then $L_{\epsilon^\ast}\in\mathcal L_{\pi_B}(s_0,d)$ and $\mathcal F_{L_{\epsilon^\ast}}$ contains an optimal policy (Theorem 1); the safe Bellman operator $T[W](s)=\max_{\pi\in\mathcal F_{L_{\epsilon^\ast}}(s)}T_{\pi,r}[W](s)$ is a monotone contraction whose fixed point at $s_0$ is the optimal constrained return, attained by a greedy policy (Theorem 2).
Caveat. Assumption 1 involves $D_{TV}(\pi^\ast\|\pi_B)$, i.e. knowledge of $\pi^\ast$; the authors state that verifying it "is still challenging". It motivates the algorithm, it does not certify it.
Derivation — Why the safe Bellman operator is a contraction

Use the paper's assumption that every policy (including time-dependent ones) terminates within $T$ steps, and let $w(s):=\sup_\pi\mathbb E^\pi_s[T^\ast]$ be the longest expected remaining time from $s$. Then $1\le w(s)\le T$, and playing $a$ first and continuing with near-optimal policies shows $1+\sum_{s'}P(s'|s,a)w(s')\le w(s)$ for every action $a$. In the weighted norm $\|v\|_w:=\max_s|v(s)|/w(s)$, every policy operator satisfies

$$\big|T_{\pi,r}[v](s)-T_{\pi,r}[v'](s)\big|\le\sum_a\pi(a|s)\sum_{s'}P(s'|s,a)\,w(s')\,\|v-v'\|_w\le\big(w(s)-1\big)\|v-v'\|_w ;$$

dividing by $w(s)$ gives the factor $1-1/w(s)\le1-1/T$, the same for all policies. The maximum over $\mathcal F_{L_{\epsilon^\ast}}(s)$ inherits it, because $|\max_iA_i-\max_iB_i|\le\max_i|A_i-B_i|$. This is the weighted-norm argument of walkthrough Step 1, with one weight common to all policies.

The LP for the auxiliary cost. In practice $\epsilon$ is the largest auxiliary cost that keeps $L_\epsilon$ in the Lyapunov set, a linear program (their eq. 5; LPs and vertices: Primer B):

$$\tilde\epsilon\in\operatorname*{arg\,max}_{\epsilon:S'\to\mathbb R_{\ge0}}\Big\{\sum_{s\in S'}\epsilon(s):\ \ d-V_c^{\pi_B}(s_0)\ \ge\ \mathbf 1(s_0)^\top\big(I-P_{\pi_B}\big)^{-1}\epsilon\Big\},$$

where $P_{\pi_B}$ is the transition matrix among transient states and $\mathbf 1(s_0)^\top(I-P_{\pi_B})^{-1}\mathbf 1(s)$ the expected number of visits to $s$. These visit counts must be positive: a transient state that $s_0$ never reaches under $\pi_B$ has coefficient $0$, so its $\epsilon$ is unconstrained and the LP is unbounded (fix or bound $\epsilon$ at such states; the paper only notes that the counts are non-negative). With one constraint and positive counts, an optimal vertex puts all the slack on the least visited state $\underline s$: $\tilde\epsilon(s)=(d-V_c^{\pi_B}(s_0))\mathbb 1\{s=\underline s\}/\mathbb E[\#\text{visits to }\underline s]$. A constant $\epsilon$ gives $\tilde\epsilon=(d-V_c^{\pi_B}(s_0))/\mathbb E[T^\ast\mid s_0,\pi_B]$ ($T$ may replace the expected stopping time). Exercise 11.5 shows why the vertex can be useless: it may leave zero slack where the policy should change.

Background — Transient chains, expected visits and where the LP comes from

Let $\mathbf 1(s)$ be the coordinate vector with a 1 at state $s$. On transient states, $P_\pi(s,s')=\sum_a\pi(a|s)P(s'|s,a)$; row sums may be below 1 because the probability of terminating is omitted. For a finite chain that terminates with probability one, $P_\pi^t\to0$, and the telescoping identity $(I-P_\pi)\sum_{t=0}^mP_\pi^t=I-P_\pi^{m+1}$ gives $N_\pi:=(I-P_\pi)^{-1}=I+P_\pi+P_\pi^2+\cdots$. The entry $\mathbf 1(s_0)^\top P_\pi^t\mathbf 1(s)$ is the probability of being in $s$ at time $t$, so $\mathbf 1(s_0)^\top N_\pi\mathbf 1(s)$ sums them: the expected number of visits. Example: one transient state that survives each step with probability $1/2$ has $N=1+\tfrac12+\tfrac14+\cdots=2$ expected visits, although no deterministic bound on termination exists.

The termination time $T^\ast=\min\{t:s_t\text{ is terminal}\}$ is a stopping time: whether it has occurred is decided by the states seen so far. Now $L_\epsilon$ solves $L=c+\epsilon+P_{\pi_B}L$, so $L_\epsilon=N_{\pi_B}(c+\epsilon)=V_c^{\pi_B}+N_{\pi_B}\epsilon$, and the budget condition $L_\epsilon(s_0)\le d$ is exactly the LP constraint $\mathbf 1(s_0)^\top N_{\pi_B}\epsilon\le d-V_c^{\pi_B}(s_0)$.

Derivation — Why the LP puts all slack on the least visited state

Write the positive visit counts as $a_s$, the unused budget as $b=d-V_c^{\pi_B}(s_0)$, and $a_{\min}=\min_sa_s$. Every feasible $\epsilon\ge0$ satisfies $a_{\min}\sum_s\epsilon(s)\le\sum_sa_s\epsilon(s)\le b$, hence $\sum_s\epsilon(s)\le b/a_{\min}$, with equality when all of $b/a_{\min}$ goes to one least-visited state (with ties, any split among the least-visited states is optimal too). For a constant $\epsilon(s)\equiv e$ the constraint reads $e\sum_sa_s\le b$, and $\sum_sa_s=\mathbb E[T^\ast\mid s_0,\pi_B]$, the expected number of steps before termination.

Algorithm 2: Safe Policy Iteration (SPI; Chow et al. 2018, Alg. 1, in our notation)
  1. Input: a feasible initial policy $\pi_0$
  2. for $k=0,1,2,\dots$ do
  3. with $\pi_B=\pi_k$: solve the LP for $\epsilon_k$ and evaluate $L_k:=L_{\epsilon_k}$ // Lyapunov function of the current policy
  4. evaluate the task value $V^{\pi_k}$
  5. for every $s\in S'$: $\pi_{k+1}(\cdot|s)\in\operatorname*{arg\,max}_{\pi\in\mathcal F_{L_k}(s)}T_{\pi,r}[V^{\pi_k}](s)$ // an LP in $\pi(\cdot|s)$

Proposition 1 (SPI): (i) consistent feasibility ($\pi_k$ feasible $\Rightarrow$ $\pi_{k+1}$ feasible); (ii) monotone improvement at every state; (iii) convergence if strictly concave/convex regularisers are added. Safe value iteration (Proposition 2) has consistent feasibility and convergence; neither has an optimality guarantee. For large or unknown MDPs, safe DQN and safe DPI (see Primer E) learn a constraint-value network $\hat Q_D$ and a stopping-time network $\hat Q_T$, set $\hat Q_L=\hat Q_D+\epsilon'\hat Q_T$ with $\epsilon'=(d-\pi_k(\cdot|s_0)^\top\hat Q_D(s_0,\cdot))/\pi_k(\cdot|s_0)^\top\hat Q_T(s_0,\cdot)$, solve the per-state LP on sampled states and distil the result into a policy network; exact feasibility needs certified error bounds and enough slack to absorb them, as below.

Small approximation error alone is insufficient at an active constraint. For example, if $\|\widehat L-L\|_\infty\le e$ and transitions are known, write $T_{\pi,c}L-L=(T_{\pi,c}\widehat L-\widehat L)+P_\pi(L-\widehat L)-(L-\widehat L)$; each of the last two terms is at most $e$ in absolute value (the rows of $P_\pi$ sum to at most one), so enforcing $T_{\pi,c}\widehat L-\widehat L\le-2e$ guarantees $T_{\pi,c}L-L\le0$. With discount $\gamma$ the middle term carries a factor $\gamma$ and the margin becomes $(1+\gamma)e$. Learned transitions or action values need their own error bounds; without such margins the deep versions have no exact feasibility guarantee.

Continuous actions: parameter projection and a Lyapunov safety layer

Chow, Nachum, Faust, Duenez-Guzman & Ghavamzadeh, arXiv 2019 carry this to policy gradients (DDPG, PPO) with discount $\gamma$: $Q_L(s,a)=c(s)+\epsilon(s)+\gamma\sum_{s'}P(s'|s,a)L_{\pi_B,\epsilon}(s')$, constraint $\int_a(\pi(a|s)-\pi_B(a|s))Q_L(s,a)\,da\le\epsilon(s)$, and $\epsilon=(1-\gamma)(d-V_c^{\pi_B}(s_0))$. $\theta$-projection linearises objective and constraint in the parameters (a minorisation–maximisation step built on the conservative-policy-iteration bound of Kakade & Langford, ICML 2002, with a second-order KL penalty whose weight is tuned in practice) and averages the constraints over sampled states; with constant $\epsilon$ this is the CPO constraint, i.e. CPO without line search (Module 9). $a$-projection linearises $Q_L$ in the action, $g_L(s):=\nabla_aQ_L(s,a)|_{a=\pi_B(s)}$, and projects the policy's action inside the network, a safety layer in the spirit of Dalal et al., arXiv 2018:

$$\begin{aligned}&a^\ast(s)=\operatorname*{arg\,min}_a\tfrac12\|a-\pi_\theta(s)\|^2\ \ \text{s.t.}\ \ (a-\pi_B(s))^\top g_L(s)\le\epsilon(s)\\ \Longrightarrow\ \ &a^\ast=\pi_\theta(s)-\lambda^\ast g_L(s),\qquad\lambda^\ast=\Big[\frac{g_L^\top(\pi_\theta(s)-\pi_B(s))-\epsilon(s)}{g_L^\top g_L}\Big]_+ ,\end{aligned}$$

by the KKT conditions, for $g_L(s)\ne0$: exactly the single-constraint CBF-QP closed form of Module 10. If $g_L(s)=0$ the linearised constraint reads $0\le\epsilon(s)$: vacuous when $\epsilon(s)\ge0$ (always the case for a feasible baseline), so $a^\ast=\pi_\theta(s)$ and $\lambda^\ast=0$; infeasible otherwise.

Background — Minorisation–maximisation (MM) and majorise–minimise

To maximise $J(\theta)$, an MM method maximises at step $k$ a surrogate $m_k$ that lies below $J$ everywhere and touches it at the current point: $m_k(\theta)\le J(\theta)$ for all $\theta$ and $m_k(\theta_k)=J(\theta_k)$ (Primer B). Then $J(\theta_{k+1})\ge m_k(\theta_{k+1})\ge m_k(\theta_k)=J(\theta_k)$: the true objective never decreases. Majorise–minimise is the mirror image for minimisation, with an upper surrogate. Example: to minimise $J(\theta)=\theta^2$, the surrogate $M_k(\theta)=\theta^2+(\theta-\theta_k)^2\ge J$ touches $J$ at $\theta_k$; its minimiser is $\theta_{k+1}=\theta_k/2$, which divides $J$ by four.

The monotonicity needs a global bound. The conservative-policy-iteration bound behind $\theta$-projection is one, but its linearisation with a second-order KL penalty and a tuned weight is not automatically below $J$ everywhere, so the practical step inherits no monotone-improvement guarantee.

Caveat — what is (and is not) guaranteed
  • Sign in the 2019 paper. Its Proposition 1 prints $\pi_\theta(s)+\lambda^\ast g_L(s)$. For the half-space $(a-\pi_B)^\top g_L\le\epsilon$ and $\lambda^\ast\ge0$, stationarity of $\tfrac12\|a-\pi_\theta\|^2+\lambda((a-\pi_B)^\top g_L-\epsilon)$ gives $a=\pi_\theta-\lambda g_L$. Check: $\pi_B=0$, $\pi_\theta=(1,0.5)$, $g_L=(2,1)$, $\epsilon=1$ give $\lambda^\ast=0.3$, $a^\ast=(0.4,0.2)$, $g_L^\top a^\ast=1$; the printed sign gives $(1.6,0.8)$ with $g_L^\top a=4\gt\epsilon$.
  • Exact versus learned. With a known tabular model every SPI/SVI iterate is feasible. Linearisations, averaged constraints and errors in $\hat Q_L$ break this: the deep-RL versions are safe only up to these errors.
  • What "safe" means. An expected cumulative cost from $s_0$, not state-wise safety; contrast Sections 2 and 5 and the viability view of Module 7.

4. Neural Lyapunov, Barrier and Contraction Certificates

Sections 1–2 take the Lyapunov function as given. For nonlinear systems with input limits a good certificate is as hard to find as a good controller, and sums-of-squares methods need polynomial dynamics and scale badly. The neural-certificate programme, surveyed by Dawson, Gao & Fan, T-RO 2023, parameterises the certificate (often with the controller) by a network, trains it on the certificate conditions, and then verifies it formally or accepts statistical evidence.

CertificateConditions (continuous time, $\dot x=f_c(x)$ or $\dot x=f(x)+g(x)u$)Proves
Lyapunov$V(x_g)=0$, $V\gt0$ elsewhere, $\dot V=\nabla V^\top f_c\le0$ (strict away from $x_g$)stability (asymptotic if strict)
exponential Lyapunov (Primer D)$k_1\|x-x_g\|^a\le V\le k_2\|x-x_g\|^a$, $\dot V\le-k_3\|x-x_g\|^a$exponential stability
CLF (exponential)$V(x_g)=0$, $V\gt0$ elsewhere, $\inf_{u\in U}[L_fV+L_gVu]\lt0$ for $x\ne x_g$; exponential: $k_1\|x-x_g\|^a\le V\le k_2\|x-x_g\|^a$ and $\inf_{u\in U}[L_fV+L_gVu+cV]\le0$, $c\gt0$asymptotic (exponential) stabilisability by a regular (e.g. Lipschitz) feedback choosing such inputs
barrier (Dawson: $C=\{h\le0\}$)$\dot h\le-\alpha(h(x))$, $\alpha$ extended class-$\mathcal K$forward invariance of $C$
CBF$\inf_{u\in U}[L_fh+L_ghu+\alpha(h(x))]\le0$existence of safe inputs
contraction metric $M(x)\succ0$$\underline m I\preceq M(x)\preceq\overline m I$ uniformly on a convex domain containing the trajectories and connecting geodesics; smooth dynamics and metric, with trajectories defined for all future times; $\dot M+\mathrm{sym}\big(M(A+BK)\big)+2\lambda M\prec0$, $A=\tfrac{\partial f}{\partial x}$, $B=\tfrac{\partial f}{\partial u}$, $K=\tfrac{\partial\pi}{\partial x}$$\|x(t)-x^\ast(t)\|\le Re^{-\lambda t}\|x(0)-x^\ast(0)\|$, $R=\sqrt{\overline m/\underline m}$
discrete-time Lyapunov$V$ continuous, $V(F(x))-V(x)\lt0$ (or $\le-\kappa V$) on the setSection 1; ReLU networks allowed

The CLF row only says that at every state some admissible input decreases $V$; a closed-loop conclusion also needs a feedback that selects such inputs regularly enough (e.g. Lipschitz) for the closed-loop trajectories to exist, hence "by a regular feedback" in the last column.

Background — Contraction metrics and geodesics

A contraction metric measures a small displacement $\delta x$ between two neighbouring trajectories by the squared length $\delta x^\top M(x)\delta x$. Along the closed loop the displacement obeys the linearised dynamics $\delta\dot x=(A+BK)\delta x$, and $M$ changes along the flow as $\dot M=\sum_i(\partial M/\partial x_i)\,f_{c,i}$. With $\mathrm{sym}(H):=H+H^\top$, the product rule and the table's inequality give

$$\tfrac{d}{dt}\big(\delta x^\top M\delta x\big)=\delta x^\top\big(\dot M+\mathrm{sym}(M(A+BK))\big)\delta x\lt-2\lambda\,\delta x^\top M\delta x,$$

so squared lengths shrink at least like $e^{-2\lambda t}$ and lengths like $e^{-\lambda t}$. Example: $\dot x=-x$, $M=1$: $\delta\dot x=-\delta x$ and $\tfrac{d}{dt}\delta x^2=-2\delta x^2$. For two trajectories, join $x(0)$ and $x^\ast(0)$ by the straight segment (it stays in the convex domain); its $M$-length is at most $\sqrt{\overline m}\,\|x(0)-x^\ast(0)\|$. Let every point of the segment flow with the dynamics: at time $t$ the curve joins $x(t)$ and $x^\ast(t)$, its $M$-length is at most $e^{-\lambda t}$ times the initial one, and its Euclidean length is at most $1/\sqrt{\underline m}$ times its $M$-length. The distance $\|x(t)-x^\ast(t)\|$ is at most that Euclidean length, which gives the table's bound with $R=\sqrt{\overline m/\underline m}$. A geodesic is the shortest connecting curve measured in $M$; starting from it gives the sharper version with the initial metric distance. The argument needs smooth dynamics and metric and connecting curves that stay in the certified domain.

The survey's $\{h\le0\}$ is the opposite sign of Module 10's $\{h\ge0\}$ (Module 10). Two rows are sharper than the survey's wording: it writes the CLF condition with $\le0$, which with a regular feedback and a non-strict derivative bound gives only stability in the sense of Lyapunov, and the exponential CLF without the norm bounds, which makes $V$ but not $\|x-x_g\|$ decay exponentially (the bounds are part of the exponential-CLF definition of Ames et al., TAC 2014); and its "$R$ = square root of the condition number of $M$" must be read with uniform bounds, as in Manchester & Slotine, TAC 2017 (Theorem 1): a pointwise condition number does not suffice, since a scalar metric $M(x)=m(x)I$ has $\kappa\equiv1$. Synthesis is a feasibility problem, find $V$ with $c_i(x,V)\le0$ for all $x\in X$, $i\in I$, relaxed twice: constraints become penalties and "for all $x$" becomes "for sampled $x_j$", giving the empirical loss $\mathcal L_V=\sum_{i\in I}\frac{\alpha_i}{N}\sum_{j=1}^N\max(c_i(x_j,V),0)$ (their eq. 19). The survey is explicit that zero empirical loss "does not guarantee that the certificate is valid; it provides only statistical evidence of validity". A verifier closes the gap, either in the loop or after training.

Learner and falsifier: Neural Lyapunov Control

Definition — Lyapunov risk and falsification constraint (Chang, Roohi & Gao 2019, Defs. 4–6)
For a sampling distribution $\rho$ on the domain $D$ and closed-loop vector field $f_u$, $$L_\rho(\theta,u)=\mathbb E_{x\sim\rho(D)}\Big[\max\big(0,-V_\theta(x)\big)+\max\big(0,L_{f_u}V_\theta(x)\big)+V_\theta^2(0)\Big];$$ true Lyapunov functions have zero risk (not conversely). The falsifier searches for a solution of $$\Phi_\varepsilon(x):=\Big(\textstyle\sum_ix_i^2\ge\varepsilon\Big)\wedge\Big(V(x)\le0\ \vee\ L_{f_u}V(x)\ge0\Big),\qquad x\in D,$$ with dReal, a $\delta$-complete solver: it proves $\Phi_\varepsilon$ unsatisfiable, which certifies the Lyapunov conditions on $D$ outside the ball $\sum_ix_i^2\lt\varepsilon$ of radius $\sqrt\varepsilon$ (the paper's "$\varepsilon$-ball"), or returns a point satisfying a $\delta$-weakening; spurious counterexamples only add training samples. (Here $\wedge$ means "and", $\vee$ "or": the falsifier looks for a point outside the excluded ball where positivity or strict decrease fails. A $\delta$-weakening relaxes every inequality by $\delta$, e.g. $g(x)\ge0$ becomes $g(x)\ge-\delta$; so "unsatisfiable" is a proof, while a returned point may be spurious.)

Chang, Roohi & Gao, NeurIPS 2019 alternate gradient steps on the empirical risk in $(\theta,u)$ with SMT queries, adding counterexamples until $\Phi_\varepsilon$ is unsatisfiable, and return the largest sublevel set inside $D$. The network must be smooth so that its Lie derivative exists and can be written in closed form for dReal (ReLU networks are excluded); the authors use tanh networks, other smooth activations would also do. A penalty $\frac1N\sum_i\|x_i\|_2-\alpha V_\theta(x_i)$ enlarges the certified region, which on their benchmarks exceeds LQR and SOS/SDP designs. Caveats: the SMT query is NP-hard in general (low-dimensional demos), and with the ball of radius $\sqrt\varepsilon$ excluded the certificate strictly gives invariance and convergence to the smallest sublevel set containing the ball; the inside needs another argument (the authors deem it physically insignificant; a linearisation as in Section 1 closes it).

Verify only what you claim: Yang et al. (ICML 2024)

Verifying the decrease on a whole box $B$ and then taking the largest sublevel set inside it is valid but wasteful. Yang, Dai, Shi, Hsieh, Tedrake & Zhang, ICML 2024 work in discrete time, $\xi_{t+1}=f_{cl}(\xi_t)$ with $\xi=x$ (state feedback) or $\xi=(x,\hat x-x)$ (output feedback with a neural observer), build positive definiteness into $V(\xi)=|\phi_V(\xi)-\phi_V(\xi^\ast)|+\|(\epsilon I+R^\top R)(\xi-\xi^\ast)\|_1$ (or a quadratic), and certify $S:=\{\xi\in B:V(\xi)\lt\rho\}$.

Derivation — Why this $V$ is positive definite for any network weights

Here $\phi_V$ is a scalar-valued network, $R$ a trainable matrix with as many columns as $\xi$ has entries, and $\epsilon\gt0$ fixed. For $y=\xi-\xi^\ast$ let $Q=\epsilon I+R^\top R$. It is symmetric with $y^\top Qy=\epsilon\|y\|_2^2+\|Ry\|_2^2\ge\epsilon\|y\|_2^2$, so all its eigenvalues are at least $\epsilon$ and $\|Qy\|_2\ge\epsilon\|y\|_2$ (Primer A). Since $\|v\|_1\ge\|v\|_2$, the second term of $V$ is at least $\epsilon\|y\|_2\gt0$ for $y\ne0$; the first term is nonnegative, and both vanish at $y=0$. So $V(\xi^\ast)=0$ and $V\gt0$ elsewhere, whatever the trained weights.

Background — Observers and output feedback

In output feedback the controller measures an output $y=Cx$ rather than all of $x$. An observer updates an estimate $\hat x$ from measurements and past inputs, and the controller acts on that estimate. The combined state $\xi=(x,e)$, with estimation error $e=\hat x-x$, evolves by the plant and observer equations together, so a Lyapunov certificate for it must control both the physical state and the estimation error. Example: for a linear plant $x^+=Ax+Bu$, $y=Cx$, the observer $\hat x^+=A\hat x+Bu+L_o(y-C\hat x)$ gives, after subtracting the plant equation, $e^+=(A-L_oC)e$; with $u=K\hat x=K(x+e)$ the plant becomes $x^+=(A+BK)x+BKe$. Both $x$ and $e$ enter the dynamics, hence the certificate.

Theorem — Verifiable ROA condition (Yang et al. 2024, Theorem 3.3)
Setting. $B$ a compact box with $\xi^\ast$ in its interior; $f_{cl}$ continuous and defined on $B$, $f_{cl}(\xi^\ast)=\xi^\ast$; $V$ continuous and positive definite with $V(\xi^\ast)=0$, $\rho\gt0$, $0\lt\kappa\lt1$, $F(\xi):=V(f_{cl}(\xi))-(1-\kappa)V(\xi)$.
Statement. If $\ \big(-F(\xi)\ge0\ \wedge\ f_{cl}(\xi)\in B\big)\ \vee\ \big(V(\xi)\ge\rho\big)\ $ for all $\xi\in B$, then $S$ is forward invariant, $V(\xi_{t+1})\le(1-\kappa)V(\xi_t)$ on $S$, and $S$ is an inner estimate of the region of attraction.
Proof. For $\xi\in B$ with $V(\xi)\lt\rho$ the disjunction gives $V(\xi_{t+1})\le(1-\kappa)V(\xi_t)\lt\rho$ and $\xi_{t+1}\in B$, i.e. $\xi_{t+1}\in S$: invariance. Iterating, $V(\xi_t)\le(1-\kappa)^tV(\xi_0)\to0$; the iterates stay in the compact $B$ and $V$ vanishes only at $\xi^\ast$, so $\xi_t\to\xi^\ast$ by the compactness argument of Section 1. Geometric decay of $V$ gives geometric decay of $\|\xi_t-\xi^\ast\|$ only with norm bounds as in the exponential-Lyapunov row above.
Why it matters. The condition is posed on the explicit box, which branch-and-bound verifiers accept, but constrains only the implicit set $S$; the largest verifiable $\rho$ is found by bisection (Primer B).
Fact — What a sound verifier proves (the interface used here)

Assumptions. A compact box $B$ is covered by finitely many verifier cells $B_j$. For each scalar violation function $g_i$, the verifier returns an enclosing bound $g_i(\xi)\le b_{ij}$ for every $\xi\in B_j$, including all nonlinearities and numerical rounding errors. The functions encode the exact claimed condition.

Statement. If every $b_{ij}\le0$, then $g_i\le0$ throughout $B$. A positive bound is inconclusive; an actual point with $g_i\gt0$ refutes the claim. A timeout proves nothing.

Intuition. Bound propagation replaces each layer by enclosing affine inequalities. Branch-and-bound splits difficult cells and repeats. A mixed-integer encoding instead represents piecewise-linear choices such as ReLU by binary variables; only a sound global bound from that solver proves the inequality. PGD samples or searches points and has no all-points conclusion. For Yang's disjunction, use the zero condition of the hinge below, together with the stated Lyapunov assumptions. See Module 15 for the algorithms.

Yang et al., ICML 2024

Bisection starts with a verified level and an unverified larger level, tests their midpoint, and keeps the corresponding half interval; larger $\rho$ imposes the condition on more states. A verifier timeout means only "not verified", so the result is the largest level found by this procedure, not necessarily the mathematically largest level. Projected gradient ascent (PGD) instead searches for violations by $\xi^+=\operatorname{proj}_B(\xi+s\nabla g(\xi))$ with a step size $s\gt0$, clipping coordinates back into $B$ after each ascent step (Primer B). It supplies counterexamples, never an absence proof.

In their Example 3.2 ($x_{t+1}=x_t+0.1u_t$, $\pi(x)=-x$, $V=x^\top x$, $B=[-1,1]^2$, $\kappa=0.1$) the condition holds on all of $B$; choose $\rho\gt2=\max_B V$ to obtain $S=B$, whereas the two-step approach certifies only the disc $x^\top x\le1$. Verification uses $\alpha,\beta$-CROWN (bound propagation plus branch-and-bound; see Module 15); training uses cheap PGD falsification of the hinge $\mathrm{ReLU}\big(\min(\mathrm{ReLU}(F(\xi))+c_0H(\xi_{t+1}),\ \rho-V(\xi))\big)$, with $H$ measuring how far $\xi_{t+1}$ leaves $B$. On pendulum and path tracking their certified regions strictly contain those of the baselines (MIP-verified DITL, SMT-verified NLC, UNL); they report faster verification than MIP on the harder systems (cart-pole 129 s versus 448 s, PVTOL 217 s versus 1935 s; MIP wins on the small pendulum), and, to their knowledge, the first formally verified neural output-feedback controllers with Lyapunov certificates (the quadrotor case takes 8.9 hours to verify).

Derivation — The training hinge vanishes exactly where the condition holds

For $B=\prod_i[\underline b_i,\overline b_i]$ one choice is $H(y)=\sum_i\big[\max(\underline b_i-y_i,0)+\max(y_i-\overline b_i,0)\big]$, which is zero exactly on $B$. With $c_0\gt0$, the term $\mathrm{ReLU}(F(\xi))+c_0H(\xi_{t+1})$ is zero exactly when decrease and box membership both hold, and $\mathrm{ReLU}\big(\min(\cdot,\ \rho-V(\xi))\big)$ is zero exactly when that happens or $V(\xi)\ge\rho$. So the hinge vanishes at $\xi$ if and only if the disjunction of the theorem holds there. PGD drives sampled violations to zero during training; only the verifier proves the condition on all of $B$.

Caveat — what a neural certificate certifies
  • The model, not the robot. All conditions are evaluated on $f$; model error must enter explicitly, as a bounded residual (robust CLF/CBF, Module 10) or through GP confidence bounds as in Section 2.
  • Statistical evidence is not verification. Zero loss on samples says nothing between samples; a sound Lipschitz bound or successful formal verification is needed; a falsifier that finds no violation is inconclusive.
  • Scaling. Complete verification is exponential in the worst case, and verified examples are low-dimensional; learned barrier filters face the same trade-off (Module 10).

5. Learning-Based and Robust MPC

MPC solves a constrained finite-horizon problem (see Primer D) at every step and applies the first input. Its safety argument is Section 1 in disguise: an invariant terminal set with a terminal controller makes the shifted previous plan a feasible candidate (recursive feasibility), and the optimal cost acts as a Lyapunov function. Learning enters in three places (taxonomy of Hewing, Wabersich, Menner & Zeilinger, Annu. Rev. 2020):

CategoryWhat is learnedWhere safety comes fromExamples
(i) the prediction modelmodel error $g$ (GP, NN, set-membership)robust or stochastic MPC with the learned uncertaintyAswani 2013, Soloperto 2018, Koller 2018, cautious MPC
(ii) the controller designcost, constraints, terminal ingredientsthe MPC structure (constraints stay explicit)Gros & Zanon 2020, Zanon & Gros 2021
(iii) MPC for safe learningany learning controller (e.g. RL)an MPC-based safety filterpredictive safety filters (Module 10)

Tube MPC in one page

Robust MPC is the backbone of most results below (invariance: Blanchini, Automatica 1999; tubes: Mayne, Seron & Raković, Automatica 2005).

Definition — Set algebra and robust positive invariance
Minkowski sum $A\oplus B=\{a+b:a\in A,b\in B\}$; Pontryagin difference $A\ominus B=\{x:\ x+b\in A\ \forall b\in B\}$; $TA=\{Ta:a\in A\}$. Rules: $(A\ominus B)\oplus B\subseteq A$ and $(A\ominus(B\oplus C))\oplus C\subseteq A\ominus B$. For $e^+=A_Ke+w$, $w\in W$, a set $Z$ is robust positively invariant (RPI) if $A_KZ\oplus W\subseteq Z$. When $A_K$ is Schur stable and $W$ is compact with $0\in W$, the minimal RPI set is $F_\infty:=\mathrm{cl}\bigcup_{k\ge0}\bigoplus_{i=0}^{k}A_K^iW$ (cl = closure): it is RPI and contained in every closed RPI set.
Derivation — Why $F_\infty$ is the minimal RPI set

Start the error at $e_0=0$. After $k+1$ steps $e_{k+1}=\sum_{i=0}^kA_K^iw_{k-i}$, so the reachable errors form $Z_k:=\bigoplus_{i=0}^kA_K^iW$. Because $0\in W$ the sets are nested, $Z_k\subseteq Z_{k+1}$, and because $A_K$ is Schur stable, $\|A_K^i\|\le C\varrho^i$ for some $C$ and $\varrho\lt1$ (Primer D), so all of them lie in the ball of radius $C\max_{w\in W}\|w\|/(1-\varrho)$. Invariance: $A_KZ_k\oplus W=Z_{k+1}$, and taking limits keeps the inclusion, so $A_KF_\infty\oplus W\subseteq F_\infty$. Minimality: if $Z$ is a nonempty closed RPI set and $z\in Z$, induction gives $A_K^{m+1}z+Z_m\subseteq Z$ for every $m$. A point $y\in Z_k$ lies in $Z_m$ for all $m\ge k$, so $A_K^{m+1}z+y\in Z$; letting $m\to\infty$ ($A_K^{m+1}z\to0$, $Z$ closed) gives $y\in Z$. Hence $F_\infty\subseteq Z$.

Lemma — Tube constraint satisfaction (Mayne et al. 2005)
Setting. $x^+=Ax+Bu+w$, $w\in W$ compact; nominal $z^+=Az+Bv$; $u=v+K(x-z)$ with $A_K=A+BK$ Schur stable; $Z$ RPI for $e^+=A_Ke+w$.
Statement. If $x_0-z_0\in Z$, then $x_t-z_t\in Z$ for all $t$; if also $z_t\in X\ominus Z$ and $v_t\in U\ominus KZ$, then $x_t\in X$, $u_t\in U$ for every disturbance sequence.
Proof. $e=x-z$ obeys $e^+=A_Ke+w\in A_KZ\oplus W\subseteq Z$; induction. Then $x=z+e\in(X\ominus Z)\oplus Z\subseteq X$ and $u=v+Ke\in(U\ominus KZ)\oplus KZ\subseteq U$.

Aswani et al. (2013): one model for safety, one for performance

Aswani, Gonzalez, Sastry & Tomlin, Automatica 2013 consider $x_{n+1}=Ax_n+Bu_n+g(x_n,u_n)$ with polytopes $X,U$ (bounded intersections of finitely many half-spaces; Primer B), $g(x,u)\in W$ (a polytope) and full state measurement. Learning-based MPC (LBMPC) drives two predictions with the same inputs:

$$\begin{aligned} \min_{c,\theta}\ \ &\psi_n\big(\theta,\tilde x_n,\dots,\tilde x_{n+N},\check u_n,\dots,\check u_{n+N-1}\big)\\ \text{s.t.}\ \ &\tilde x_{n+i+1}=A\tilde x_{n+i}+B\check u_{n+i}+\mathcal O_n(\tilde x_{n+i},\check u_{n+i})\qquad\text{(learned model, cost only)}\\ &\bar x_{n+i+1}=A\bar x_{n+i}+B\check u_{n+i},\qquad \check u_{n+i}=K\bar x_{n+i}+c_{n+i}\qquad\text{(nominal model)}\\ &\bar x_{n+i}\in X\ominus R_i,\quad \check u_{n+i}\in U\ominus KR_i,\quad(\bar x_{n+N},\theta)\in\Omega\ominus(R_N\times\{0\}),\\ &R_0=\{0\},\qquad R_i=\textstyle\bigoplus_{j=0}^{i-1}(A+BK)^jW\qquad\text{(tube cross-sections)}, \end{aligned}$$

with $\tilde x_n=\bar x_n=x_n$. The "oracle" $\mathcal O_n$ is any learned, time-varying model; $\Omega$ is a disturbance-invariant, constraint-admissible set of the augmented state $(x,\theta)$, $\theta$ parameterising the tracked steady state $x_s=\Lambda\theta$, $u_s=\Psi\theta$, under the terminal tracking law $u=K(x-x_s)+u_s=Kx+(\Psi-K\Lambda)\theta$ (their eqs. 3–4).

Derivation — The steady-state parameters, and why the tube absorbs one disturbance

Steady states of the nominal model satisfy $x_s=Ax_s+Bu_s$, i.e. $(I-A)x_s-Bu_s=0$. If this solution space has dimension $r$, stack a basis as the columns of $[\Lambda^\top\ \Psi^\top]^\top$ with $\Lambda\in\mathbb R^{q\times r}$, $\Psi\in\mathbb R^{p\times r}$; then $\theta\in\mathbb R^r$ selects one steady state and stays fixed during the terminal tracking step. The vector $c_{n+i}$ is the optimised correction added to the feedback $K\bar x_{n+i}$. After the true step, $x_{n+1}=\bar x^{\rm old}_{n+1}+w_n$ with $w_n=g(x_n,u_n)\in W$, and the new nominal prediction starts at $x_{n+1}$. With the same corrections $c$, the difference $\delta_i:=\bar x^{\rm new}_{n+1+i}-\bar x^{\rm old}_{n+1+i}$ obeys $\delta_0=w_n$ and $\delta_{i+1}=A\delta_i+BK\delta_i=(A+BK)\delta_i$ (the $c$ terms cancel), so $\delta_i=(A+BK)^iw_n\in(A+BK)^iW$: exactly the growth of the tube cross-sections $R_i$ used in the proof below.

Theorem — Robust feasibility and constraint satisfaction of LBMPC (Aswani et al. 2013, Theorem 1 and Corollary 1)
Assumptions. $g(x,u)\in W$ on $X\times U$; $A+BK$ Schur stable; $\Omega$ constraint-admissible and disturbance-invariant; LBMPC feasible at $x_n$ with $M_n=\{c_n,\dots,c_{n+N-1},\theta_n\}$.
Statement. Applying $u_n=Kx_n+c_n$ gives (a) a feasible $M_{n+1}$ at $x_{n+1}$ and (b) $x_{n+1}\in X$. Hence feasibility at $n=0$ implies constraint satisfaction, feasibility and boundedness for all $n$, whatever the oracle does.
Proof idea. The candidate $M_{n+1}=\{c_{n+1},\dots,c_{n+N-1},(\Psi-K\Lambda)\theta_n,\theta_n\}$, whose appended term makes the last input $K\bar x+(\Psi-K\Lambda)\theta_n$, the terminal tracking law of $\Omega$ (the paper appends $0$, which is valid only when $(\Psi-K\Lambda)\theta_n=0$, but its next step uses exactly this input), shifts the nominal predictions by $(A+BK)^ig(x_n,u_n)\in(A+BK)^iW$. Since $R_{i+1}=R_i\oplus(A+BK)^iW$, a prediction in $X\ominus R_{i+1}$ lands in $(X\ominus(R_i\oplus(A+BK)^iW))\oplus(A+BK)^iW\subseteq X\ominus R_i$, the next constraint; $\Omega$ handles the terminal step. And $x_{n+1}=\bar x_{n+1}+w_n\in(X\ominus W)\oplus W\subseteq X$.

Safety is deterministic and decoupled from learning. The authors also prove robust asymptotic stability (Theorem 2) and, under sufficient excitation, convergence in probability of the LBMPC law to that of an MPC with the true model (Section 4, Theorems 5–7; a simplified version is stated below). That second result needs identification assumptions that the robust-feasibility theorem does not supply. The price of the robust design is conservatism: the worst case $W$ is used everywhere. Soloperto, Müller, Trimpe & Allgöwer, NMPC 2018 address exactly this with robust MPC for linear systems with bounded state-dependent uncertainty sets, which can shrink where a learned model is confident (in their numerical example the sets come from GP regression).

Theorem — The learned MPC law converges (simplified from Aswani et al. 2013, Sec. 4)

Assumptions. The cost is time-invariant (their simplification $\psi_n\equiv\psi_0$). Fix the measured state and the horizon; let $F$ be the nonempty compact set of decisions $m=(c,\theta)$ feasible for the robust constraints (it does not depend on the oracle). Let $J_n(m)$ and $J_*(m)$ be the continuous LBMPC costs with the learned model $\mathcal O_n$ and with the true $g$. Assume $\sup_{m\in F}|J_n(m)-J_*(m)|\to0$ in probability, $m_n$ an exact minimiser of $J_n$ over $F$, and a unique minimiser $m_*$ of $J_*$ over $F$.

Statement. $m_n\to m_*$ in probability: for every tolerance $a\gt0$, $\Pr(\|m_n-m_*\|\ge a)\to0$. Hence the applied input $Kx_n+c_n$, a continuous function of $m$, converges in probability to that of the MPC with the true model. Without uniqueness the minimisers approach the set of true minimisers.

Proof. Fix $a\gt0$. The set $\{m\in F:\|m-m_*\|\ge a\}$ is compact and $J_*$ is continuous with unique minimiser $m_*$, so $\Delta:=\min\{J_*(m):m\in F,\ \|m-m_*\|\ge a\}-J_*(m_*)\gt0$. If $\sup_F|J_n-J_*|\lt\Delta/2$, every such $m$ has $J_n(m)\gt J_*(m_*)+\Delta/2\gt J_n(m_*)$, so no minimiser of $J_n$ lies there. Hence $\Pr(\|m_n-m_*\|\ge a)\le\Pr(\sup_F|J_n-J_*|\ge\Delta/2)\to0$.

Where excitation enters. Uniform convergence of the costs needs the oracle to converge to $g$ where the predictions go. Aswani et al. obtain it under sufficient excitation: for parametric oracles the standard identification condition (for a model linear in parameters with regressors $\varphi_t$: $\tfrac1n\sum_{t\lt n}\varphi_t\varphi_t^\top\succeq aI$ for some $a\gt0$ and all large $n$, which with a correct model and averaging noise makes least squares consistent), for nonparametric oracles data covering $X\times U$ ever more densely (a finite sample cover of radius $h_n\to0$, their Theorems 6–7). A loop that settles at an equilibrium stops exciting the system, which is why this is an assumption, not a consequence. They derive convergence of minimisers from epi-convergence (their Theorem 4), a weaker requirement than the uniform convergence used here.

Aswani et al., Automatica 2013

Koller et al. (2018): GP confidence tubes and a safe terminal set

Koller, Berkenkamp, Turchetta & Krause, CDC 2018 combine Section 2's GP model with MPC: $x_{t+1}=h(x_t,u_t)+g(x_t,u_t)$, $h$ twice continuously differentiable, $\|g\|_k\le B_g$, polytopic constraints. Their Lemma 1 (Chowdhury–Gopalan type, noise parameter $\lambda$) gives $|\mu_{n-1,j}(z)-g_j(z)|\le\beta_n\sigma_{n-1,j}(z)$ for all output dimensions and all $z=(x,u)$ with probability at least $1-\delta$. Under affine feedback $\pi_t(x)=K_t(x-p_t)+u_t$, the linearised prior (with a Lipschitz-gradient remainder) and the GP confidence box map an ellipsoid $R_t$ to an ellipsoid $R_{t+1}=\tilde m(R_t,\pi_t)$ containing all reachable states.

Lemma — Ellipsoidal enclosure of the reachable states (Koller et al. 2018, Lemma 2 and Corollary 1)
Assumptions. Those of their Lemma 1 above ($\|g\|_k\le B_g$, sub-Gaussian noise, $\beta_n$ as there); $\tilde m$ over-approximates the linearised prior $h$ plus its remainder bound plus the GP confidence box $\mu_n\pm\beta_n\sigma_{n-1}$ over the whole ellipsoid.
Statement. With probability at least $1-\delta$, simultaneously for all ellipsoids $R\subset X$ and all admissible inputs (constant inputs in their Lemma 2, affine feedbacks $\pi$ in its extension): $f(x,\pi(x))\in\tilde m(R,\pi)$ for every $x\in R$. Hence, on this one event, $x_0\in R_0$ and $R_{t+1}=\tilde m(R_t,\pi_t)$ give $x_t\in R_t$ for all $t$ (induction).
In words. Propagate a set that encloses every possible next state, not just a mean prediction; because the event is uniform over sets and inputs, every plan the MPC ever considers is covered at once. If the model is updated, a fallback plan certified with the old model must be re-checked (or kept) before it is replaced.
Theorem — SafeMPC is $\delta$-safe (Koller et al. 2018, Assumption 2, Definition 1, Theorem 2)
Assumptions. $\|g\|_k\le B_g$, sub-Gaussian noise, $\beta_n$ from their Lemma 1; a polytopic $X_{\mathrm{safe}}\subseteq X$ that is robust control positively invariant under a known backup $\pi_{\mathrm{safe}}$ with $\pi_{\mathrm{safe}}(x)\in U$ there; $x_0\in X_{\mathrm{safe}}$.
Scheme. Minimise $J_t$ s.t. $R_{t+1}=\tilde m(R_t,\pi_t)$, $R_t\subset X$, $\pi_t(R_t)\subset U$ ($t\lt T$) and $R_T\subset X_{\mathrm{safe}}$; apply the first feedback; if infeasible, shift the previous sequence and append $\pi_{\mathrm{safe}}$.
Statement. $\Pr\big[\forall t:\ x_{t+1}\in X,\ \pi(x_t)\in U\big]\ge1-\delta$.
Proof idea. Induction: every feasible plan has a return path into $X_{\mathrm{safe}}$, valid uniformly with high probability; if a new plan is infeasible, the old one still leads there and $\pi_{\mathrm{safe}}$ takes over.

A performance plan with any model may run alongside if its first $r$ inputs coincide with the safe return plan. The experiments again fix $\beta_n=2$ "following" Berkenkamp et al., outside the theorem. The terminal-set-plus-backup argument is also the core of predictive safety filters (Module 10). The cautious MPC of Hewing, Kabzan & Zeilinger, TCST 2020 is the stochastic alternative: mean and variance of a GP residual are propagated by linearisation and $\Pr(h^\top x_k\le b)\ge p$ becomes $h^\top\mu_k+\Phi^{-1}(p)\sqrt{h^\top\Sigma_kh}\le b$; fast (autonomous racing), but the propagation, and hence the guarantee, is approximate.

Derivation — A Gaussian chance constraint is a deterministic constraint

If $x_k\sim\mathcal N(\mu_k,\Sigma_k)$, then $Y=h^\top x_k$ is normal with mean $m=h^\top\mu_k$ and variance $s^2=h^\top\Sigma_kh$ (Primer C). For $s\gt0$, $(Y-m)/s$ is standard normal, so $\Pr(Y\le b)=\Phi((b-m)/s)$, where $\Phi$ is the standard normal CDF. Since $\Phi$ is strictly increasing, $\Phi((b-m)/s)\ge p$ is equivalent to $(b-m)/s\ge\Phi^{-1}(p)$, i.e. $m+\Phi^{-1}(p)\,s\le b$. If $s=0$, $Y=m$ and the requirement is $m\le b$. When $\mu_k,\Sigma_k$ come from an approximate propagation, the equivalence holds for the approximation only.

Gros & Zanon (2020): MPC as the function approximator of RL

Category (ii) keeps a possibly wrong model and learns the MPC's cost so that the closed loop is optimal. Gros & Zanon, TAC 2020 show this is possible in principle.

Theorem — An MPC on a wrong model can be optimal (Gros & Zanon 2020, Theorem 1)
Setting. True MDP with stage cost $L$ (minimised, discount $\gamma$) and optimal $V^\star,Q^\star,\pi^\star$; a possibly wrong model $P[\hat s^+\mid s,a]$; modified stage cost $\hat L(s,a)=Q^\star(s,a)-\gamma\,\mathbb E[V^\star(\hat s^+)\mid s,a]$ ($+\infty$ where the expectation is infinite); $\hat V_N(s)=\min_\pi\mathbb E\big[\gamma^NV^\star(\hat s_N)+\sum_{k\lt N}\gamma^k\hat L(\hat s_k,\pi(\hat s_k))\big]$ on the model.
Statement. On the set $S$ of states whose model trajectories under $\pi^\star$ keep $\mathbb E[V^\star(\hat s_k)]$ finite for all $k$: $\hat V_N=V^\star$, the optimal first-action sets agree wherever all expectations are finite (selected policies agree with consistent tie-breaking), and $\hat Q_N=Q^\star$ for inputs with finite $\mathbb E[V^\star(\hat s^+)\mid s,a]$.
Proof idea. Telescoping (stage $k$ contributes $-\gamma^{k+1}V^\star(\hat s_{k+1})$, which pairs with stage $k+1$'s $Q^\star$ into an advantage; the last one cancels the terminal cost) turns the $N$-step cost of any policy $\pi$ into $Q^\star(s,\pi(s))+\mathbb E\sum_{k=1}^{N-1}\gamma^kA^\star(\hat s_k,\pi(\hat s_k))$ with $A^\star=Q^\star-V^\star\ge0$; every term is minimised by $\pi^\star$ (their eq. 11).
Derivation — The telescoping for $N=2$

Take a policy that plays $a_0$ at $s$ and $a_1$ at $\hat s_1$. By definition $\hat J_2=\mathbb E\big[\hat L(s,a_0)+\gamma\hat L(\hat s_1,a_1)+\gamma^2V^\star(\hat s_2)\big]$. Insert $\hat L(s,a)=Q^\star(s,a)-\gamma\,\mathbb E[V^\star(\hat s^+)\mid s,a]$ and use the tower property:

$$\hat J_2=Q^\star(s,a_0)-\gamma\mathbb EV^\star(\hat s_1)+\gamma\mathbb EQ^\star(\hat s_1,a_1)-\gamma^2\mathbb EV^\star(\hat s_2)+\gamma^2\mathbb EV^\star(\hat s_2)=Q^\star(s,a_0)+\gamma\,\mathbb E\big[A^\star(\hat s_1,a_1)\big].$$

For cost minimisation $V^\star(s)=\min_aQ^\star(s,a)$, so $A^\star=Q^\star-V^\star\ge0$ with equality exactly at optimal actions. Minimising over $a_1$ makes the second term zero, and minimising over $a_0$ gives $\min_{a_0}Q^\star(s,a_0)=V^\star(s)$. Longer horizons cancel the same way stage by stage. Where $Q^\star$ has several minimisers the sets of optimal actions agree; the selected actions agree only under a consistent tie-break rule.

Since $\hat L$ is unknown, an economic NMPC (their eq. 21; soft constraints and slack penalties: Primer B) is parameterised, with cost $\lambda_\theta(x_0)+\gamma^NV^f_\theta(x_N)+w_f^\top\sigma_N+\sum_{k\lt N}\gamma^k(l_\theta(x_k,u_k)+w^\top\sigma_k)$, model $f_\theta$, input constraints and softened state constraints $h_\theta(x_k,u_k)\le\sigma_k$; its optimal cost $V_\theta$ and first-input-fixed version $Q_\theta$ serve as function approximators, updated by Q-learning or deterministic policy gradients with NLP sensitivities:

$$\begin{aligned}&\tau_k=L_\theta(s_k,a_k)+\gamma V_\theta(s_{k+1})-Q_\theta(s_k,a_k),\qquad\theta\leftarrow\theta+\alpha\,\tau_k\nabla_\theta Q_\theta(s_k,a_k),\\&L_\theta(s,a)=l_\theta(s,a)+w^\top\max\big(0,h_\theta(s,a)\big).\end{aligned}$$
Background — Differentiating the optimal value of an optimisation problem

An NLP (nonlinear program) is a finite-dimensional optimisation problem with nonlinear objective and constraints. The update needs $\nabla_\theta Q_\theta$, the derivative of an optimal value with respect to parameters. At a regular local solution (smooth data, linearly independent active-constraint gradients, strict complementarity, second-order sufficiency) two tools give it: the envelope formula $\nabla_\theta V_\theta=\nabla_\theta\mathcal L(z^\ast,\lambda^\ast;\theta)$, the partial derivative of the Lagrangian at the optimal primal–dual pair; and, for the optimiser itself, the implicit function theorem applied to the active KKT equations $G(z,\theta)=0$: $dz^\ast/d\theta=-G_z^{-1}G_\theta$ when the Jacobian $G_z$ is invertible.

Example: $J(u,\theta)=(u-\theta)^2+\theta^2$ has minimiser $u^\ast=\theta$ and optimal value $\theta^2$, with derivative $2\theta$; the envelope formula gives the same, $\partial_\theta J(u^\ast,\theta)=-2(u^\ast-\theta)+2\theta=2\theta$, without differentiating $u^\ast$. When the active set changes or the solution is not unique, the optimal value can have kinks and these formulas fail: differentiability is an assumption on the MPC, not something backpropagation provides.

The constraints stay explicit but softened, so violations remain possible; the authors state that safety-critical constraints need robust NMPC. Zanon & Gros, TAC 2021 take this step: the policy is a robust (tube) linear MPC whose parameters $\theta$ include the uncertainty set $\hat S^+(s,a,\theta)$ it robustifies against, and the RL update is a constrained problem whose "safe-design constraint" demands that every observed transition lies in $\hat S^+(s_k,a_k,\theta)$. Given that constraint, the robust MPC satisfies the constraints by construction; but the constraint is checked only on data, so safety is $\eta$-safety, with $\eta(\mathcal D)$ the probability, given the data, that the learned set really contains the true dispersion. Their "Fundamental Limitation 1" states that $\eta(\mathcal D)=1$ cannot be guaranteed from finite data without further assumptions on the system.

6. Certified Approximate MPC with Neural Networks

A nonlinear MPC needs an NLP solve every sampling period, too slow or unpredictable for fast embedded systems. Fitting a network $\pi_{\mathrm{approx}}\approx\pi_{\mathrm{MPC}}$ offline throws away what made MPC attractive: any approximation error can break recursive feasibility and constraint satisfaction. The papers below keep guarantees, but in different senses, and conflating them is the most common mistake here. This is also where the two research lines behind this site meet topically: the Trimpe group's approximate MPC (Hertneck, Nubert, Hose, Tokmak, with Johannes Köhler as robust-MPC link) and Patricia Pauli's quadratic-constraint view (Drummond et al., Chatzikiriakos et al.), both rooted in Allgöwer's Stuttgart group; no joint paper links them.

PaperApproximatedGuaranteeMechanism
Hertneck et al. 2018robust MPC law (any regressor)statistical: confidence $1-\delta_h$ that a fraction $\mu_{\mathrm{crit}}$ of random initial conditions is safeMPC robust to input errors $\le\eta$ + Hoeffding validation
Nubert et al. 2020robust tracking MPC, 7-DOF armstatistical as above (not reached on hardware)tube budget split: model error + approximation error
Hose et al. 2025whole input sequence (NN)deterministic, relative to the prediction modelonline feasibility check + safe shifted candidate
Tokmak et al. 2025robust MPC law (kernel interpolation)deterministic error $\le\epsilon$, given an RKHS-norm extrapolationpower-function bound + robust MPC
Drummond et al. 2022given NN versus given linear MPCworst-case bound on a boxquadratic constraints + S-procedure (SDP)
Chatzikiriakos et al. 2024value function of a soft-constrained MPCISS in the error; constraints, for starts in the tightened hard-MPC feasible set, under a uniform error boundLipschitz value function + one-step controller

Hertneck et al. (2018): robustness by design, validation by Hoeffding

Hertneck, Köhler, Trimpe & Allgöwer, L-CSS 2018 make the MPC tolerate the error it will later suffer: treat $\pi_{\mathrm{approx}}(x)=\pi_{\mathrm{MPC}}(x)+d$, $\|d\|_\infty\le\eta$, as an input disturbance. The robust MPC is a nominal MPC with terminal set $X_f$ and tightened constraints $\bar X_k=(1-\epsilon_k)X$, $\bar U_k=(1-\epsilon_k)U_t$, $U_t=U\ominus\{\|d\|_\infty\le\eta\}$, $\epsilon_k=\epsilon(1-\sqrt\rho^{\,k})/(1-\sqrt\rho)$ growing like a tube. Assumptions 1–4: local incremental stabilisability (a $\delta$-Lyapunov function contracting at rate $\rho\lt1$), $\|f(x,u+d)-f(x,u)\|\le\lambda\|d\|_\infty$, a bound on $\eta$, and a robustly invariant terminal set with a local control Lyapunov function. Written out:

Definition — The robust-design assumptions of Hertneck et al. (2018, Assumptions 1–4 and eq. 9)
Assumption 1 (local incremental stabilisability). There are a feedback $\kappa(x,z,v)$, a $\delta$-Lyapunov function $V_\delta(x,z,v)\ge0$ with $V_\delta(z,z,v)=0$, and constants $c_{\delta,l},c_{\delta,u},\delta_{\rm loc},k_{\max}\gt0$, $\rho\in(0,1)$ such that, whenever $V_\delta(x,z,v)\le\delta_{\rm loc}$, $$c_{\delta,l}\|x-z\|^2\le V_\delta(x,z,v)\le c_{\delta,u}\|x-z\|^2,\qquad\|\kappa(x,z,v)-v\|\le k_{\max}\|x-z\|,\qquad V_\delta(x^+,z^+,v^+)\le\rho\,V_\delta(x,z,v),$$ with $x^+=f(x,\kappa(x,z,v))$ and $z^+=f(z,v)$: a feedback that lets the true state $x$ track a nominal trajectory $z$ driven by inputs $v$, the squared tracking error shrinking by the factor $\rho$ per step.
Assumption 2. $\|f(x,u+d)-f(x,u)\|\le\lambda\|d\|_\infty$ for $x\in X$ and $u,\,u+d\in U$.
Assumption 3. $\eta\le\lambda^{-1}\sqrt{\delta_{\rm loc}/c_{\delta,u}}$.
Tightening (eq. 9). With $X=\{x:Hx\le\mathbf 1\}$ and $U_t=\{u:L_tu\le\mathbf 1\}$: $\epsilon:=\eta\lambda\sqrt{c_{\delta,u}/c_{\delta,l}}\,\max\{\|H\|_\infty,\|L_t\|_\infty k_{\max}\}$ and $\epsilon_k=\epsilon\sum_{j=0}^{k-1}\sqrt\rho^{\,j}$.
Assumption 4 (terminal set). $V_f(x)=x^\top Px$, $X_f=\{V_f\le\alpha_f\}$ and a controller $k_f$ with, for all $x\in X_f$: $f(x,k_f(x))+w\in X_f$ for every $\|w\|\le\lambda\eta\sqrt{\rho^Nc_{\delta,u}/c_{\delta,l}}$; $V_f(f(x,k_f(x)))\le V_f(x)-\big(x^\top Qx+k_f(x)^\top Rk_f(x)\big)$; $(x,k_f(x))\in\bar X_N\times\bar U_N$.
In words. An input error $\|d\|_\infty\le\eta$ moves the next state by at most $\lambda\eta$ (Assumption 2), so the re-planned trajectory starts within $V_\delta\le c_{\delta,u}\lambda^2\eta^2\le\delta_{\rm loc}$ of the old plan: Assumption 3 is exactly what keeps this inside the region where the tracking certificate holds. Afterwards $\sqrt{V_\delta}$ shrinks by $\sqrt\rho$ per step, so $k$ steps later the state is within $\sqrt\rho^{\,k}\lambda\eta\sqrt{c_{\delta,u}/c_{\delta,l}}$ of the old plan; in constraint units (factors $\|H\|_\infty$, and $\|L_t\|_\infty k_{\max}$ for inputs) this is at most $\epsilon\sqrt\rho^{\,k}=\epsilon_{k+1}-\epsilon_k$, the increment of the tightening. What is left after $N$ steps is the ball the terminal set must absorb. All of this is checked before learning; validation only estimates how often the network stays within $\eta$.
Theorem — Robust MPC and its learned approximation (Hertneck et al. 2018, Theorems 5 and 8)
Theorem 5. Under Assumptions 1–4, from every $x(0)\in X_{\mathrm{feas}}$ the closed loop with inputs $\pi_{\mathrm{MPC}}(x)+d$, $\|d\|_\infty\le\eta$, satisfies $(x(t),u(t))\in X_{\mathrm{feas}}\times U$ for all $t$ and converges to a robust positively invariant set around the origin. So any $\pi_{\mathrm{approx}}$ with $\|\pi_{\mathrm{approx}}-\pi_{\mathrm{MPC}}\|_\infty\le\eta$ on the visited states inherits this.
Validation. Draw initial conditions i.i.d. from $\Omega$ on $X_{\mathrm{feas}}$, simulate $\pi_{\mathrm{approx}}$ until its first entry into $X_f$, and set $I(X_i)=1$ only if that entry occurs in finite time and the error bound held at every step before entry. From $X_f$ on, the terminal controller $k_f$ of Assumption 4 takes over: the validated trajectories end in $X_f$, "where we can guarantee stability and constraint satisfaction with the terminal controller" (their Sec. IV-B and Remark 6), and $\pi_{\mathrm{approx}}$ is never checked inside $X_f$. With $\mu=\mathbb P[I=1]$ and the empirical mean $\tilde\mu$ of $p$ runs, Hoeffding (their Lemma 7) gives $\mathbb P[|\tilde\mu-\mu|\ge\epsilon_h]\le2e^{-2p\epsilon_h^2}=:\delta_h$; the test passes if $$\mu_{\mathrm{crit}}\ \le\ \tilde\mu-\sqrt{\ln(2/\delta_h)/(2p)} .$$ Theorem 8. If validation succeeds, then with confidence at least $1-\delta_h$ the approximate MPC ensures stability and constraint satisfaction for a fraction $\mu_{\mathrm{crit}}$ of the initial conditions distributed according to $\Omega$. This holds for the hybrid controller that switches to $k_f$ on entering $X_f$; keeping $\pi_{\mathrm{approx}}$ inside $X_f$ would need a separate guarantee there.
In words. Robustness makes an error bound sufficient; statistics estimates how often it holds. The guarantee concerns a random initial condition, not every initial condition.
Validation protocol. A run counts as a success only if it reaches $X_f$ and the error bound held at every step before; in finite computation, fix a maximal horizon and count non-entry by then as failure (conservative). Freeze the controller before drawing each validation batch: after retraining, or when the sample is enlarged until the test passes, use a fresh batch with a split confidence budget (step 6 below) or a time-uniform bound.
Derivation — The Hoeffding sample bound for validating an approximate MPC

1. Random variables. The indicators $I(X_1),\dots,I(X_p)$ are i.i.d. (i.i.d. initial conditions, deterministic closed loop), with values in $[0,1]$ and mean $\mu$.

2. Hoeffding. For i.i.d. variables in $[a,b]$, $\mathbb P[\tilde\mu-\mu\ge\epsilon]\le e^{-2p\epsilon^2/(b-a)^2}$ and likewise for the lower tail; for $[0,1]$ and both tails, $\mathbb P[|\tilde\mu-\mu|\ge\epsilon_h]\le2e^{-2p\epsilon_h^2}$.

3. Margin. Setting this to $\delta_h$ gives $\epsilon_h=\sqrt{\ln(2/\delta_h)/(2p)}$: with probability at least $1-\delta_h$ over the validation data, $\mu\ge\tilde\mu-\epsilon_h$. Only the lower tail matters, so the one-sided $\sqrt{\ln(1/\delta_h)/(2p)}$ would do; the paper uses the two-sided one.

4. Sample size. With an expected success rate $\tilde\mu\gt\mu_{\mathrm{crit}}$ the test passes once $\epsilon_h\le\tilde\mu-\mu_{\mathrm{crit}}$: $$p\ \ge\ \frac{\ln(2/\delta_h)}{2(\tilde\mu-\mu_{\mathrm{crit}})^2},$$ which explodes as $\tilde\mu\downarrow\mu_{\mathrm{crit}}$: certifying $99\%$ needs an approximator visibly better than $99\%$.

5. The paper's example. A stirred-tank reactor, horizon $N=180$, $\eta=5.1\cdot10^{-3}$, $\delta_h=0.01$, $\mu_{\mathrm{crit}}=0.99$: validation passed with $\tilde\mu=0.9987$ on $p=34\,980$ trajectories. Here $\epsilon_h=\sqrt{\ln200/69\,960}=0.00870$, so at most 45 failures are allowed. The whole procedure took roughly 500 hours on a quad-core PC.

6. Two probability levels, one subtlety. $\delta_h$ refers to the validation sample, $\mu_{\mathrm{crit}}$ to a fresh initial condition, and the confidence to one validation run of one fixed controller. If the test is repeated (more samples, or retraining) until it passes, as the algorithm allows, a union bound over at most $K$ attempts (see Primer C) at level $\delta_h/K$ each restores confidence $1-\delta_h$.

Nubert, Köhler, Berenz, Allgöwer & Trimpe, RA-L 2020 scale this to a KUKA LBR4+ arm: a robust setpoint-tracking MPC (nonlinear constraint tightening, obstacle avoidance, artificial steady states) approximated by a 20-layer ReLU network with about $5\cdot10^6$ parameters. Their Proposition 2 splits the tube budget, $w_d=w_{d,\mathrm{model}}+w_{d,\mathrm{approx}}$, and Hertneck's lemma is reused for validation. Honestly reported: the validation criterion held for only about 90% of sampled points, so the statistical guarantee was not obtained on hardware, and input constraints were occasionally violated under measurement noise. The network ran in about 1 ms (a 200-fold speed-up), and in thousands of runs the robot never came close to an obstacle.

Hose et al. (2025): check online, fall back deterministically

Hose, Köhler, Zeilinger & Trimpe, TCST 2025 let the network predict the whole input sequence $\hat{\mathbf u}=\Pi_{\mathrm{NN}}(x)\in U^N$, which one forward simulation can check: $\mathbf u\in\mathcal U^N(x)=\{\mathbf u\in U^N:\ \phi(k;x;\mathbf u)\in X\ (k\lt N),\ \phi(N;x;\mathbf u)\in X_f\}$, with $\phi(k;x;\mathbf u)$ the predicted state and $V(x,\mathbf u)$ the MPC cost.

Algorithm 3: Approximate MPC with safety-augmented NN (Hose et al. 2025, Alg. 1)
  1. Input: $\Pi_{\mathrm{NN}}$, $\mathcal U^N(\cdot)$, terminal controller $K_f$, cost $V$, safe initial candidate $\tilde{\mathbf u}(0)=\mathbf u_{\mathrm{init}}$
  2. for $t=0,1,\dots$: measure $x=x(t)$; evaluate $\hat{\mathbf u}(t)=\Pi_{\mathrm{NN}}(x)$
  3. if $\hat{\mathbf u}(t)\in\mathcal U^N(x)$: $\mathbf u(t)=\arg\min_{\mathbf u\in\{\tilde{\mathbf u}(t),\hat{\mathbf u}(t)\}}V(x,\mathbf u)$, else $\mathbf u(t)=\tilde{\mathbf u}(t)$
  4. apply $u(t)=\mathbf u_0(t)$; set $\tilde{\mathbf u}(t+1)=\{\mathbf u_{1:N-1}(t),\,K_f(\phi(N;x;\mathbf u(t)))\}$ // shift and append
Theorem — Deterministic guarantees despite approximation errors (Hose et al. 2025, Theorem 1)
Assumptions. (1) Terminal ingredients: $\ell(x,u)\ge\alpha_\ell(\|x\|)$, $\alpha_\ell\in\mathcal K_\infty$ (class-$\mathcal K_\infty$: Primer D); $X_f$ positively invariant under $K_f$; $V_f(f(x,K_f(x)))-V_f(x)\le-\ell(x,K_f(x))$ on $X_f$. (2) $\mathbf u_{\mathrm{init}}\in\mathcal U^N(x(0))$. Costs are finite and nonnegative; $X_f\subseteq X$ and $K_f(x)\in U$ on $X_f$. State evolution is the model $f$. Nothing about the network.
Statement. Under Algorithm 3, $(x(t),u(t))\in X\times U$ for all $t$ and $x(t)\to0$.
Proof idea. The applied sequence is always feasible, and Assumption 1 makes its shift feasible at the next state (induction). The shifted candidate costs at most $V(x(t),\mathbf u(t))-\ell(x(t),u(t))$ and the NN sequence is taken only if cheaper, so telescoping gives $\sum_t\alpha_\ell(\|x(t)\|)\le V(x(0),\mathbf u_{\mathrm{init}})\lt\infty$ and $x(t)\to0$.
Scope. "Deterministic" is relative to the model $f$; mismatch needs a robust formulation. Convergence becomes asymptotic stability if the terminal controller takes over inside $X_f$ (their Remark 2).
Derivation — The cost decrease and why $x(t)\to0$

Write $V(x,\mathbf u)=\sum_{i=0}^{N-1}\ell(x_i,u_i)+V_f(x_N)$ with $x_i=\phi(i;x;\mathbf u)$. Feasibility puts $x_N$ in $X_f$. The shifted candidate keeps stages $1,\dots,N-1$, drops stage $0$ and appends $K_f(x_N)$, so the old terminal cost $V_f(x_N)$ is replaced by $\ell(x_N,K_f(x_N))+V_f(f(x_N,K_f(x_N)))\le V_f(x_N)$ (Assumption 1). Hence the candidate costs at most $V(x,\mathbf u)-\ell(x,u_0)$, and since the network's sequence is used only when it is cheaper,

$$V\big(x(t+1),\mathbf u(t+1)\big)\le V\big(x(t),\mathbf u(t)\big)-\ell\big(x(t),u(t)\big).$$

Summing from $t=0$ to $T-1$ and using $V\ge0$: $\sum_{t\lt T}\alpha_\ell(\|x(t)\|)\le\sum_{t\lt T}\ell(x(t),u(t))\le V(x(0),\mathbf u_{\mathrm{init}})$ for every $T$. If $\|x(t)\|\ge a\gt0$ held infinitely often, the sum would contain infinitely many terms of at least $\alpha_\ell(a)\gt0$ and diverge; so $x(t)\to0$.

How often the network is accepted is a performance question. Their Lemmas 2–3 give a sufficient condition: approximate a robust MPC with $\Pi_{\mathrm{RMPC}}(x)\oplus\mathcal B_\epsilon^N\subseteq\mathcal U^N(x)$; then any network with error at most $\epsilon$ is feasible. The online cost, one network evaluation and one forward simulation (typically within 0.2 ms), is one to four orders of magnitude below online optimisation on three nonlinear benchmarks on which the naive network violates constraints: a predictive safety filter whose optimisation is replaced by a single check.

Tokmak et al. (2025): certify the approximation error itself

Tokmak, Fiedler, Zeilinger, Trimpe & Köhler, TAC 2025 replace statistics by an interpolation bound. For samples $X$ of the function $f$ to approximate (here the MPC law), the kernel interpolant $h_X(x)=k_X(x)^\top K_X^{-1}f_X$ and the power function $P_X(x)=\sqrt{1-k_X(x)^\top K_X^{-1}k_X(x)}$ satisfy the bound below (their Prop. 1, compare the noise-free bound of Module 3). It needs their Assumption 1, a continuous, strictly positive definite, radial kernel normalised to $k(x,x)=1$ (for a general kernel the $1$ becomes $k(x,x)$), pairwise distinct samples, so that $K_X$ is invertible, and $f$ in the RKHS of $k$:

$$|f(x)-h_X(x)|\ \le\ P_X(x)\sqrt{\|f\|_k^2-\|h_X\|_k^2}\qquad\text{for all }x .$$
Derivation — Where $\sqrt{\|f\|_k^2-\|h_X\|_k^2}$ comes from

Module 3 shows that the interpolant $h_X$ is the orthogonal projection $\Pi f$ of $f$ onto $\mathrm{span}\{k(x_i,\cdot)\}$. Pythagoras gives $\|f\|_k^2=\|h_X\|_k^2+\|f-h_X\|_k^2$. The residual $f-h_X$ is orthogonal to that span, so by the reproducing property and Cauchy–Schwarz

$$f(x)-h_X(x)=\big\langle f-h_X,\ k(x,\cdot)-\Pi k(x,\cdot)\big\rangle_k\le\|f-h_X\|_k\,\|k(x,\cdot)-\Pi k(x,\cdot)\|_k=\sqrt{\|f\|_k^2-\|h_X\|_k^2}\;P_X(x),$$

since the second norm is the power function (Module 3), and the same bound holds for $h_X(x)-f(x)$. This is sharper than $P_X(x)\|f\|_k$. For a vector-valued MPC law apply it to each input component and combine the bounds in the norm the robust MPC uses.

ALKIA-X refines an adaptive partition into cubes sampled at their vertices, assuming the power function peaks at the cube centre (Assumption 2), until the bound is below $\epsilon$; a robust MPC for input errors up to $\epsilon$ then gives closed-loop guarantees (Theorem 2, Corollary 2). The catch is $\|f\|_k$: its bound is extrapolated exponentially from the interpolants' norms on two refinement levels (Assumption 5), "only a heuristic" that "may fail on some occasions" in the authors' words; finite samples never bound the RKHS norm of an unknown function (their Remark 1; Module 5).

The Pauli line: worst-case bounds and value functions

Drummond, Duncan, Turner, Pauli & Allgöwer, L4DC 2022 compare a given NN controller with a given input-constrained linear MPC. The MPC law solves a QP, $z_{\mathrm{MPC}}(x)=\arg\min_uu^\top Hu+u^\top hx$ s.t. $Lu\le b$, whose solution satisfies quadratic constraints derived from the QP's structure (their Theorem 6, Lemma 7); the NN, in implicit form $z=\phi(Wz+W_0x+\beta)$ (see Primer E), satisfies activation QCs; the state box satisfies a diagonal QC. The S-procedure (Module 2, also the engine of LipSDP; see Module 12) turns this into an SDP certifying $\|u_{\mathrm{NN}}(x)-u_{\mathrm{MPC}}(x)\|_2^2\le\gamma_x\|x\|_2^2+\gamma$ for all $x$ in the box (their Theorem 10); the same machinery synthesises an MPC that robustly approximates a given NN (Theorem 11).

Theorem — What the SDP certificate of Drummond et al. (2022, Theorem 10) proves
Setting. On the given state box the QP is feasible with $H\succ0$, so its minimiser $u_{\rm MPC}(x)$ is unique; its optimality conditions give valid quadratic constraints. The network output is $u_{\rm NN}=W_{\rm out}z+b_{\rm out}\in\mathbb R^p$, where $z$ stacks the hidden activations; for a feed-forward network $W$ is block-triangular, so $z=\phi(Wz+W_0x+\beta)$ is evaluated layer by layer (a general implicit network needs its own well-posedness argument). A selector picks the first $p$ entries of the MPC input sequence. Stack $x$, $z$ and the QP's primal and dual variables into $\zeta$; every actual controller pair satisfies known quadratic inequalities $q_i(\zeta)\ge0$ (state box, activations, QP structure).
Statement. If there are $\lambda_i\ge0$ such that $\gamma_x\|x\|^2+\gamma-\|u_{\rm NN}-u_{\rm MPC}\|^2-\sum_i\lambda_iq_i(\zeta)\ge0$ for all $\zeta$ (a quadratic function of $(1,\zeta)$, so this is an LMI in $\gamma_x,\gamma,\lambda$), then $\|u_{\rm NN}(x)-u_{\rm MPC}(x)\|^2\le\gamma_x\|x\|^2+\gamma$ on the whole box. Equality constraints may take multipliers of either sign; an infeasible SDP is inconclusive.
Why. At an actual controller pair every subtracted term $\lambda_iq_i(\zeta)$ is nonnegative, so the error bound itself must be nonnegative: the S-procedure implication of Module 2.

Chatzikiriakos, Wabersich, Berkel, Pauli & Iannelli, L4DC 2024 learn the value function $\tilde V^s$ of a tightened soft-constrained MPC instead of its policy: soft constraints make it locally Lipschitz (the policy may be discontinuous), and $\tilde u(x)\in\arg\min_{u\in U}\ell(x,u)+\tilde V^s(f(x,u))$ is a one-step problem, solvable by gridding a low-dimensional input. Given a uniform error bound $|\tilde V^s-V^{\ast,s}|\le\hat\varepsilon$ (their Assumption 3, assumed, not certified) and invariance of the feasible set, the closed loop is input-to-state stable with respect to the error, $|x(k)|\le\beta(|x(0)|,k)+\sigma(\sup_{i\lt k}|\bar\varepsilon(x(i))|)$ with $\beta\in\mathcal{KL}$, $\sigma\in\mathcal K$ (Theorem 3, for all $x(0)$ in the soft-constrained feasible set $X_{N,s}$). Constraint satisfaction needs more (Theorem 4): let $X_N^\eta$ be the feasible set of the hard-constrained MPC with state constraints tightened by $\eta\in(0,1)$; if $\ell(x,0)\gt2\hat\varepsilon$ for all $x\in X\setminus X_N^\eta$ (their Assumption 5) and the slack penalty $\rho$ satisfies $\eta\rho\ge V_{\max}+4\hat\varepsilon$ with $V_{\max}\ge V^{\ast,s}(x)$ for all $x\in X_N^\eta$, then trajectories starting in $X_N^\eta$ satisfy the original constraints. Starting anywhere in the larger $X_{N,s}$ gives ISS, not this guarantee.

Background — Input-to-state stability and the error signal $\bar\varepsilon$

A class-$\mathcal{KL}$ function $\beta(r,k)$ is increasing in $r$ with $\beta(0,k)=0$ and tends to zero as $k\to\infty$ for fixed $r$; $\sigma\in\mathcal K$ is continuous, strictly increasing and zero at zero (Primer D). The ISS estimate therefore says: the effect of the initial state dies out, and what remains is bounded by an increasing function of the largest error seen so far. Example: $x^+=x/2+d$ with $|d|\le b$ gives $|x_k|\le2^{-k}|x_0|+b(1+\tfrac12+\tfrac14+\cdots)\le2^{-k}|x_0|+2b$, a neighbourhood of zero of size proportional to the disturbance, not convergence to zero.

Here the "disturbance" is the value-function error $e(x):=\tilde V^s(x)-V^{\ast,s}(x)$, $|e|\le\hat\varepsilon$, evaluated at the two candidate successors: $\bar\varepsilon(x):=|e(f(x,\tilde u(x)))|+|e(f(x,\pi^s_{\rm MPC}(x)))|\le2\hat\varepsilon$ (their eq. 13). Since $\tilde u$ minimises $\ell(x,u)+\tilde V^s(f(x,u))$, comparing with $u=\pi^s_{\rm MPC}(x)$, switching between $\tilde V^s$ and $V^{\ast,s}$ at both successors (each costs at most $|e|$) and using the MPC decrease $\ell(x,\pi^s_{\rm MPC})+V^{\ast,s}(f(x,\pi^s_{\rm MPC}))\le V^{\ast,s}(x)$ gives $V^{\ast,s}(f(x,\tilde u))-V^{\ast,s}(x)\le-\ell(x,\tilde u)+\bar\varepsilon(x)$: a persistent error competes with the decrease, and the ISS estimate yields $\limsup_k|x(k)|\le\sigma(2\hat\varepsilon)$.

7. Statistical Guarantees Meet Robust Control (Fiedler, Scherer, Trimpe)

Robust control certifies a controller for every member of an uncertainty set; learning produces sets that are correct only with high probability. Composing the two gives closed-loop guarantees with the probability of the learning step, provided the learned set is expressed in the language of robust control. Three statistical tools recur.

Definition — Hoeffding's inequality and sub-Gaussianity
For i.i.d. $Z_1,\dots,Z_p\in[a,b]$ with mean $\mu$ and $\bar Z=\frac1p\sum_iZ_i$: $\ \mathbb P[\bar Z-\mu\ge t]\le\exp\big(-2pt^2/(b-a)^2\big)$, and twice that for $|\bar Z-\mu|$. For indicators, $\mu\ge\bar Z-t$ at confidence $1-\delta$ needs $p\ge\ln(1/\delta)/(2t^2)$. The proof rests on Hoeffding's lemma: a variable in $[a,b]$ is $\tfrac{b-a}2$-sub-Gaussian, the noise model of the GP bounds in Module 3. Use: a-posteriori validation of a fixed design.
Theorem — The scenario approach (Calafiore & Campi 2006; exact bound of Campi & Garatti 2008)
Setting. A convex program with $d$ decision variables and uncertain convex constraints $\theta\in\Theta_\delta$, $\delta\sim P$ (in this box $\delta$ is the uncertain parameter, as in the scenario literature, not a failure probability; $\theta$ is the decision variable, not a policy parameter). Solve the scenario program $\min c^\top\theta$ s.t. $\theta\in\Theta_{\delta^{(i)}}$ for $N$ i.i.d. samples; the required finite instances, including singleton problems, constraint removals and one-constraint augmentations, are assumed feasible with nonempty feasible interiors and attained unique minimisers of the original linear objective; the optimiser and violation event are measurable. Merely selecting one point from a nonunique optimum set does not supply these assumptions: a tie-break variant needs its own support/compression argument before using the same dimension-$d$ bound. Violation probability: $V(\theta)=P\{\delta:\theta\notin\Theta_\delta\}$.
Statement. For any $P$ (binomial tails: Primer C), $$P^N\big\{V(\theta_N^\ast)\gt\epsilon\big\}\ \le\ \sum_{i=0}^{d-1}\binom Ni\epsilon^i(1-\epsilon)^{N-i},$$ with equality for fully-supported problems (Campi & Garatti, SIAM J. Optim. 2008), sharpening the earlier $\binom Nd(1-\epsilon)^{N-d}$ (Calafiore & Campi, TAC 2006, their eq. 11, from which their explicit sample-size formulas are derived). A simple sufficient size, which follows from a Chernoff bound on the binomial tail and is the form von Rohr et al. use (after Campi, Garatti & Prandini, Annu. Rev. Control 2009), is $N\ge\frac2\epsilon(\ln\frac1\beta+d)$.
In words. At most $d$ sampled constraints support the solution, so violating more than an $\epsilon$-fraction has binomial-tail probability. Use: a-priori design of convex (e.g. LMI) controllers. For $\epsilon=0.05$, $\beta=10^{-5}$, $d=10$: exact $N=581$, older bound 1344, sufficient formula 861 (our computation).
Reading the statement. A support constraint is a sampled constraint whose removal changes the solution; convexity together with those original-objective uniqueness assumptions allows at most $d$ of them; arbitrary selection from a nonunique optimum set need not preserve this count. Fully supported means that, with probability one, the solution has exactly $d$ of them; only the equality case needs this. There are two probability levels: $V(\theta)$ is the probability that one fresh uncertain system violates the design, while $P^N\{V(\theta_N^\ast)\gt\epsilon\}$ is the probability, over the $N$ design samples, of drawing a sample that produces a design with $V\gt\epsilon$. The sample-size calculator below applies only to feasible convex scenario programs with the solution assumptions above.

The third tool is the uniform GP error bound of Module 3: a tube $|\varphi(x)-\mu_D(x)|\le\beta_D\sigma_D(x)$ for all $x$ simultaneously, with probability $1-\delta$ over the data, for a fixed unknown function. This frequentist reading is what robust control needs.

Fiedler, Scherer & Trimpe (CDC 2021): from a GP tube to IQC synthesis

Fiedler, Scherer & Trimpe, CDC 2021 build a pipeline in which every stage carries a guarantee. The plant is a linear fractional representation: a known LTI part $G$ in feedback with $p=\Delta(q)$, here a diagonal static nonlinearity $p_i=\varphi_i(q_i)$, $i\le n_p$, whose unknown parts are "pulled out" so that all prior knowledge stays in $G$.

Background — Linear fractional representations and the star product

A linear fractional representation (LFR) separates a known linear system from an uncertain block. With an external input $w$ (e.g. a disturbance) and a performance output $z$, the known part is $q=G_{qp}p+G_{qw}w$, $z=G_{zp}p+G_{zw}w$, and the loop is closed by $p=\Delta(q)$. If $\Delta$ is a matrix and $I-G_{qp}\Delta$ is invertible, eliminating $p$ and $q$ gives

$$z=\Big(G_{zw}+G_{zp}\Delta\big(I-G_{qp}\Delta\big)^{-1}G_{qw}\Big)w,$$

and this closed map is written $\Delta\star G$ (the star product; below, $\Delta_0\star P$). Well-posedness means that the loop equations have a unique solution for every $w$; for a matrix $\Delta$ it is the invertibility of $I-G_{qp}\Delta$. Example: scalar $G_{qp}=\tfrac12$, $\Delta=1$ gives $q=\tfrac q2+G_{qw}w$, so $q=2G_{qw}w$; with $G_{qp}=\Delta=1$ the equation $q=q+G_{qw}w$ has no solution unless $G_{qw}w=0$, an ill-posed loop, which must be excluded before any stability claim.

  1. Learn. $y_n=\varphi(x_n)+\epsilon_n$ with independent $R$-sub-Gaussian noise (Assumption 1), $\varphi\in\mathcal H_k$, $\|\varphi\|_k\le B$ (Assumption 2); prior knowledge such as $\varphi(0)=0$ goes into the kernel, $k_0(x,x')=k(x,x')-k(x,0)k(x',0)/k(0,0)$ (for $k(0,0)\gt0$). This is the covariance of a zero-mean GP conditioned on the exact observation $\varphi(0)=0$ (Gaussian conditioning, Primer C), so $k_0(0,x)=0$ and every sample vanishes at $0$; in the frequentist reading used here, $\mathcal H_{k_0}$ consists exactly of the functions in $\mathcal H_k$ with $\varphi(0)=0$, with the same norm, so Assumption 2 carries over.
  2. Bound. Proposition 1: $\mathbb P[|\varphi(x)-\mu_D(x)|\le\beta_D\sigma_D(x)\ \forall x]\ge1-\delta$ with a data-dependent $\beta_D$.
  3. Sector. $\varphi$ is in the sector $[\kappa_1,\kappa_2]$ if $\kappa_1x^2\le x\varphi(x)\le\kappa_2x^2$ (Module 2). Lemma 2 gives the tightest sector compatible with the tube on $[a,b]$: $\hat\kappa_1=\inf_{x\in[a,b]\setminus\{0\}}\min\{x(\mu_D-\beta_D\sigma_D),\,x(\mu_D+\beta_D\sigma_D)\}/x^2$ and $\hat\kappa_2$ likewise with outer supremum and inner maximum, so with probability $1-\delta$ the true $\varphi$ lies in the sector $[\hat\kappa_1,\hat\kappa_2]$ on $[a,b]$: $\hat\kappa_1x^2\le x\varphi(x)\le\hat\kappa_2x^2$ there. Equivalently, $[\kappa_1,\kappa_2]\subseteq[\hat\kappa_1,\hat\kappa_2]$ for the tightest local sector, $\kappa_1=\inf_{[a,b]\setminus\{0\}}\varphi(x)/x$ and $\kappa_2$ likewise with sup. The paper states the containment for any sector of $\varphi$, which fails for a loose one: $\varphi(x)=x$ lies in $[-10,10]$, but an exact tube gives $[1,1]$. (For $x\ne0$ the tube gives an interval for $\varphi(x)/x$; dividing by a negative $x$ swaps its endpoints, hence the inner $\min$/$\max$. The punctured interval is not compact, hence $\inf$/$\sup$, and the ratios can blow up near $0$: the sector is finite only if, e.g., $|\mu_D(x)|+\beta_D\sigma_D(x)\le C|x|$ near $0$.)
  4. Multipliers. With $\hat\kappa_1^{(i)}\le0\le\hat\kappa_2^{(i)}$ (Assumption 3), full-block multipliers on the vertices of the estimated sector box cover all channels; step 2 with $\delta/n_p$ per channel and a union bound makes the IQC description valid with probability $1-\delta$ (Lemma 3).
  5. Synthesise. IQC analysis LMIs (via the KYP lemma, Module 2) alternate with controller synthesis until the robust performance level $\tilde\gamma$ stops improving (later use for neural networks: Module 14).
Going deeper — Why vertex conditions give a valid multiplier

Write the diagonal nonlinearity as $p=\Theta q$ with $\Theta=\mathrm{diag}(\theta_i)$ and $\theta_i=\varphi_i(q_i)/q_i\in[\hat\kappa_1^{(i)},\hat\kappa_2^{(i)}]$ (any value in the interval if $q_i=0$). A symmetric $\Pi$ is a valid static multiplier if $[q;p]^\top\Pi[q;p]\ge0$ for all such pairs. The full-block class of Fiedler et al. (their eq. 18) requires

$$\Pi_{pp}:=\begin{bmatrix}0\\I\end{bmatrix}^\top\Pi\begin{bmatrix}0\\I\end{bmatrix}\preceq0,\qquad\begin{bmatrix}I\\\Theta\end{bmatrix}^\top\Pi\begin{bmatrix}I\\\Theta\end{bmatrix}\succeq0\ \ \text{at every vertex }\theta_i\in\{\hat\kappa_1^{(i)},\hat\kappa_2^{(i)}\}.$$

This suffices: for fixed $q$, $\theta\mapsto[q;\Theta q]^\top\Pi[q;\Theta q]$ is quadratic in $\theta$ with second-order part $(\Theta q)^\top\Pi_{pp}(\Theta q)\le0$, hence concave, and a concave function on a box takes its minimum at a vertex, where it is nonnegative. With probability at least $1-\delta$ every true $\varphi_i$ lies in its estimated sector (Lemma 3), and on that event any controller certified for the whole class (well-posedness, stability, and a performance level such as $\|z\|_{L_2}\le\tilde\gamma\|w\|_{L_2}$ from zero initial state) is certified for the true plant. Because the sectors are learned on $[a_i,b_i]$ only, the loop signals must also be shown to stay there (caveat below).

Theorem — Learning-enhanced robust synthesis (Fiedler, Scherer & Trimpe 2021, Theorem 1)
Assumptions. Assumptions 1–3; learning per channel with $\delta/n_p$; multipliers as in Lemma 3 (full-block, vertex conditions above); the synthesis certifies well-posedness, stability and the performance level for the whole sector class and returns $K$ after $M\ge1$ iterations with robust performance level $\tilde\gamma_{M+1}$. Since the sectors are learned on $[a_i,b_i]$ only, the loop signals must stay in these intervals (our addition; the paper omits the details, see the caveat).
Statement. With probability at least $1-\delta$, $K$ stabilises the true system and achieves robust performance level $\tilde\gamma_{M+1}$.
In words. The ground truth lies in the learned set with probability $1-\delta$ and the robust guarantee covers the whole set; more data shrink the set and improve performance.
Caveat — reading the CDC 2021 result correctly
  • The constant. Proposition 1 prints $\beta_D=B+2R\sqrt{\log\det(K_D+\bar\lambda I_N)-2\log\delta}$, $\bar\lambda=\max\{1,\lambda\}$: a pre-correction form. Like the AAAI 2021 print version (which has $R$ where this has $2R$) it lacks the factor $1/\sqrt\lambda$ and the $\bar\lambda/\lambda$ inside the determinant that the correction added. For $\lambda\ge1$ it dominates the corrected bound and stays valid, only conservative; for $\lambda\lt1$ use the corrected Theorem 1 of Module 3.
  • Local sectors. The sector is computed on $[a,b]$ but used as a global condition; the loop signals must be shown to stay in $[a,b]$ (the paper says this can be done with robust-control techniques and omits details), or the certificate is local.
  • Known $B$ and frequentist reading. The RKHS-norm bound must be given (Module 5). The authors contrast their setting with randomised robust control such as the scenario approach: there probability refers to the algorithm's samples, here only the data are random and the ground truth is fixed.

Exploration with a-priori robustness. Holicki, Scherer & Trimpe, L-CSS 2021 take the complementary route for a partially unknown linear plant $P_0=\Delta_0\star P$, $\Delta_0\in\boldsymbol\Delta$ known. They partition $\boldsymbol\Delta=\bigcup_k\boldsymbol\Delta_k$ and, for each partition member, synthesise by IQC/LMI relaxation (their Theorem 1) a controller $F(k)$ that robustly stabilises all of $\boldsymbol\Delta$ while optimising the worst-case performance over $\boldsymbol\Delta_k$ only. Every controller of the family therefore stabilises every plant in $\boldsymbol\Delta$, so experiments can explore the index with any black-box method without risk of instability. The paper separates two gaps. Refining the partition around the best member shrinks the parameterisation gap, their (P). The "cost of safety", their gap (S), comes from demanding stabilisation of all of $\boldsymbol\Delta$, and a finer partition of the same set leaves it unchanged. It shrinks only when the members inconsistent with the measurements, those whose measured cost $L(k)$ of $F(k)$ exceeds the level $\gamma_\varepsilon^\ast(k)$ that Theorem 1 certifies for $F(k)$ over $\boldsymbol\Delta_k$, are discarded (their eq. 13, since $\Delta_0\in\boldsymbol\Delta_k$ implies $L(k)\le\gamma_\varepsilon^\ast(k)$) and the family is redesigned for the smaller remaining set. It is the robust-control answer to safe BO (Module 4): certify candidates a priori, not statistically.

Fact — Safe experiments and sound elimination in the controller family (Holicki et al. 2021)
Assumptions. The true plant is $\Delta_0\star P$ with $\Delta_0\in\boldsymbol\Delta=\bigcup_k\boldsymbol\Delta_k$; for each tested index $k$ the synthesis certifies that $F(k)$ well-posedly stabilises $\Delta\star P$ for every $\Delta\in\boldsymbol\Delta$ and that $\|\Delta\star P\star F(k)\|_\infty\le\gamma_\varepsilon^\ast(k)$ for every $\Delta\in\boldsymbol\Delta_k$; the measured cost is the same quantity, $L(k)=\|\Delta_0\star P\star F(k)\|_\infty$, or a valid lower bound on it.
Statement. Every experiment with a fixed $F(k)$ is stable. If $L(k)\gt\gamma_\varepsilon^\ast(k)$, then $\Delta_0\notin\boldsymbol\Delta_k$ (otherwise the certificate would give $L(k)\le\gamma_\varepsilon^\ast(k)$), so discarding such members never discards the true plant. With a measurement error $|L_{\rm measured}(k)-L(k)|\le e_k$, discard only if $L_{\rm measured}(k)-e_k\gt\gamma_\varepsilon^\ast(k)$.
Scope. The certificates cover each controller used on its own; switching between controllers during one experiment is not covered, and a redesign for the remaining set must again certify all of it.

Probabilistic robust LQR with GPs (von Rohr et al. 2021)

von Rohr, Neumann-Brosig & Trimpe, L4DC 2021 differentiate a GP dynamics model at the operating point, obtaining a Bayesian (matrix-normal) posterior over $S=[\bar A\ \bar B]$, truncated to its credible region of mass $c$ (a region of posterior probability $c$; Primer C). With $\phi_K=\mathbb 1[\rho(\bar A+\bar BK)\lt1]$ they seek $\mathbb E[\phi_K]\ge1-\epsilon$ at minimal expected LQR cost. A priori: the initial controller solves a common-Lyapunov LMI over $M$ sampled systems, a convex scenario program with $n_k=2d_x^2+d_xd_u$ variables; their Theorem 1 gives $\mathbb E[\phi_{K_{\mathrm{init}}}]\ge1-\epsilon$ with probability greater than $1-\beta$ if $M\ge\lceil\frac2\epsilon(\ln\frac1\beta+n_k)\rceil$. A posteriori: the improvement steps (a majorise–minimise scheme averaging over samples; see the MM background in Section 3) are not a scenario program, so the final $K^\ast$ is returned only if its empirical stability rate on $M_{\mathrm{val}}$ fresh samples satisfies $\hat\phi\ge1-\epsilon$.

Background — Derivatives of a GP, matrix-normal posteriors and the common-Lyapunov LMI

Derivatives of a GP. If the mean $m_a$ and covariance $k_{ab}(z,z')$ of a vector-valued GP have continuous mixed second derivatives near the operating point, difference quotients are linear combinations of jointly Gaussian values, and their mean-square limits, the partial derivatives, are again jointly Gaussian with mean $\partial_im_a(z)$ and covariance

$$\operatorname{Cov}\big(\partial_if_a(z),\partial_jf_b(z')\big)=\partial_{z_i}\partial_{z'_j}k_{ab}(z,z').$$

Conditioning on data with Gaussian noise (Primer C) gives a Gaussian posterior over these derivatives, i.e. over the entries of $S=[\bar A\ \bar B]$. Matrix-normal is a special, separable structure: $\operatorname{vec}(S)\sim\mathcal N(\operatorname{vec}(M),C\otimes R)$, with $R$ the covariance between rows, $C$ between columns, and $C\otimes R$ the block matrix with blocks $C_{ij}R$; general derivative posteriors need not have it. Either way this is uncertainty about a local linearisation, not a nonlinear stability certificate.

The common-Lyapunov LMI. For sampled systems $(A_i,B_i)$ look for $X\succ0$ and $Y$ with

$$\begin{bmatrix}X&(A_iX+B_iY)^\top\\A_iX+B_iY&X\end{bmatrix}\succ0\quad\text{for every }i,\qquad K:=YX^{-1}.$$

With $Y=KX$ the off-diagonal block is $(A_i+B_iK)X$; the Schur complement (Module 2) turns the LMI into $X-X(A_i+B_iK)^\top X^{-1}(A_i+B_iK)X\succ0$, and multiplying by $X^{-1}$ on both sides gives $(A_i+B_iK)^\top P(A_i+B_iK)-P\prec0$ with $P=X^{-1}$: one Lyapunov matrix for all sampled systems. The conditions are linear in $(X,Y)$, so this is a convex scenario program; its $n_k$ must count all scalar decision variables, including any performance variables.

Caveat — corrected statement of the validation theorem (von Rohr et al. 2021, Theorem 2)
Theorem 2 prints $M_{\mathrm{val}}\ge\frac1{2\epsilon_{\mathrm{val}}}\log\frac1\alpha$, but its proof uses $\mathbb P(\hat\phi-\mathbb E\phi\ge\epsilon_{\mathrm{val}})\le e^{-2M_{\mathrm{val}}\epsilon_{\mathrm{val}}^2}$, which is at most $\alpha$ only if $$M_{\mathrm{val}}\ \ge\ \frac{\log(1/\alpha)}{2\,\epsilon_{\mathrm{val}}^2}:$$ for $\alpha=0.01$, $\epsilon_{\mathrm{val}}=0.02$, 5757 samples instead of the printed 116. Eq. (8) prints $\mathbb P(\mathbb E[\phi_{K^\ast}]\ge1-\epsilon_{\mathrm{pr}})\gt1-\alpha$ with $\epsilon_{\mathrm{pr}}=c-(\epsilon+\epsilon_{\mathrm{val}})$, whereas the proof concludes $\mathbb P(\mathbb E[\phi]\le\epsilon_{\mathrm{pr}})\le\alpha$. Consistent statement: with probability at least $1-\alpha$ over the validation samples, the posterior probability that $K^\ast$ stabilises the true linearisation is at least $c-\epsilon-\epsilon_{\mathrm{val}}$, since Hoeffding gives $\mathbb E_{P_c}[\phi]\ge1-\epsilon-\epsilon_{\mathrm{val}}$ on the truncated posterior and truncation removes mass $1-c$: $\mathbb E_P[\phi]\ge c(1-\epsilon-\epsilon_{\mathrm{val}})\ge c-\epsilon-\epsilon_{\mathrm{val}}$. Both slips are visible in the full text (statement versus proof); arXiv v2 fixes a different typo in Algorithm 1. This guarantee is Bayesian, unlike Fiedler et al.'s, and the algorithm repeats until the test passes, so the union-bound remark of Section 6 applies.
Sample-Size Calculator: Scenario Design versus Hoeffding Validation
Limitations and open problems
  • Dimensionality. Grid certificates (Section 2) scale exponentially with the state dimension; neural certificates move the burden to verification, exponential in the worst case (Section 4).
  • Unverifiable constants. RKHS-norm bounds, Lipschitz constants and $\beta_n$ are assumed known; practice substitutes heuristics ($\beta_n=2$, local constants, extrapolated norms) that void the theorems (Module 5).
  • Guarantee types do not compose automatically. A posterior probability (von Rohr), a confidence over data (Fiedler), a confidence over algorithmic samples (scenario) and one over validation runs (Hertneck) are different objects; chaining them needs union bounds and care about which randomness is meant.
  • Model versus reality. "Deterministic" guarantees are relative to a model (Hose) or to a model with a bounded error set (Aswani, tube MPC); that error bound is itself the learning problem.
  • Learning while certified. Efficient certified exploration (Theorem 4 of Section 2 is finite-sample, not rate-optimal) and online certification of high-dimensional learned policies remain open; safety filters (Module 10) are the pragmatic answer.
How the pieces connect
Sections 1–2 are the stability counterpart of the invariance certificates of Module 10 and reuse the GP machinery of Modules 3–4. Section 3 is constrained RL (Module 8, Module 9) with one linear Lyapunov constraint per state instead of a Lagrange multiplier (the guarantee is still the expected-cost one of the CMDP, not state-wise safety). For the verification algorithms used in Section 4, see Module 15. Sections 5–7 are where learning meets the LMI/IQC toolkit of Module 2, which the Pauli modules (12–14) turn on neural networks themselves.

Walkthrough: Certifying a Region of Attraction from Samples

The heart of Section 2 is a Lipschitz interpolation argument that turns finitely many uncertain evaluations into a statement about a continuum of states. Step through it with the notation of Section 2: $u_n$ is the upper confidence bound on $V(f(x,\pi(x)))$, $X_\tau$ a grid with $\|x-[x]_\tau\|_1\le\tau$, and $E$ the event of probability at least $1-\delta$ on which the model is calibrated.

Interactive: Certified Invariant Set Under Model Uncertainty

Compare a certified invariant set for an uncertain pendulum with simulated trajectories. Increase the uncertainty or change the grid resolution and predict how the certificate changes. The certificate concerns forward invariance and ultimate boundedness; the shaded capture region is empirical, and convergence to the origin is not certified.

Background — Pendulum model, grid certificate and interpretation

A torque-limited pendulum, $\ddot\theta=\frac gl\sin\theta-b\dot\theta+u$ with $g/l=10$ (gravity over length, not the unknown $g$ below), $b=0.1$, $|u|\le5$, Euler-discretised with $\Delta t=0.1$ (Primer D), is stabilised by an LQR designed on its linearisation, and $V=x^\top Px$ solves $A_{cl}^\top PA_{cl}-P=-I$. The true system adds an unknown $g(x)$ to the velocity update with $|g|\le\sigma_g$ (a stand-in for a GP confidence width) and Lipschitz constant $2\sigma_g$ in the normalised coordinates $(\theta/0.5,\ \omega/1.5)$ in which the grid and all constants live. At each point of an $n_g\times n_g$ grid the explorer computes the exact worst case $u(\bar x)=\max_{|g|\le\sigma_g}V(f(\bar x))$ and tests $u(\bar x)\lt V(\bar x)-L(\bar x)\tau$, with $L(\bar x)$ a rigorous bound on the Lipschitz constant of $\Delta V$ over the grid cell of $\bar x$ (the square of side $\tau$ around it), times $\vartheta$. In the 1-norm that constant is the supremum of $\|\nabla\Delta V\|_\infty$ over the cell (dual norms: Primer B): for each saturation branch of the input that meets the cell, the explorer takes the largest gradient on a $5\times5$ sub-grid, adds an interval bound on the Hessian times the sub-grid's covering radius, and adds an analytic bound for the $g$-part. (The largest sampled gradient alone is not a bound: on this pendulum it falls short of the cell supremum by up to about 20% on 4 to 7% of the checked cells.) It finds the largest passing $c$ among 300 tested levels such that all grid points of $\mathcal V(c+L_V\tau)$ outside a small inner set pass (Step 4), checks that the inner set cannot be left beyond $\mathcal V(c)$, and simulates one sampled $g$ from the certified boundary. The certificate is forward invariance of $\mathcal V(c)$ plus ultimate boundedness, not a region of attraction of the origin: the class $|g|\le\sigma_g$ does not force $g(0)=0$, and a sampled $g$ generally shifts the equilibrium (for the default draw to $\theta\approx-0.016$ rad). The shading is empirical: the cells whose simulated 150-step trajectory ends within $0.15$ (normalised units) of the origin, a finite-horizon capture region, not a certificate.

Fact — Inner-set escape bound and ultimate boundedness
Assumptions. $V\ge0$ is continuous, $\{V\le c\}$ is compact and all successors are defined. For every allowed disturbance: every state with $c_{\rm in}\lt V(x)\le c$ has $V(x^+)\le V(x)-\eta$ for a fixed $\eta\gt0$, and every state with $V(x)\le c_{\rm in}$ has $V(x^+)\le c_{\rm in}+b$. Let $c_{\rm ult}:=c_{\rm in}+\max(0,b)\le c$.
Statement. $\{V\le c\}$ and $\{V\le c_{\rm ult}\}$ are forward invariant, and every trajectory starting in $\{V\le c\}$ enters $\{V\le c_{\rm ult}\}$ after at most $(c-c_{\rm ult})/\eta$ steps.
Proof. From $V(x)\le c_{\rm in}$ the next value is at most $c_{\rm ult}$; from $c_{\rm in}\lt V(x)\le c$ it is smaller than $V(x)$. So from $V\le c$ one stays in $V\le c$, and from $V\le c_{\rm ult}$ in $V\le c_{\rm ult}$. While a trajectory is above $c_{\rm ult}\ge c_{\rm in}$, each step lowers $V$ by at least $\eta$, which can happen at most $(c-c_{\rm ult})/\eta$ times.
Scope. This elementary extension of the decrease argument of Section 1 says where trajectories end up; it does not certify convergence to the origin. Decrease outside an inner set without the escape bound $b$ would not even give invariance of $\{V\le c\}$.
Derivation — How the explorer computes its certificate

Coordinates. With $x=(\theta,\omega)=Dz$, $D=\operatorname{diag}(0.5,1.5)$, Euler's method gives $\theta^+=\theta+0.1\,\omega$ and $\omega^+=\omega+0.1\,\big(10\sin\theta-0.1\,\omega+\operatorname{clip}(Kx,-5,5)\big)+g(x)$. In normalised coordinates $V=z^\top P_zz$ with $P_z=D^\top PD$, and the velocity disturbance becomes $d=g/1.5$, $|d|\le s:=\sigma_g/1.5$. The grid on $[-1,1]^2$ has spacing $h=2/(n_g-1)$; every point is within $h/2$ of its nearest node in each coordinate, so the 1-norm covering radius is $\tau=h$. The level search tests 300 levels and reports the largest one that passes, not an exact continuous optimum; the physical area of the certified ellipse is $\pi c\det(D)/\sqrt{\det P_z}$.

Exact worst case over $g$. Let $y$ be the nominal successor of $\bar x$ (without $g$) in normalised coordinates and $e_2=(0,1)^\top$. Then $V(y+e_2d)=V(y)+2d\,(P_zy)_2+(P_z)_{22}\,d^2$, a convex quadratic in $d$, so its maximum over $|d|\le s$ sits at an endpoint:

$$u(\bar x)=V(y)+2s\,|(P_zy)_2|+(P_z)_{22}\,s^2 .$$

Lipschitz bound of $\Delta V$ on a cell. By Hölder, $|\nabla D(x)^\top v|\le\|\nabla D(x)\|_\infty\|v\|_1$, so the 1-norm Lipschitz constant of a smooth $D$ on a convex cell is $\sup\|\nabla D\|_\infty$ there. If every Hessian entry of $D$ is at most $H$ in magnitude and every point of the cell is within 1-norm distance $r$ of a sample point ($r=\tau/4$ for the $5\times5$ sub-grid), the mean-value theorem applied to each partial derivative gives $\sup\|\nabla D\|_\infty\le\max_{\rm samples}\|\nabla D\|_\infty+Hr$. $H$ is bounded by interval arithmetic, which propagates ranges that enclose every possible value (for $x\in[-1,2]$, $x^2\in[0,4]$, not the endpoint range $[1,4]$). This is done on every saturation branch of the input that meets the cell; since $\Delta V$ is continuous, the largest branch bound is a Lipschitz constant on the whole cell. The code evaluates these enclosures in ordinary floating point, so the bound is mathematically rigorous but not a machine-checked rounding certificate.

Inner set and escape bound. The inner level $c_{\rm in}$ is the largest $V$ of a failing node plus $L_V\tau$, so every point whose nearest node fails lies in $\{V\le c_{\rm in}\}$. On the nodes with $V\le c_{\rm in}+L_V\tau$, which include the nearest node of every inner point, let $b$ be the largest value of $u(\bar x)-V(\bar x)+L(\bar x)\tau$. For an inner point $x$ with nearest node $\bar x$, $\Delta V(x)\le\Delta V(\bar x)+L(\bar x)\tau\le u(\bar x)-V(\bar x)+L(\bar x)\tau\le b$, so $V(x^+)\le c_{\rm in}+b$. The explorer reports $c_{\rm ult}=c_{\rm in}+\max(0,b)$ and accepts the level $c$ only if $c_{\rm ult}\le c$; the strict inequality at the finitely many passing nodes supplies the uniform margin $\eta$ of the Fact above.

Certified Invariant Sublevel Set of an Uncertain Pendulum

What to try. (1) Raise $\sigma_g$: the certified ellipse (green) shrinks away from the grey $\sigma_g=0$ certificate and the inner set grows; at $n_g=40$ nothing is certified beyond $\sigma_g=0.055$, a finer grid recovers it. (2) Switch to the composite constant: about four times the largest cell bound on the box and, at the default settings, about 6.5 to 60 times the cell bounds on the certified set; nothing is certified at any grid size offered, which is why fine grids and local constants are used in practice. (3) Set $\vartheta\lt1$: the certificate grows but is no longer backed by a bound, and the trajectories usually still behave, which is exactly why such shortcuts are tempting. (4) Compare with the shaded capture region, simulated for one sampled $g$ (empirical, not a certificate): the certificate is limited by the shape of $V$, not only by uncertainty.

From the mathematics to a real decision

What you will be able to do
  • Distinguish nominal convergence, robust invariance and convergence to a disturbance-dependent neighborhood.
  • Allocate a physical margin between disturbance and controller approximation error.
  • Separate deterministic certificates from statistical validation of a fixed controller.

Replacing a thermal controller with a fast approximation

A temperature-control example connects stability, approximation and validation. Let $x_k$ be temperature error from a target, in degrees Celsius. A hypothetical identified one-step model is $x_{k+1}=0.8x_k+u_k+w_k$, where $u_k$ is the temperature change attributed to the control action over one sample and $|w_k|\le0.05$. The permitted band is $|x_k|\le1$. This model is normalized to a fixed sampling interval; changing that interval requires re-identification or a justified discretization.

Take the reference feedback $u_k=-0.3x_k$. A fast approximation returns $u_k=-0.3x_k+\eta_k$, with a uniformly certified error $|\eta_k|\le0.02$ over the operating region. These are assumptions for the example, not claims that a neural network automatically has a uniform error bound. The combined error $e_k=w_k+\eta_k$ obeys $|e_k|\le0.07$ by the triangle inequality.

Worked certificate: a neighborhood replaces exact convergence

Without either error, the closed-loop map is $x_{k+1}=0.5x_k$. For $V(x)=x^2$, the change is $V(x_{k+1})-V(x_k)=(0.25-1)x_k^2=-0.75x_k^2$. Every nonzero state decreases its value, and $x_k=0.5^kx_0$ tends to zero. Both statements are exact for this nominal model.

With errors, absolute values give $|x_{k+1}|\le0.5|x_k|+0.07$. Repeated substitution produces

$$|x_k|\le0.5^k|x_0|+0.07\sum_{j=0}^{k-1}0.5^j=0.5^k|x_0|+0.14(1-0.5^k).$$

The interval $[-0.14,0.14]$ is robustly invariant, because $0.5(0.14)+0.07=0.14$. States outside it have a strict decrease in this absolute-value upper bound. The displayed sequence bound approaches 0.14, rather than zero. This is not merely a loose conceptual distinction: constant error $e_k=0.07$ makes $x=0.14$ a fixed point, so exact convergence cannot be guaranteed from these assumptions.

The larger permitted interval $[-1,1]$ is also invariant: a state in it has next magnitude at most $0.5+0.07=0.57\le1$. Invariance answers whether the band can be violated; the smaller interval describes the eventual accuracy allowed by persistent errors. A controller can satisfy one goal without satisfying the other.

Worked design: spend a margin before training

Let a nominal trajectory obey $z_{k+1}=0.5z_k$ and initialize $z_0=x_0$. The deviation $q_k=x_k-z_k$ obeys $q_{k+1}=0.5q_k+e_k$. Thus $|q_k|\le0.14$ for all $k$. Planning nominal states inside $|z_k|\le0.86$ ensures $|x_k|\le0.86+0.14=1$. This subtraction of the tube radius from the physical band is constraint tightening.

Suppose the design instead requires a tube of radius at most 0.1 degree. Write the allowable approximation error as $a\ge0$. The invariant-radius inequality is $0.5r+0.05+a\le r$. At $r=0.1$, it demands $a\le0$. The disturbance already consumes the whole margin. Training a network with a “small” but positive uniform error cannot meet this particular requirement without changing the feedback gain, disturbance bound or required radius.

The calculation provides a specification for approximation before training. It also has a local-to-global condition: the error bound must hold on every state the invariant-set argument can visit. An approximation evaluated only along nominal training trajectories does not establish that condition.

Validation is evidence about a distribution, not a replacement for a uniform bound

For a fixed controller and fixed test horizon, draw IID complete test scenarios from a stated deployment distribution. Each scenario includes the initial condition, disturbance realization and any controller randomness; independent initial states alone do not ensure independent failures. Zero failures in those tests can bound failure probability under that scenario distribution. It does not prove a worst-case error bound over the full temperature interval or an infinite-horizon guarantee. Selecting or retuning the controller using those same test outcomes changes the statistical question and requires separate treatment.

Application exercises

Exercise 11.B1 — Medium: Diagnose a larger learned-controller error

The approximation error bound worsens to 0.04 while the disturbance remains 0.05. Find the invariant deviation radius and the tightened nominal state interval. Does the physical interval $[-1,1]$ remain invariant under the given feedback?

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Use the combined error 0.09 in both the radius equation and the one-step band check.

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The radius solves $0.5r+0.09=r$, hence $r=0.18$. The nominal interval becomes $[-0.82,0.82]$. Physical invariance still holds because $0.5(1)+0.09=0.59\le1$. The larger approximation error reduces the useful nominal planning region and worsens the accuracy neighborhood even though the broad safety band remains invariant.

Exercise 11.B2 — Hard: Design to an accuracy requirement

Retain $|w_k|\le0.05$ and nominal closed-loop coefficient 0.5, but require invariant deviation radius at most 0.16. Derive the largest allowable uniform approximation error and an associated nominal state bound for the physical interval $[-1,1]$. Explain why a mean squared prediction error would not directly satisfy this specification.

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Insert $r=0.16$ into $0.5r+0.05+a\le r$.

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The inequality is $0.08+0.05+a\le0.16$, giving $a\le0.03$. A radius 0.16 permits nominal states in $[-0.84,0.84]$. Mean squared error is an average over a distribution and can be small while a rare state has error above 0.03. The invariant-tube proof requires a pointwise bound throughout the operating region, or a different analysis explicitly designed for probabilistic errors.

Exercise 11.B3 — Hard: A rare approximation error defeats an average-error argument

Evaluate controller approximation error under a uniform test distribution on $[-1,1]$. Suppose $\eta(x)=0.1$ for $x\in[0.99,1]$ and zero elsewhere. Compute its mean squared error, root mean squared error and maximum error. Does the root mean squared error establish the uniform 0.03 specification above? Find the probability that 200 independent test states all miss the exceptional interval.

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The exceptional interval occupies 0.01 of a total length 2. A square must be averaged before taking its square root.

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The exceptional probability is 0.005. Mean squared error is $0.005(0.1)^2=0.00005$, and root mean squared error is $\sqrt{0.00005}\approx0.007071$. The maximum is 0.1, exceeding 0.03. Thus the small average does not establish the stated uniform bound. The chance of missing the interval in 200 independent tests is $0.995^{200}\approx0.3670$. The failure of this error-bound test alone does not prove that the actual closed loop violates its state band; it means the particular uniform-error certificate cannot be invoked. A different certificate using the error's location and state dynamics might establish more.

A useful handoff between control and learning

The control analysis states an allowable error and its domain; learning attempts to meet that specification; verification checks a bound; independent validation studies a sampling distribution. Reproduce that chain without looking at the formulas. Then continue to Module 12 to study bounds on network sensitivity, or work through the tank project to combine sensor uncertainty, action filtering and model checks.

Exercises

Graded practice

Each level has four problems. Write your steps before opening the answer; use the hint when you need a first move. Close the answer and retry after checking it.

Easy practice

Exercise 11.P1 — Easy: Compute decrease and a sublevel set

Let $F(x)=0.6x$ and $V(x)=x^2$. Find $\Delta V$, the set $\mathcal V(4)=\{x:V(x)\le4\}$ and its image under $F$. Is the decrease strict at every point?

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Square the coefficient $0.6$ before subtracting $V(x)$.
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$\Delta V=(0.6x)^2-x^2=(0.36-1)x^2=-0.64x^2$. The sublevel set is $[-2,2]$ and its image is $[-1.2,1.2]$, contained in the original set. Decrease is strict when $x\ne0$, and zero at the equilibrium $x=0$. This exception is expected: a fixed equilibrium cannot strictly decrease a function on every step.

Exercise 11.P2 — Easy: Read a quadratic certificate geometrically

Let $P=\mathrm{diag}(4,1)$ and $V(x)=x^\top Px$. Find the axis radii of $\{V\le1\}$. Test $x=(0.4,0.5)$ and $(0.6,0)$.

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Expand $V=4x_1^2+x_2^2$ and compare it with $1$.
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The radii are $1/2$ along the first axis and $1$ along the second. At $(0.4,0.5)$, $V=4(0.16)+0.25=0.89\le1$, so the point is inside. At $(0.6,0)$, $V=4(0.36)=1.44\gt1$, so it is outside. A Euclidean distance check alone would miss the different weights in the certificate.

Exercise 11.P3 — Easy: Check uncertain decrease at one point

Let $x^+=0.5x+w$, $|w|\le0.1$ and $V=x^2$. At $x=0.4$, find all possible successors and the largest next value of $V$. Does this prove decrease there? What happens at $x=0$?

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Find an interval for the successor and maximize its square at the endpoint of largest magnitude.
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At $x=0.4$, the successor lies in $[0.1,0.3]$. Its largest squared value is $0.09$, whereas current $V=0.16$, so $\Delta V\le-0.07$: every permitted disturbance decreases $V$ at this point.

At $x=0$, successors lie in $[-0.1,0.1]$ and can have positive $V$. The origin is not invariant under arbitrary nonzero disturbances. A pointwise decrease check away from zero does not prove asymptotic convergence to zero under this absolute disturbance bound.

Exercise 11.P4 — Easy: Check a local expected-budget inequality

A proposed budget function has $L(s)=0.7$, immediate expected cost $0.2$, discount $1/2$ and expected next value $0.8$. Check $L(s)\ge c(s)+\gamma\mathbb E[L(s^+)]$. What is the slack?

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Discount the next value, then add the immediate cost.
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The right side is $0.2+(1/2)(0.8)=0.6$, so $0.7\ge0.6$ holds with slack $0.1$. The function allocates enough current budget to pay the immediate cost and the discounted expected future budget. This is an expected-cost argument, distinct from a state-space Lyapunov function proving every trajectory returns to the origin.

Medium practice

Exercise 11.P5 — Medium: Carry a sample certificate between grid points

At a grid point, the estimated decrease is at most $-0.12$ and its error is bounded by $0.02$. The true decrease function is Lipschitz with constant $5$, and every queried state is within distance $0.01$ of a grid point. Bound its decrease. Repeat with grid radius $0.03$.

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Add estimation error and the Lipschitz interpolation margin to the sampled upper estimate.
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For radius $0.01$, the true off-grid decrease is at most $-0.12+0.02+5(0.01)=-0.05\lt0$. Thus the assumed error bounds certify strict decrease at those covered states. For radius $0.03$, the same expression is $-0.12+0.02+0.15=0.05$, which cannot certify decrease.

The positive upper bound does not prove the true decrease is positive. It means the grid and uncertainty margins are too large for this certificate. All constants must use the same norm and cover the region being certified.

Exercise 11.P6 — Medium: Build an invariant error tube and tighten constraints

An error evolves as $e^+=0.5e+w$, $|w|\le0.1$. Prove $|e|\le0.2$ is invariant. If actual state is $x=z+e$ and $|x|\le1$, find a sufficient nominal bound on $z$. For feedback correction $-0.5e$ and input bound $|u|\le1$, give a sufficient bound on nominal input $v$.

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Use the triangle inequality separately for the error, the actual state and the actual input.
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If $|e|\le0.2$, then $|e^+|\le0.5(0.2)+0.1=0.2$, proving invariance. For the state, $|x|\le|z|+|e|\le|z|+0.2$, so $|z|\le0.8$ suffices. The correction magnitude is at most $0.5(0.2)=0.1$, so $u=v-0.5e$ satisfies $|u|\le1$ whenever $|v|\le0.9$. Tightening reserves room for every error in the tube, rather than only its expected value.

Exercise 11.P7 — Medium: Solve a one-step MPC objective with an input limit

For $x^+=x+u$, minimize $x^2+u^2+2(x+u)^2$ with $|u|\le1/2$. At $x=1$, find the unconstrained minimizer, constrained minimizer, successor and objective value.

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Differentiate with respect to $u$ while holding $x$ fixed, then project onto the input interval.
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The derivative is $2u+4(x+u)=6u+4x$, giving $u=-2x/3$. At $x=1$, this is $-2/3$, outside the input interval. The objective is strictly convex, so the constrained optimum is the nearest allowed value $u^\ast=-1/2$. The successor is $1/2$, and the cost is $1+1/4+2(1/4)=7/4=1.75$.

This solves the stated finite-horizon optimization. A guarantee of recursive feasibility or stability still needs the relevant terminal and model assumptions.

Exercise 11.P8 — Medium: Interpret zero failures through Hoeffding

A fixed controller is tested on $1000$ independent identically distributed Bernoulli failure trials with no observed failures. Using one-sided Hoeffding at confidence $0.95$, find the bound $\sqrt{\log(20)/(2\cdot1000)}$ on its failure probability. Does it certify a $1\%$ target?

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Zero empirical failure rate leaves the confidence radius as the upper bound.
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The radius is $\sqrt{\log(20)/2000}\approx0.038702$. With probability at least $0.95$ over repeated validation samples, the failure probability is no more than empirical rate plus this radius; for the observed zero rate, the bound is about $3.87\%$. It does not certify $1\%$ using this inequality.

Zero observations are evidence, not proof of zero population failure. This confidence statement requires the controller and sample size to be fixed independently of the validation outcomes.

Hard practice

Exercise 11.P9 — Hard: Find the strict-decrease region of a nonlinear loop

For $F(x)=x-x^3$ and $V=x^2$, derive $\Delta V$ and find which sublevel sets $\{V\le c\}$ with $c\gt0$ have strict decrease at all nonzero points. Prove convergence on $\{V\le1\}$ and explain failure of strict decrease at $c=2$.

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Factor the decrease as $x^4(x^2-2)$. If a positive limiting value of $V$ existed on $\{V\le1\}$, decrease would stay bounded away from zero.
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$\Delta V=(x-x^3)^2-x^2=-2x^4+x^6=x^4(x^2-2)$. It is negative for $0\lt|x|\lt\sqrt2$, so exactly the sublevels with $0\lt c\lt2$ have the requested strict decrease.

On $V\le1$, decrease keeps the trajectory in the sublevel set. The nonnegative decreasing sequence $V_t$ has a limit $\ell\ge0$. If $\ell\gt0$, then $x_t^2\ge\ell$ and $x_t^2\le1$, giving $\Delta V_t=x_t^4(x_t^2-2)\le-\ell^2$. Repeatedly subtracting this positive constant would eventually make $V_t$ negative, a contradiction. Hence $\ell=0$ and $x_t\to0$.

At $x=\sqrt2$, $F(x)=-x$, so the two boundary points alternate and $V$ stays at $2$. This explains the strict endpoint exclusion.

Exercise 11.P10 — Hard: Distinguish relative and absolute policy approximation errors

For plant $x^+=x+u$, the exact controller is $u=-x/2$. An approximation adds error $e(x)$. Prove geometric convergence if $|e(x)|\le0.1|x|$. Can the absolute bound $|e(x)|\le0.1$ alone prove convergence to zero?

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Bound $|0.5x+e(x)|$. For the absolute-error case, try constant error $e(x)=0.1$.
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With relative error, $|x^+|\le0.5|x|+0.1|x|=0.6|x|$. Induction gives $|x_t|\le0.6^t|x_0|$, so the state converges to zero. Equivalently, $V(x^+)\le0.36V(x)$ and $\Delta V\le-0.64x^2$.

With only an absolute bound, the permitted constant error $e(x)=0.1$ gives $x^+=0.5x+0.1$, whose equilibrium is $0.2$. It satisfies the error bound but need not converge to zero. Vanishing error near the equilibrium is a substantive assumption, not just a smaller training loss.

Exercise 11.P11 — Hard: Choose a validation sample size and account for selection

For a fixed controller, take $n$ independent identically distributed Bernoulli failure trials under a fixed validation distribution, with common failure probability $p$. Observing zero failures rejects every $p\ge0.01$ at level $0.05$ when $(0.99)^n\le0.05$. Find the smallest $n$. If selecting among $20$ controllers fixed before testing on the validation data, find a sufficient zero-failure sample size per controller using a union bound.

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Take logarithms and remember $\log(0.99)$ is negative. Allocate failure probability $0.05/20$ to each controller for the simultaneous guarantee.
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For one fixed controller, $n\ge\log(0.05)/\log(0.99)\approx298.073$, so the smallest integer is $299$. If its failure probability were at least $0.01$, the chance of observing zero failures would be at most $(0.99)^{299}\lt0.05$.

For simultaneous validity over $20$ preselected controllers, use level $0.0025$ for each, giving $n\ge\log(0.0025)/\log(0.99)\approx596.146$, hence $597$. The union bound then limits any false zero-failure acceptance to $0.05$; independence between controllers’ tests is not needed. Reusing validation outcomes to invent arbitrary new controllers requires another argument.

Exercise 11.P12 — Hard: Compose statistical coverage with a robust certificate

A robust controller is safe for every model in a learned set $\mathcal M$. Suppose $\mathbb P(f_{\rm true}\in\mathcal M)\ge0.99$. State the inherited guarantee. If two learned sets are simultaneously needed, with failure probabilities $0.01$ and $0.02$, give a bound without independence. Why are pointwise coverage statements at individual states insufficient by themselves?

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On the coverage event, the robust theorem applies to the true model. Combine missing-coverage events with the union bound.
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Whenever $f_{\rm true}\in\mathcal M$, the deterministic robust guarantee covers the actual system. Therefore the closed-loop guarantee holds with probability at least $0.99$ over the data generating that set, provided all other theorem assumptions hold.

If both learned sets are needed, their simultaneous coverage is at least $1-0.01-0.02=0.97$. Robustness on their joint uncertainty description inherits that bound. No independence is needed.

A separate $99\%$ statement at each state does not give $99\%$ coverage of an entire trajectory or region. The uncertainty event must be uniform over every state and model error to which the robust certificate is applied, or be converted to such an event using an appropriate finite or uniform argument.

Original longer exercises

Exercise 11.1 — Strict decrease on a compact sublevel set (proof)

Let $F$ be continuous with $F(0)=0$, $V$ continuous and positive definite, $\mathcal V(c)$ compact, and $\Delta V(x)=V(F(x))-V(x)\lt0$ on $\mathcal V(c)\setminus\{0\}$. (a) Prove that $\mathcal V(c)$ is forward invariant. (b) Prove that $x_t\to0$ for every $x_0\in\mathcal V(c)$. (c) Show with $V(x)=x^2$ and a discontinuous $F$ that (b) fails without continuity, and locate the step that breaks.

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(a) Induction: if $x_t\in\mathcal V(c)\setminus\{0\}$ then $V(x_{t+1})\lt V(x_t)\le c$; if $x_t=0$ then $x_{t+1}=F(0)=0$.

(b) $V(x_t)$ is non-increasing and bounded below, so $V(x_t)\to v^\ast\ge0$. If $v^\ast\gt0$, all $x_t$ lie in $A:=\{x\in\mathcal V(c):V(x)\ge v^\ast\}$, compact (closed in a compact set) and not containing $0$. $\Delta V$ is continuous (composition of continuous maps) and negative on $A$, so it attains a maximum $-\eta\lt0$; then $V(x_t)\le V(x_0)-t\eta\to-\infty$, a contradiction. So $V(x_t)\to0$. If $x_t\not\to0$, a subsequence stays in the compact $\mathcal V(c)\cap\{\|x\|\ge\varepsilon\}$, where $V$ has a positive minimum, contradicting $V(x_t)\to0$.

(c) $F(x)=x/2$ for $x\le1$, $F(x)=(1+x)/2$ for $x\gt1$: for $0\ne x\le1$, $x^2/4\lt x^2$; for $x\gt1$, $(1+x)/2\lt x$. Sublevel sets are compact and $F(0)=0$, yet from $x_0=2$, $x_{t+1}=(1+1+2^{-t})/2=1+2^{-(t+1)}$, so $x_t=1+2^{-t}\to1$. The broken step is "$\Delta V$ attains a negative maximum on $A$": on $A=\{1\le|x|\le2\}$, $\Delta V(1)=-3/4$ but $\Delta V(x)\to0$ as $x\downarrow1$, so $\sup_A\Delta V=0$ is not attained.

Exercise 11.2 — The Lipschitz interpolation bound and why it is loose (derivation)

(a) Derive $|\Delta V(x)-\Delta V(\bar x)|\le L_{\Delta V}\|x-\bar x\|_1$ with $L_{\Delta V}=L_VL_f(L_\pi+1)+L_V$. (b) With $L_V=2$, $L_f=1.1$, $L_\pi=4$, $\tau=0.01$: compute the margin. Does a grid point with $V(\bar x)=0.5$, $u_n(\bar x)=0.35$ pass? Can one with $V(\bar x)=0.1$ ever pass? (c) Let $F_\pi(x)=x+\Delta t\,F(x)$ with $F$ $L_F$-Lipschitz, $\|F\|\le F_{\max}$, and $\nabla V$ $M$-Lipschitz with $\|\nabla V\|\le L_V$ (Euclidean norms). Show $\mathrm{Lip}(\Delta V)\le\Delta t\,(MF_{\max}(1+\Delta tL_F)+L_VL_F)$ and compare with the composite bound for $\Delta t=0.01$, $M=2$, $F_{\max}=1$, $L_F=3$, $L_V=2$.

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(a) $\|z-\bar z\|_1=\|x-\bar x\|_1+\|\pi(x)-\pi(\bar x)\|_1\le(1+L_\pi)\|x-\bar x\|_1$, then $|\Delta V(x)-\Delta V(\bar x)|\le L_V\|f(z)-f(\bar z)\|_1+L_V\|x-\bar x\|_1\le\big(L_VL_f(1+L_\pi)+L_V\big)\|x-\bar x\|_1$.

(b) $L_{\Delta V}=2\cdot1.1\cdot5+2=13$, margin $0.13$. The test needs $u_n\lt0.5-0.13=0.37$: $0.35$ passes. For $V(\bar x)=0.1$ the threshold is $-0.03\lt0\le u_n$: it can never pass, whatever the data (the inner region of Step 4).

(c) With $\psi(s):=V(x+s\Delta tF(x))$, the chain rule gives $\psi'(s)=\nabla V(x+s\Delta tF(x))^\top\Delta t\,F(x)$, and the fundamental theorem of calculus, $\Delta V(x)=\psi(1)-\psi(0)$, gives $\Delta V(x)=\int_0^1\nabla V(x+s\Delta tF(x))^\top\Delta t\,F(x)\,ds$ (the bounds on $\nabla V$ must hold along these segments). For two points, add and subtract $\nabla V(y+s\Delta tF(y))^\top F(x)$: $$\begin{aligned}\Delta V(x)-\Delta V(y)=\Delta t\int_0^1\Big[&\big(\nabla V(x+s\Delta tF(x))-\nabla V(y+s\Delta tF(y))\big)^\top F(x)\\&+\nabla V(y+s\Delta tF(y))^\top\big(F(x)-F(y)\big)\Big]ds.\end{aligned}$$ The first term is at most $M(1+\Delta tL_F)F_{\max}\|x-y\|$, the second at most $L_VL_F\|x-y\|$, which gives the claim: $0.01\,(2\cdot1.03+6)=0.0806$. The composite bound, even in its closed-loop form $L_V(\mathrm{Lip}(F_\pi)+1)\le2(1.03+1)=4.06$, is about 50 times larger, and the ratio diverges as $\Delta t\to0$.

Exercise 11.3 — Validating an approximate MPC with Hoeffding

Take $\delta_h=0.01$, $\mu_{\mathrm{crit}}=0.99$ as in Hertneck et al. (a) Your network violates the error bound on about $0.2\%$ of trajectories ($\tilde\mu\approx0.998$). How many validation trajectories are needed with the two-sided bound, and with the one-sided one? (b) With the paper's $p=34\,980$, how many failures can the test tolerate? (c) You may retrain and re-validate up to $K=5$ times, using fresh independent IID validation trajectories for each trained controller: which per-attempt confidence keeps the overall one at $0.99$, and what happens to $p$ in (a)? (d) State precisely what a passed test guarantees.

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(a) Need $\epsilon_h\le0.998-0.99=0.008$. Two-sided: $p\ge\ln200/(2\cdot0.008^2)=\ln200/(1.28\cdot10^{-4})\approx41\,393.1$, so $41\,394$. One-sided: $\ln100/(1.28\cdot10^{-4})\approx35\,977.9$, so $35\,978$.

(b) $\epsilon_h=\sqrt{\ln200/(2\cdot34\,980)}\approx0.0087025$, so the exact test requires $\tilde\mu\ge0.99+\epsilon_h\approx0.9987025$, i.e. at least $34\,935$ successes: at most 45 failures. (The reported $0.9987$ is rounded; exactly $0.9987$ would fail by about $3\cdot10^{-6}$.)

(c) Union bound: $0.01/5=0.002$ per attempt, so $p\ge\ln(1000)/(1.28\cdot10^{-4})\approx53\,966.8$, i.e. $53\,967$ (two-sided): about 30% more samples buy honesty about repeated testing.

(d) With probability at least $1-\delta_h$ over the validation data, with probability at least $\mu_{\mathrm{crit}}$, a fresh initial condition from the same $\Omega$ gives a trajectory that enters $X_f$ in finite time and satisfies $\|\pi_{\mathrm{approx}}-\pi_{\mathrm{MPC}}\|_\infty\le\eta$ at every step before entry; on it Theorem 5 gives constraint satisfaction up to that entry, after which the terminal controller $k_f$ keeps the state in $X_f$ and gives convergence (Remark 6). The guarantee is for this switched controller: $\pi_{\mathrm{approx}}$ itself is not validated inside $X_f$. Nothing is said for other initial distributions.

Exercise 11.4 — Scenario design and validation, numerically

(a) For $d=1$ show that the Campi–Garatti bound reduces to $(1-\epsilon)^N\le\beta$; compute $N$ for $\epsilon=0.05$, $\beta=10^{-3}$. (b) A common-Lyapunov LMI with $d_x=2$, $d_u=1$ has $n_k=2d_x^2+d_xd_u$ variables. For $\epsilon=0.05$, $\beta=10^{-5}$ compare the exact scenario sample size (calculator) with the sufficient formula. (c) For validation with $\alpha=0.01$, $\epsilon_{\mathrm{val}}=0.02$, compute $M_{\mathrm{val}}$ with the corrected and the misprinted rule; what confidence does the misprinted size give? (d) With $c=0.95$, $\epsilon=0.05$ and (c), what does the corrected theorem guarantee?

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(a) Only the term $i=0$ remains: $(1-\epsilon)^N\le\beta\iff N\ge\ln(1/\beta)/(-\ln(1-\epsilon))$. For the given values, $\ln1000/\ln(20/19)\approx134.7$ (with $\ln1000\approx6.9078$ and $\ln(20/19)\approx0.051293$), so $N=135$. Under the convex scenario-program assumptions, at most one sample can support the one-variable solution, and a violation above $\epsilon$ needs all $N$ samples to miss an $\epsilon$-mass region.

(b) $n_k=10$. The binomial tail first drops below $10^{-5}$ at $N=581$ ($9.67\cdot10^{-6}$; at $580$ it is $1.003\cdot10^{-5}$); the sufficient formula gives $\lceil40(\ln10^5+10)\rceil=861$, about 48% more.

(c) Corrected: $\ln100/(2\cdot0.02^2)\approx5756.5$, so $5757$. Misprinted: $\ln100/(2\cdot0.02)\approx115.1$, so $116$, for which $e^{-2\cdot116\cdot0.02^2}=e^{-0.0928}\approx0.911$. Hoeffding supplies the lower confidence bound $1-e^{-0.0928}\approx9\%$, rather than $99\%$ (a lower bound from this inequality, not the actual confidence of the procedure).

(d) For a fixed $K^\ast$ and one fresh independent batch of at least $5757$ validation samples from the normalized $c=0.95$ credible posterior, with probability at least $0.99$ over that batch, acceptance (empirical stability at least $0.95$) implies that the full posterior probability that $K^\ast$ stabilises the true linearisation is at least $0.95-0.05-0.02=0.88$: a Bayesian statement, as good as the GP posterior. This is a fixed-batch implication; repeated validation attempts need a separate failure budget.

Exercise 11.5 — Chow's LP on a three-state MDP

Transient states $s_1$ (start), $s_2$ (hazard), $s_3$, terminal state $T$; constraint cost $c(s_2)=1$, $c(s_1)=c(s_3)=0$, budget $d=0.5$. The baseline plays $a_B$ in $s_1$: to $s_2$ w.p. $0.2$, to $s_3$ w.p. $0.8$. Single actions: $s_2\to s_3$ w.p. $0.5$, $\to T$ w.p. $0.5$; $s_3\to s_2$ w.p. $0.1$, $\to T$ w.p. $0.9$. A faster action $a'$ in $s_1$ goes to $s_2$ w.p. $0.6$, to $T$ w.p. $0.4$. (a) Compute $V_c^{\pi_B}$ and the expected visits from $s_1$. (b) Write and solve the LP for $\epsilon$. (c) Repeat with a constant $\epsilon$. (d) For each, find the largest probability $p$ of $a'$ in $s_1$ allowed by the Lyapunov constraint and compare with the largest feasible $p$.

Show answer

(a) From $V=c+P_{\pi_B}V$: $V(s_3)=0.1V(s_2)$, $V(s_2)=1+0.05V(s_2)$, so $V(s_2)=20/19\approx1.0526$, $V(s_3)=2/19\approx0.1053$, $V_c^{\pi_B}(s_1)=28/95\approx0.2947$: feasible, slack $39/190\approx0.2053$. The first row of $(I-P_{\pi_B})^{-1}$, with $P_{\pi_B}=\begin{bmatrix}0&0.2&0.8\\0&0&0.5\\0&0.1&0\end{bmatrix}$, gives the visits: they solve $v_1=1$, $v_2=0.2v_1+0.1v_3$, $v_3=0.8v_1+0.5v_2$, so $v_2=28/95$, $v_3=18/19$, i.e. $(1,\ 28/95,\ 18/19)\approx(1,\ 0.2947,\ 0.9474)$ and $\mathbb E[T^\ast]=213/95\approx2.2421$. (The $L_\epsilon$ below solve $L=c+\epsilon+P_{\pi_B}L$: $L_2=1+\epsilon_2+0.5L_3$, $L_3=\epsilon_3+0.1L_2$, $L_1=\epsilon_1+0.2L_2+0.8L_3$.) (The loop $s_2\leftrightarrow s_3$ makes $T^\ast$ unbounded, so the paper's uniform bound $T$ fails here; every policy still terminates with probability one, $(I-P_\pi)^{-1}$ exists, and that is enough for this finite chain's LP and Steps 1–2 of the walkthrough. The uniform bound is an additional assumption of the paper; its Lemma 1 and Assumption 1 cannot be invoked directly for this chain.)

(b) The exact LP is $\max\ \epsilon_1+\epsilon_2+\epsilon_3$ s.t. $\epsilon_1+(28/95)\,\epsilon_2+(18/19)\,\epsilon_3\le39/190$, $\epsilon\ge0$: all slack on the smallest coefficient, the hazard: $\epsilon=(0,\ 39/56,\ 0)\approx(0,\ 0.6964,\ 0)$, $L_\epsilon=(1/2,\ 25/14,\ 5/28)\approx(0.5,\ 1.7857,\ 0.1786)$ with $L_\epsilon(s_1)=d$.

(c) $\epsilon\equiv(39/190)/(213/95)=13/142\approx0.09155$, $L_\epsilon=(1/2,\ 85/71,\ 15/71)\approx(0.5,\ 1.1972,\ 0.2113)$.

(d) In $s_1$: $p\,(Q_L(s_1,a')-Q_L(s_1,a_B))\le\epsilon(s_1)$. Vertex: $Q_L(s_1,a_B)=1/2$, $Q_L(s_1,a')=(3/5)\cdot(25/14)=15/14\approx1.0714$, $\epsilon(s_1)=0$, so $p=0$: no change allowed in $s_1$, because the slack went to $s_2$, where there is nothing to choose. Constant: $Q_L(s_1,a_B)=1/2$, $Q_L(s_1,a')=13/142+(3/5)\cdot(85/71)=115/142\approx0.8099$, so $p\le(13/142)/(22/71)=13/44\approx0.2955$; at this largest permitted probability, $V_c(s_1)=28/95+(32/95)\,p=412/1045\approx0.3943<1/2$. The true feasibility limit is $p\le39/64=0.609375\approx0.6094$: the Lyapunov constraint is sufficient, not necessary; $p=1$ ($V_c(s_1)=12/19\approx0.6316$) would be infeasible.

Key Papers

PaperVenueContributionWhy read it
Safe Model-based Reinforcement Learning with Stability GuaranteesNeurIPS 2017GP bounds plus Lipschitz continuity certify a Lyapunov sublevel set from a grid; safe policy optimisation and exploration inside itTemplate of GP-plus-Lipschitz safety arguments (read with the caveats of Section 2)
A Lyapunov-based Approach to Safe Reinforcement LearningNeurIPS 2018Lyapunov functions for CMDP budgets via an LP; safe policy/value iteration feasible at every iterateTurns a trajectory constraint into state-wise linear constraints
Provably safe and robust learning-based model predictive controlAutomatica 2013Tube constraints on a nominal model, learned oracle in the cost onlyThe seminal decoupling of safety from learning in MPC
Learning an Approximate Model Predictive Controller with GuaranteesIEEE L-CSS 2018Robust MPC tolerating bounded input errors, NN approximation, Hoeffding validationStart of certified approximate MPC; a statistical guarantee easy to misread
Approximate non-linear model predictive control with safety-augmented neural networksIEEE TCST 2025NN input sequences checked online, safe shifted candidate as fallbackDeterministic guarantees at the cost of one forward simulation
Learning-enhanced robust controller synthesis with rigorous statistical and control-theoretic guaranteesCDC 2021GP tube → sector bounds → IQC synthesis, guarantees with probability $1-\delta$Blueprint for composing learning bounds with robust control
Learning-Based Model Predictive Control for Safe ExplorationCDC 2018Ellipsoidal GP reachability, safe terminal set, backup controllerThe reference GP-based safe MPC
Lyapunov-stable Neural Control for State and Output Feedback: A Novel FormulationICML 2024Verifiable ROA condition on an invariant sublevel set; $\alpha,\beta$-CROWN; neural observersState of the art in verified neural Lyapunov control
Neural Lyapunov ControlNeurIPS 2019Lyapunov risk plus SMT falsifier in a learner–verifier loopOrigin of counterexample-guided neural certificates
Safe Control With Learned Certificates (survey)IEEE T-RO 2023Lyapunov, barrier and contraction certificates: conditions, losses, verificationEntry point to learned certificates (mind the sign convention)
Learning-Based Model Predictive Control: Toward Safe Learning in ControlAnnu. Rev. 2020Taxonomy: learned model, learned design, MPC as safety filterMap of learning-based MPC
Automatic nonlinear MPC approximation with closed-loop guaranteesIEEE TAC 2025ALKIA-X: kernel interpolation with a uniform error bound plus robust MPCDeterministic alternative to validation, with an honest RKHS caveat
Probabilistic Robust Linear Quadratic Regulators with Gaussian ProcessesL4DC 2021Scenario LMI initialisation plus Hoeffding validation for GP-linearised dynamicsBoth statistical tools in one example (read Theorem 2 corrected)
The Exact Feasibility of Randomized Solutions of Uncertain Convex ProgramsSIAM J. Optim. 2008Exact binomial-tail bound for scenario programsThe sharp sample complexity of scenario design
Data-driven Economic NMPC using Reinforcement LearningIEEE TAC 2020An MPC on a wrong model can be optimal after a cost modification; MPC as RL approximatorFoundation of MPC-based RL (pair with Zanon & Gros 2021)

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