Open Problems & Research Gaps

What is still unsolved in safe learning as of autumn 2026, organised by research line. Each problem comes with a plain-language explanation, its notation and a worked example, and links to the place in these notes where its background is taught

Before you start

Every problem below explains its own symbols, and Section 2 collects all notation in one table. For the full background, these sections are the best entry points:

Contents
1. How to Read This Page 2. Notation Used on This Page Interactive: Problem Map 3. Safe Exploration with Checkable Assumptions (Trimpe line) 4. Certified Neural Networks via Robust Control (Pauli line) 5. Constrained & Safe Deep RL 6. Safety Filters & Learned Certificates 7. Statistical Guarantees & Trustworthy Certificates 8. Where the Two Lines Meet 9. Gaps in These Notes Discussion Questions Key Papers

1. How to Read This Page

Reading this for the first time? Choose one problem connected to a module you have studied. Read its plain-language explanation and worked example, then follow its background links before tackling the formal question. These are research directions; use the graded exercises in the primers and modules to check your understanding first.

The fifteen modules present what is known. This page collects what is not known: the problems that the papers behind the modules, and the surveys and reviews that summarise them, name as open.

Every problem has the same parts, in the same order:

Problems are numbered by research line:

Caveat
"Open" means open in the literature reviewed for these notes, up to September 2026. The field moves quickly, so check for newer work before you build on any of these. Section 8 (bridges) is our own synthesis, not a published claim.
One thread runs through almost everything
Most open problems come back to the same tension. A guarantee is only as good as its assumptions, and the assumptions that make proofs work are often ones an engineer cannot check: a bound on an RKHS norm, calibrated uncertainty, exchangeable data, a known Lipschitz constant. Making those assumptions checkable, or replacing them with ones that are, is the frontier in both research lines this section follows.

2. Notation Used on This Page

The problems come from five literatures, and each has its own notation. The table lists every symbol used below, what it means, a tiny example, and where it is taught. Keep this section open in a second tab while reading.

SymbolMeaningExampleTaught in
General
$x \in D$, $x \in X$An input that you may try (in BO), or the state of a system (in control). $D$ is the set of allowed inputs, $X$ the state space.$x$ = a controller gain, $D = [0, 5]$Primer 0 §1
$f$The unknown function in BO, the dynamics in control, or the network in verification. Each problem says which.$f(x)$ = tracking error of gain $x$Module 4 §1
$\|v\|_2$, $\|v\|_\infty$Length of a vector: square root of the sum of squares; largest absolute entry.$v = (3, -4)$: $\|v\|_2 = 5$, $\|v\|_\infty = 4$Primer A §1
$O(\cdot)$, $\tilde O(\cdot)$"Grows at most like", up to a constant factor. The tilde also ignores logarithmic factors.$3T^2 + 5T = O(T^2)$, $\ T\log T = \tilde O(T)$Primer 0 §7
$\mathbb E[\cdot]$, $\mathbb P(\cdot)$Expectation (the average value) and probability.Fair die: $\mathbb E[X] = 3.5$, $\mathbb P(X \ge 5) = 1/3$Primer C §1
$\delta$Allowed failure probability of a guarantee: with probability at least $1 - \delta$ the statement holds.$\delta = 0.01$: fails with probability at most 1%Primer C §6
$\varepsilon$A small tolerance ("$\varepsilon$-optimal" = within $\varepsilon$ of the best value), or the size of an input perturbation.$\varepsilon$-optimal with $\varepsilon = 0.1$Primer 0 §4
Gaussian processes and safe BO
$k(x, x')$Kernel: how similar the function values at $x$ and $x'$ are expected to be.SE kernel $e^{-(x - x')^2/2}$: $k(0, 1) = 0.61$Module 3 §1
$\mathcal H_k$, $\|f\|_k$The RKHS: functions built from kernel bumps $k(\cdot, x_i)$. Its norm measures how large and how wiggly $f$ is, in the kernel's terms.$f = 2\,k(\cdot, 0)$ has $\|f\|_k = 2$ when $k(0,0)=1$Module 3 §1
$B$An assumed upper bound $\|f\|_k \le B$. It bounds the norm, not the squared norm.$B = 2$Module 3 §5
$\mu_t(x)$, $\sigma_t(x)$GP posterior mean and standard deviation after $t$ measurements.—Module 3 §2
$\beta_t$Width of the confidence band in standard deviations: $f(x) \in [\mu_t(x) - \beta_t\sigma_t(x),\ \mu_t(x) + \beta_t\sigma_t(x)]$.$\mu = 1$, $\sigma = 0.2$, $\beta = 2$: $[0.6, 1.4]$Module 3 §5
$l_t(x)$, $u_t(x)$Lower and upper confidence bounds. In SafeOpt they are the ends of the running intersection of all bands so far, so they can only tighten over time.Bands $[0, 2]$, then $[1, 3]$: $l = 1$, $u = 2$Module 4 §2
$h$ (in BO)Safety threshold: input $x$ is safe if $f(x) \ge h$.$h = 0$Module 4 §1
$S_t$, $G_t$, $M_t$Safe set (inputs certified safe after $t$ steps); expanders (safe inputs whose measurement could certify new ones); maximisers (safe inputs that could be optimal).—Module 4 §2
$L$Lipschitz constant: $|f(x) - f(x')| \le L\,d(x, x')$, a speed limit on how fast $f$ can change.$f(x) = 3x$: $L = 3$Primer B §3
$E$LoSBO's noise bound: every measurement error satisfies $|\epsilon_t| \le E$.$E = 0.1$Module 5 §3
$\gamma_T$Maximum information gain: the most a GP can learn about $f$ from $T$ noisy measurements. Grows like $(\log T)^{d+1}$ for the SE kernel in $d$ dimensions.$\gamma_T \approx 85$ at $T = 10^4$, $d = 1$ (constants ignored)Module 3 §4
$R_T$Cumulative regret $\sum_{t=1}^{T}\big(f(x^\star) - f(x_t)\big)$: total shortfall against the best input $x^\star$. "Sublinear" means $R_T/T \to 0$.$R_T = 2\sqrt T$ is sublinearPrimer E §1
Reinforcement learning and CMDPs
$s$, $a$, $\pi$State, action, and policy (the rule that picks $a$ given $s$).—Primer E §2
$r$, $c$, $\gamma$Reward, cost, and discount factor in $(0, 1)$: a reward $t$ steps ahead counts $\gamma^t$ times as much.$\gamma = 0.99$: 100 steps ahead counts $0.99^{100} = 0.37$Primer E §2
$V^\pi$, $J_r(\pi)$, $J_c(\pi)$, $d$Value function; expected discounted reward $J_r(\pi) = \mathbb E\big[\sum_t \gamma^t r(s_t, a_t)\big]$ and cost $J_c(\pi)$ (the same with $c$ in place of $r$); the cost limit in the constraint $J_c(\pi) \le d$.$d = 0.05$ crashes per episodeModule 8 §1
$\lambda$ (in RL)Lagrange multiplier: the price per unit of cost in $J_r(\pi) - \lambda\,(J_c(\pi) - d)$.$\lambda = 1$Module 8 §3
$\mathrm{CVaR}_\alpha$Conditional value at risk: the average of the worst $(1 - \alpha)$ fraction of outcomes.$\alpha = 0.9$: average of the worst 10%Primer C §9
$X_V$, $X_U$, $T_f$Viability kernel (states from which failure can be avoided forever); the unviable states outside it; maximum time to failure once in $X_U$.—Module 7 §1
$p$, $p^\star$, $\tau$Failure penalty, its sufficient threshold, and the discount time constant in continuous time (weight $e^{-t/\tau}$).$\tau = 1$ sModule 7 §4
Control
$\dot x = f(x) + g(x)u$Control-affine dynamics: the input $u$ enters linearly. $\dot x$ is the time derivative of $x$.$\dot x = u$Primer D §1
$h(x)$, $\mathcal C$ (in control)Barrier function and safe set $\mathcal C = \{x : h(x) \ge 0\}$.$h(x) = x$: stay at $x \ge 0$Module 10 §2
$\alpha(\cdot)$Class-$\mathcal K$ function: continuous, strictly increasing, $\alpha(0) = 0$.$\alpha(h) = 2h$Primer D §4
$V(x)$Lyapunov function: an energy-like function that decreases along trajectories.$V(x) = x^2$Primer D §3
$A$, $B$, $C$, $D$State-space matrices: $x_{t+1} = Ax_t + Bu_t$, $y_t = Cx_t + Du_t$.See P5Primer D §8
$M \succeq 0$, $M \preceq 0$, $M \succ 0$Matrix inequalities for symmetric $M$: all eigenvalues $\ge 0$, $\le 0$, or $\gt 0$.$\mathrm{diag}(1, 2) \succ 0$Primer A §4
LMI, BMI, SDPLinear matrix inequality: $M(z) \preceq 0$ with $M$ affine in the unknowns $z$ (convex, solvable). Bilinear: products of unknowns appear (non-convex). SDP: an optimisation problem with LMI constraints.See P3Module 2 §2
Certified neural networks
$W_i$, $b_i$, $\varphi$Weights, biases and activation function of layer $i$.$\mathrm{ReLU}(v) = \max(v, 0)$Primer E §7
$[\alpha, \beta]$ (slope)Slope restriction: every chord of $\varphi$ has slope between $\alpha$ and $\beta$.ReLU: $[0, 1]$Module 12 §3
$\Delta v$, $\Delta z$Increments between two inputs: $\Delta v = v - \bar v$ before and $\Delta z = \varphi(v) - \varphi(\bar v)$ after the activation.See P2Module 12 §3
$T = \mathrm{diag}(\lambda_i)$Multiplier matrix: one nonnegative weight per neuron in LipSDP.—Module 12 §4
$L^\star$The true Lipschitz constant of a network: the smallest valid $L$.$\sqrt5$ in P1Module 12 §1
Clean / certified accuracyFraction of test inputs classified correctly / classified correctly and provably unchanged for every perturbation up to radius $\varepsilon$.57.0% / 41.2% (P4)Module 13 §6
$\varepsilon = 36/255$Perturbation radius for images with pixel values in $[0, 1]$: 36 of the 255 brightness levels.$36/255 = 0.141$Module 13 §6
Statistics
$R^{(p)}$The $p$-th smallest of a list of scores (an order statistic).Scores $3, 1, 2$: $R^{(2)} = 2$Module 15 §5
ExchangeableThe joint distribution of the data does not change when they are reordered. Independent, identically distributed data are exchangeable.—Primer C §10
$\Phi$, $\Phi^{-1}$Cumulative distribution function of the standard normal distribution, and its inverse (the quantile function).$\Phi^{-1}(0.9) = 1.28$Primer C §4
$\sigma$ (smoothing)Standard deviation of the noise added in randomized smoothing.$\sigma = 0.5$Module 15 §4
Symbols that mean two things
The literatures reuse letters. On this page:
  • $h$ is a safety threshold in BO ($f(x) \ge h$) but a barrier function $h(x)$ in control.
  • $\gamma$ is the discount factor in RL, while $\gamma_T$ is the maximum information gain in BO.
  • $\beta_t$ is the confidence width in BO, while $[\alpha, \beta]$ is the slope range of an activation.
  • $\lambda$ is the Lagrange multiplier in constrained RL, while $\lambda_i$ are the neuron multipliers inside $T$ in LipSDP.
  • $T$ is a number of rounds or steps, the multiplier matrix in LipSDP, and (with subscript) the time to failure $T_f$.
  • $\sigma_t(x)$ is the GP standard deviation; $\sigma$ alone is the smoothing noise level.
  • $B$ and $\mathcal C$ are the RKHS bound and the safe set; plain $B$ and $C$ in a state-space model are matrices.

Interactive: Problem Map

Each row is an open problem, each column a module. A dot means the problem builds on that module. Filter by research line; click a row label to jump to the problem, or a column header to open the module.

Problem Map

3. Safe Exploration with Checkable Assumptions (Trimpe line)

The Trimpe group's recent work (RWTH Aachen, DSME) argues that a safety guarantee should rest only on assumptions a user can check. Their "On Safety in Safe Bayesian Optimization" paper made that concrete. They tested 100 functions with 10,000 runs each. SafeOpt with the common heuristic $\beta \equiv 2$ violated the safety constraint in 3.95% of runs on average, and in 28.62% of runs on the worst function (Fiedler et al., TMLR 2024). The problems below are what remains after that analysis.

T1Checkable regularity and noise assumptions

The question
SafeOpt-type guarantees need a known bound $B \ge \|f\|_k$ on the RKHS norm of the unknown function, and a correctly chosen kernel. Can these be replaced by, or derived from, quantities an engineer can actually verify?

In plain words. Safe BO decides that an untested input is safe by reasoning: "the function cannot change too much between the points I measured and this new point". To make "too much" precise, SafeOpt assumes the function is not too wiggly, in a sense set by the kernel, and that you know a number $B$ that bounds this wiggliness. Nobody can look at a robot and read off $B$. Guess it too small and the guarantee silently fails; guess it too large and the algorithm is so cautious that it barely explores.

Symbols in this problem

What is known.

Why it is hard. No finite data set can bound $\|f\|_k$ from above. If the RKHS is infinite-dimensional, some nonzero function $g$ vanishes at every sampled point: take any direction orthogonal to $\operatorname{span}\{k(\cdot,x_1),\dots,k(\cdot,x_n)\}$, since $g(x_i) = \langle g, k(\cdot,x_i)\rangle = 0$. Then $f + c\,g$ fits the data exactly for every number $c$, and its norm grows without bound as $c$ grows. Any norm estimate therefore needs an extra assumption about which functions are plausible.

Worked example — two functions that agree on every measurement

Use the SE kernel and two noise-free measurements, at $x_1 = 0$ and $x_2 = 1$. We build a function $g$ in the RKHS that is zero at both points but not in between.

Step 1: the shape. Start from a kernel bump centred between the data and subtract bumps at the data points, with the same weight $a$ on both (by symmetry):

$$g = k(\cdot, 0.5) - a\,k(\cdot, 0) - a\,k(\cdot, 1).$$

Step 2: make it vanish at the data. Using $k(0, 0.5) = e^{-1/8} = 0.8825$ and $k(0, 1) = e^{-1/2} = 0.6065$,

$$g(0) = e^{-1/8} - a\,\big(1 + e^{-1/2}\big) = 0 \quad\Longrightarrow\quad a = \frac{0.8825}{1.6065} = 0.5493 .$$

By symmetry $g(1) = 0$ as well.

Step 3: the value in the middle. $g(0.5) = 1 - 2a\,e^{-1/8} = 1 - 2(0.5493)(0.8825) = 0.0305$.

Step 4: the norm. Expand $\|g\|_k^2 = \langle g, g\rangle$ along the three bumps of $g$ and use the reproducing property:

$$\|g\|_k^2 = \langle g, k(\cdot,0.5)\rangle - a\langle g, k(\cdot,0)\rangle - a\langle g, k(\cdot,1)\rangle = g(0.5) - a\,g(0) - a\,g(1) = 0.0305,$$

so $\|g\|_k = 0.1745$.

Step 5: two explanations of the same data. Suppose you measured $f(0) = f(1) = 0$, and the true function is $f \equiv 0$, with norm 0. The function $f_c = c\,g$ produces exactly the same measurements for every number $c$. For $c = -100$ it has $f_c(0.5) = -3.05$ and $\|f_c\|_k = 17.45$.

What it means for safety. With threshold $h = -1$, the data fit a function that is safe everywhere ($f \equiv 0$) and one that is unsafe at $x = 0.5$ ($-3.05 \lt -1$). Only the assumed bound $B$ can separate them. The noise-free error bound of Module 3 §3 says $|f(x) - \mu(x)| \le \|f\|_k\,\sigma(x)$. Here the posterior mean is $\mu(0.5) = 0$ and the noise-free posterior standard deviation is $\sigma(0.5) = 0.1745$. That equals $\|g\|_k$, because $g/\|g\|_k$ is exactly the worst-case function of norm 1. So the guarantee is $f(0.5) \ge -0.1745\,B$:

  • with $B = 1$: $f(0.5) \ge -0.17 \gt -1$, so $x = 0.5$ is certified safe, and $f_c$ is ruled out because its norm is above 1;
  • with $B = 20$: $f(0.5) \ge -3.49$, so $x = 0.5$ cannot be certified, and $f_c$ is still possible.

The measurements are identical in both cases. The decision rests entirely on $B$, a number the data cannot supply.

T2Exploration guarantees for Lipschitz-only safe BO

The question
LoSBO and LoS-GP-UCB never make an unsafe query, given valid $L$ and $E$. Do they also provably find an $\varepsilon$-optimal point of the reachable safe region, as SafeOpt does under its (stronger) assumptions?

In plain words. LoSBO is safe for a simple reason: a function with Lipschitz constant $L$ cannot drop faster than slope $L$, so one good measurement certifies a whole interval around it. But being safe is not the same as making progress. The safe region only grows if the algorithm measures near its edge and finds good values there. LoSBO lets a GP choose where to measure, and nothing in the safety proof forces that choice to push the edge outward.

Symbols in this problem

What is known. Safety is proved (Proposition 1 of Fiedler et al., TMLR 2024). The authors point out that SafeOpt's exploration proof relies on the GP model interacting with the safety mechanism, which is exactly what LoSBO removes. They suspect that pathological situations exist in which LoSBO fails to explore properly, although none appeared in their experiments, and they leave (conditional) exploration guarantees to future work.

Why it is hard. In SafeOpt the safe set grows exactly when the GP lower bound at an expander rises. Uncertainty sampling shrinks that bound's width, and this coupling is the engine of the proof. In LoSBO the safe set grows only from measured values minus $E$. The GP decides where to sample but no longer controls how the safe set grows, so the old argument has nothing to hold on to.

Worked example — how LoSBO's safe set grows, and how it can stall

Take inputs $D = [-3, 3]$, threshold $h = 0$, Lipschitz constant $L = 2$ and noise bound $E = 0.1$. A measurement $y_i$ at $x_i$ certifies every $x$ with $y_i - E - L|x - x_i| \ge h$, because

$$f(x) \ \ge\ f(x_i) - L|x - x_i| \ \ge\ y_i - E - L|x - x_i| .$$

That is an interval of radius $r_i = (y_i - E - h)/L$ around $x_i$.

  1. Start. $x_1 = 0$, $y_1 = 1.0$: $r_1 = (1.0 - 0.1)/2 = 0.45$, so $S = [-0.45, 0.45]$.
  2. Measure at the edge. $x_2 = 0.45$, $y_2 = 0.9$: $r_2 = 0.40$ certifies $[0.05, 0.85]$, and the safe set grows to $S = [-0.45, 0.85]$.
  3. Measure inside instead. $x = 0.2$, $y = 1.2$: $r = 0.55$ certifies $[-0.35, 0.75]$, which is already inside $S$. Nothing grows, even though this is the best value measured so far.

Growth needs measurements near the boundary with values clearly above $h + E$. A badly misspecified GP (for example, one whose length scale is far too long) can keep proposing interior points like $x = 0.2$, and then the safe set stops growing. Safety is never at risk, but optimality is: if the optimum lies at $x = 2$, LoSBO never gets there. SafeOpt's proof excludes this by assuming that the GP's confidence intervals are correct, which LoSBO deliberately does not assume.

T3Tight online confidence bounds and the price of safety

The question
Are GP-UCB- and SafeOpt-style rules order-optimal? And what is the unavoidable cumulative regret of optimizing under a hard safety constraint at every step?

In plain words. Regret measures the total price of learning: how much worse your choices were, summed over all rounds, than always picking the best input. A good algorithm has regret that grows more slowly than the number of rounds, so its average loss per round goes to zero. The open questions are how fast regret can shrink with GP models, and how much extra a hard safety constraint costs, since it may stop the learner from trying the inputs that would teach it the most.

Symbols in this problem

What is known.

Why it is hard. Even if the measurement noise is independent, each new input is chosen using earlier noisy measurements, so the inputs depend on the noise. Confidence bounds must therefore use martingale and self-normalised arguments, which pay for this dependence. Hard safety also stops the learner from sampling where information is highest, so the usual exploration arguments break.

Worked example — what "a factor $\sqrt{\gamma_T}$" means in numbers

Ignore constants and take the SE kernel in one dimension, where $\gamma_T$ grows like $(\log T)^2$. After $T = 10{,}000$ rounds, $\gamma_T \approx (\ln 10^4)^2 = 84.8$.

  • Standard GP-UCB bound: $\gamma_T\sqrt T \approx 84.8 \times 100 = 8{,}483$.
  • Better rate: $\sqrt{\gamma_T T} \approx \sqrt{848{,}000} = 921$.

The ratio is $\sqrt{\gamma_T} \approx 9.2$: the guarantee is about nine times weaker than what more complicated algorithms achieve.

For a Matérn kernel the gap is qualitative. There $\gamma_T$ grows like a power of $T$, namely $T^{d/(2\nu + d)}$ up to logarithmic factors. With $\nu = 1/2$ and $d = 1$ this is $T^{1/2}$, so

$$\gamma_T\sqrt T \approx T^{1/2}\cdot T^{1/2} = T, \qquad \sqrt{\gamma_T T} \approx \sqrt{T^{1/2}\cdot T} = T^{3/4}.$$

A regret bound of order $T$ allows a constant loss in every round: it does not prove that GP-UCB learns at all. The better rate $T^{3/4}$ is still sublinear.

For the price of safety, compare Safe-LUCB's two rates at $T = 10^6$: $\sqrt T = 1{,}000$ when a positive safety gap is known, and $T^{2/3} = 10{,}000$ otherwise. Whether the second can be improved is open.

T4Rigorous safety with learned, high-dimensional models

The question
Can safe exploration keep formal guarantees when the model is a neural network ensemble or world model in a high-dimensional state space, rather than a GP on a low-dimensional parameter space?

In plain words. GP-based safe exploration works on small problems, such as tuning five controller gains. Modern robots learn neural-network models of their own dynamics, in hundreds of dimensions. To keep a safety guarantee you must know how wrong the learned model can be everywhere the robot might go. Neural networks do not come with such error bounds. The usual stand-in, disagreement within an ensemble of networks, can be small exactly where all members are wrong in the same way.

Symbols in this problem

What is known.

Why it is hard. A safety guarantee needs a uniform error bound on the model over every state the system might visit. Neural networks come with no such bound: ensembles can agree with each other and all be wrong.

Worked example — an ensemble that is confident and wrong

The true dynamics map is $f(x) = \min(2x, 2)$: linear up to $x = 1$, then flat, as when an actuator hits its limit. Data were collected only on $[0, 1]$, where $f(x) = 2x$. Three smooth models fit those data to within $0.02$:

$$\hat f_1(x) = 2x, \qquad \hat f_2(x) = 2x + 0.02x^2, \qquad \hat f_3(x) = 2x - 0.02x^2 .$$

At $x = 3$, outside the data, they predict $6.00$, $6.18$ and $5.82$. Their spread (standard deviation) is $0.15$, so the ensemble looks confident. The truth is $f(3) = 2$: the error is $4$, more than 25 times the spread.

A certain set defined by "spread below $0.2$" would include $x = 3$. This is why UPSi needs assumptions such as negligible bias inside its certain set. Disagreement measures how differently the members extrapolate, not how far they are from the truth.

T5Safety when the system changes

The question
How should a safe learner detect and react to drift or hidden context changes, so that the safe set it relies on is still valid after the change?

In plain words. A safe set learned yesterday may be unsafe today: friction changes, a payload is added, a part wears out. A learner must either notice the change before it matters, or keep a margin that covers every change that could have happened since its last measurement. Both are possible only if you know something about how fast, or how often, the system can change.

Symbols in this problem

What is known.

Why it is hard. A change can happen where you are not sampling, and safety requires noticing it before it matters. Guarantees that let the safe set shrink exist when the rate of change is known. Drift without a known rate, and abrupt changes away from the sampled inputs, have no satisfactory treatment yet.

Worked example — a certificate that expires, and one that is wrong

You measured $c(x, 0) = 0.30$ at some input $x$ (noise-free, to keep it simple). If the drift per step is at most $L_t = 0.01$, then after $k$ steps without a new measurement the value is at least $0.30 - 0.01k$.

  • After 10 steps the guaranteed value is $0.20$: still certified safe.
  • After 30 steps it is $0$, the last step at which $x$ is still certified (safety means $c \ge 0$). From step 31 on the bound is negative, so $x$ must be measured again or dropped from the safe set. This is how a safe set can shrink over time.

Now suppose the true drift is $0.02$ per step but the learner assumed $0.01$. After 20 steps the learner's bound says $0.30 - 0.20 = 0.10$, safe. The true value may already be $0.30 - 0.40 = -0.10$, unsafe. The certificate is wrong, and nothing in the data warned about it, because $x$ was not measured again. An event trigger would catch the change only if the learner happened to sample near $x$.

T6A theory of robust safety in RL

The question
Why do entropy-regularised agents tend to stay safe under action noise, and how large must a failure penalty be when the time-to-failure is unknown?

In plain words. One way to make an RL agent safe without writing explicit constraints is to punish failure: a large negative reward $-p$ when the system crashes. If $p$ is large enough, an optimal agent never chooses an action that leads to a crash. The question is how large is large enough. Future rewards are discounted, so a crash that happens late counts for little. When the crash comes long after the decisive mistake, the penalty therefore has to be enormous.

Symbols in this problem

What is known.

Why it is hard. A penalty paid at the moment of failure is discounted by the time it takes to fail. The known sufficient thresholds therefore grow exponentially with $T_f$ relative to the discount time scale. Such large penalties make learning numerically hard and exploration conservative.

Worked example — how fast the penalty threshold grows

Module 7 gives the threshold in continuous time:

$$p^\star = \Big(R_{X_U}\,\tau - \inf_{X_V} V\Big)\, e^{T_f/\tau} - R_{X_U}\,\tau .$$

Take $\tau = 1$ s, a reward rate of at most $R_{X_U} = 1$ while doomed, and $\inf_{X_V} V = 0$. Then $p^\star = e^{T_f} - 1$:

Time to failure $T_f$2 s5 s10 s
Threshold $p^\star$6.4147.422,025.5

A robot that needs 10 s to fall after the point of no return requires a penalty 22,000 times its reward rate. In discrete time the threshold contains the factor $\gamma^{-T_f}$. With $\gamma = 0.99$ this factor is $2.7$ for $T_f = 100$ steps, $152$ for $500$ steps and about $23{,}000$ for $1{,}000$ steps.

Such a penalty dominates the value function: small relative errors in learning the penalty term are larger than the whole task reward. Note that $p^\star$ is a sufficient threshold. For a specific problem, a smaller penalty may happen to work.

T7Benchmarks for safe controller tuning

The question
How should safe and crash-aware tuning algorithms be compared on real control problems?

In plain words. Papers evaluate their algorithms on their own test problems with their own metrics, so results are hard to compare. Safe tuning adds a twist: an algorithm that finds a slightly better controller but crashes the robot twice may be worse than a cautious one. Which one is "better" depends on what you measure.

What is known. The group's 2026 tutorial and review calls the lack of standardised benchmark problems for control applications a significant gap. It starts a lightweight benchmark suite for control engineering and robotics, and proposes metrics and best practices for comparing algorithms (Stenger et al., arXiv 2026, under review).

Why it is hard. Real hardware is slow and expensive, so benchmarks must be cheap simulations that still capture what makes control hard: noise, delays, constraints and crashes. And the metric itself is a design choice, as the example shows.

Worked example — which algorithm is better? (invented numbers)

Two tuning algorithms each run 20 times on the same simulated robot:

AlgorithmFinal performance (fraction of optimum)Runs with at least one crashExperiments per run
A0.952 of 2050
B0.900 of 2080

By final performance A wins. By safety B wins. By cost A wins again. A benchmark has to fix the metrics, and the trade-off between them, before the comparison. Otherwise each paper can choose the metric that favours its own method.

4. Certified Neural Networks via Robust Control (Pauli line)

Patricia Pauli's work (PhD in Stuttgart with Frank Allgöwer, now at TU/e) treats a neural network as a feedback system. Its nonlinearities are described by quadratic constraints, and its robustness is certified with LMIs, from LipSDP-constrained training to LipKernel. The open problems here are about making these certificates tighter, larger and usable for synthesis.

P1Certificates that are both tight and scalable

The question
Can SDP- or IQC-level tightness be achieved for ImageNet-scale CNNs, for transformers, and for normalisation layers, at a cost that grows gracefully with network size?

In plain words. A Lipschitz constant $L$ of a network bounds how much its output can change when its input changes: $\|f(x) - f(x')\|_2 \le L\,\|x - x'\|_2$. Computing the smallest such $L$ exactly is NP-hard, so we compute upper bounds instead. Cheap bounds are very loose. Tight bounds such as LipSDP solve a semidefinite program whose size grows with the number of neurons, and that soon becomes too expensive. The goal is bounds that are both tight and cheap for real network sizes and architectures.

Symbols in this problem

What is known. LipSDP (Fazlyab et al., NeurIPS 2019) with diagonal multipliers remains the reference for tight polynomial-time bounds. Structure-exploiting reformulations extend its reach:

Kuang et al. (arXiv 2026) prove negative results for this whole family of certificates. They give explicit ReLU examples on which such bounds are provably loose, by factors that grow with the width, and link part of the difficulty to hard reachability questions about hidden states. Module 12 states their results precisely.

Why it is hard. An SDP over all neurons at once costs at least cubic time in the number of neurons. Attention is neither slope-restricted nor globally Lipschitz, so the quadratic constraints that describe ReLU or tanh do not apply to it.

Worked example — three bounds for one small network

Module 12's walkthrough network is $f(x) = W_1\,\mathrm{ReLU}(W_0x + b_0)$ with

$$W_0 = \begin{bmatrix}1 & 2\\ 1 & -2\end{bmatrix}, \qquad W_1 = \begin{bmatrix}1 & 1\end{bmatrix}.$$
  • Product bound. $\|W_1\|_2 = \sqrt2$, and the singular values of $W_0$ are $\sqrt2$ and $2\sqrt2$, so $\|W_0\|_2 = 2\sqrt2$. Hence $L \le \sqrt2 \cdot 2\sqrt2 = 4$.
  • True constant. Wherever the on/off pattern of the two ReLUs is fixed, $f$ is linear with gradient $W_1\,\mathrm{diag}(s)\,W_0$. The four patterns $s = (1,1), (1,0), (0,1), (0,0)$ give gradients $(2, 0)$, $(1, 2)$, $(1, -2)$ and $(0, 0)$, of lengths $2$, $\sqrt5$, $\sqrt5$ and $0$. With suitable biases all four patterns occur, so $L^\star = \sqrt5 \approx 2.236$.
  • LipSDP. $4/\sqrt3 \approx 2.309$; Module 12 derives this by hand.

LipSDP is within about 3% of the truth, while the product bound is 79% too large. The catch is cost. The LipSDP matrix has one row and column per neuron (plus the inputs and outputs), and interior-point solvers need at least cubic time in that size. For 1,000 neurons that is about $10^9$ operations; for 100,000 neurons about $10^{15}$, a million times more.

P2Richer multipliers for Lipschitz and contraction certificates

The question
Which dynamic or coupled multipliers are valid in the incremental setting, and can they tighten Lipschitz and contraction certificates the way Zames–Falb multipliers tighten stability certificates?

In plain words. LipSDP describes each neuron by one simple fact: the slope of its activation lies between $\alpha$ and $\beta$. Each fact gets a nonnegative weight (a multiplier), and the SDP searches for the best weights. Richer facts, such as facts that couple several neurons, or that look at how signals evolve over time, could make the bounds tighter. But every fact must hold for every pair of inputs, and some tempting couplings are simply false, as the example shows.

Symbols in this problem

What is known.

Why it is hard. An incremental constraint compares the network at two arbitrary inputs. A coupling term between two neurons then treats a difference of two chords of the activation, taken at different places on its graph, as if it were a single chord, which nothing justifies (Module 12, §5).

Worked example — a coupled constraint that real ReLUs violate

This is Counterexample 1 of Module 12 §5. Take two ReLU neurons and two inputs whose pre-activations are $v = (0, 1)$ and $\bar v = (-1.5, 0)$. Then

$$\Delta v = (1.5,\ 1), \qquad \Delta z = \mathrm{ReLU}(v) - \mathrm{ReLU}(\bar v) = (0, 1) - (0, 0) = (0,\ 1).$$

Each neuron on its own is fine. Neuron 1 has chord slope $0/1.5 = 0$ and neuron 2 has $1/1 = 1$, both in $[0, 1]$. So the diagonal constraints hold: $\Delta z_i(\Delta v_i - \Delta z_i) = 0 \ge 0$ for $i = 1, 2$.

The coupling is not. The coupled multiplier $T = (e_1 - e_2)(e_1 - e_2)^\top = \begin{bmatrix}1 & -1\\ -1 & 1\end{bmatrix}$ treats the differences $(\Delta v_1 - \Delta v_2,\ \Delta z_1 - \Delta z_2) = (0.5, -1)$ as if they were one chord. That "chord" has slope $-2$, outside $[0, 1]$, and the quadratic form of the lemma is negative:

$$2\,\Delta v^\top T\,\Delta z - 2\,\Delta z^\top T\,\Delta z = 2(0.5)(-1) - 2(-1)^2 = -3 \ \lt\ 0 .$$

So the "constraint" fails for a genuine ReLU network, and a certificate built on it can certify a Lipschitz bound that is too small. Any new class of multipliers has to be checked against exactly this kind of pair of inputs.

P3Synthesis, not only analysis

The question
Can neural-network controllers be trained jointly with dynamic IQC stability and robustness certificates, without bilinear matrix inequalities, while handling local regions of attraction, state and input constraints and plant uncertainty?

In plain words. Checking that a given neural controller keeps a system stable (analysis) is now routine for small systems: you search for a Lyapunov certificate with an SDP. Designing the controller together with its certificate (synthesis) is harder, because the controller's weights and the certificate's matrices multiply each other. Products of unknowns make the problem non-convex, so the efficient SDP machinery no longer applies directly.

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What is known. Analysis is mature (Module 14 §2). For synthesis, Junnarkar, Arcak & Seiler (Automatica 2026) project RL policies onto closed-loop dissipativity LMIs. Recurrent equilibrium networks are stable by design (Revay, Wang & Manchester, TAC 2024).

Why it is hard. Controller parameters and multipliers multiply each other in the certificate, which makes it a BMI. BMIs are non-convex and hard to solve in general.

Worked example — why the unknowns multiply, and the classical trick that removes the product

Plant $\dot x = ax + bu$ with controller $u = kx$, so $\dot x = (a + bk)x$. Look for $V(x) = px^2$ with $p \gt 0$ and $\dot V = 2p(a + bk)\,x^2 \lt 0$ for $x \ne 0$:

$$2p\,(a + bk) \ \lt\ 0 .$$

The unknowns $p$ and $k$ appear as the product $pk$: this is a bilinear inequality.

The classical fix. Substitute $q = 1/p$ and $y = kq$. Multiplying by $q^2 \gt 0$ gives

$$2\,(aq + by) \ \lt\ 0 ,$$

which is linear in the new unknowns $(q, y)$. For the unstable plant $a = 1$, $b = 1$, the choice $q = 1$, $y = -2$ works; then $k = y/q = -2$, and the closed loop is $\dot x = -x$.

The matrix version is the standard state-feedback LMI. With $Q = P^{-1}$ and $Y = KQ$, the condition $(A + BK)^\top P + P(A + BK) \prec 0$ becomes

$$AQ + QA^\top + BY + Y^\top B^\top \ \prec\ 0 .$$

Why this does not carry over. For a neural-network controller $u = W_1\varphi(W_0x)$, the certificate multiplies $P$, the neuron multipliers $T$ and the weights $W_0$, $W_1$. In general no single change of variables removes all of these products. Current methods either fix some variables and alternate, or project a trained controller onto LMI constraints that guarantee dissipativity, as Junnarkar et al. do.

P4The expressivity–robustness gap

The question
Can Lipschitz-by-design networks close the gap between clean and certified accuracy, and stay fast enough for real-time control?

In plain words. A network that is Lipschitz by construction comes with a robustness certificate for free: if the correct class wins by a large enough margin, no small perturbation can flip the decision. The price is expressivity. Constraining every layer makes the network less accurate on clean inputs. The open problem is to close that gap, with networks fast enough for a control loop.

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What is known.

Why it is hard. A certificate at radius $\varepsilon$ needs a margin larger than $\sqrt2\,L\,\varepsilon$. Lipschitz layers keep $L$ small, but they also limit how sharply the network can separate classes, so many correctly classified inputs end up with margins too small to certify. Larger networks relax the restriction (LipNeXt uses 2 billion parameters), at the cost of inference time.

Worked example — from a margin to a certificate

Take a classifier whose logit map is 1-Lipschitz in $\ell_2$, so $L = 1$. Module 13's margin certificate says the prediction cannot change for perturbations with

$$\|\delta\|_2 \ \lt\ \frac{M_f(x)}{\sqrt2\,L} .$$

(Each logit difference $f_y - f_j$ can change by at most $\sqrt2\,L\,\|\delta\|_2$, and that change must not use up the margin.)

  • Image A has margin $0.25$, so its certified radius is $0.25/\sqrt2 = 0.177$. Since $0.177 \gt 36/255 = 0.141$, image A counts as certified at $\varepsilon = 36/255$.
  • Image B is also classified correctly, but with margin $0.15$. Its radius is $0.106 \lt 0.141$, so it counts for clean accuracy but not for certified accuracy.

The gap between LipNeXt's 57.0% clean and 41.2% certified accuracy on ImageNet consists of images like B.

How big is $\varepsilon = 36/255$? With pixel values in $[0, 1]$, it is the $\ell_2$ size of changing a single colour value by 36 of its 255 levels. Spread evenly over all $224 \times 224 \times 3 = 150{,}528$ values of an ImageNet image, it changes each value by only $0.09$ of a level, which is invisible.

P5Extending the 2-D and N-D systems view

The question
Can the state-space view of convolutions cover strided, dilated and N-D convolutions and state-space sequence models with minimal realisations, and support task-specific robustness measures?

In plain words. A convolution slides a small filter over a signal. Control theorists noticed that this is exactly what a linear dynamical system does: it keeps a short memory of past inputs (the state) and combines it with the current input. Writing convolutional layers this way lets the whole toolbox of LMIs and dissipativity certify them. The open question is how far the trick extends: to strides, dilations, higher-dimensional data and modern sequence models, ideally with the smallest possible state.

Symbols in this problem

What is known. Convolutions can be realised as 2-D systems of Roesser type (Pauli, Gramlich & Allgöwer, 2024; Gramlich et al., Automatica 2026). Minimality is proven only when the input and output channel counts agree. LipKernel's ideas were carried over to cascaded state-space models by other authors (LipSSM, arXiv 2026).

Why it is hard. Each architectural feature changes the state-space structure: a stride skips outputs, a dilation spreads the kernel out, and N-D data need memory in N directions. Minimality matters because the state size sets the size of the LMIs, and proving it needs a separate algebraic argument for each structure.

Worked example — a 1-D convolution is a linear system

Take the kernel $w = (1, 2, 3)$, so $y_t = u_t + 2u_{t-1} + 3u_{t-2}$. Store the two past inputs in the state $x_t = (u_{t-1}, u_{t-2})$. Then

$$x_{t+1} = \underbrace{\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}}_{A} x_t + \underbrace{\begin{bmatrix}1\\ 0\end{bmatrix}}_{B} u_t, \qquad y_t = \underbrace{\begin{bmatrix}2 & 3\end{bmatrix}}_{C} x_t + \underbrace{1}_{D}\cdot u_t .$$

Check. $Ax_t + Bu_t = (u_t,\ u_{t-1}) = x_{t+1}$, and $Cx_t + Du_t = 2u_{t-1} + 3u_{t-2} + u_t = y_t$.

Run it. The input $u = (1, 1, 0, 2, 0)$ from a zero initial state gives $y = (1, 3, 5, 5, 4)$, exactly the convolution of $u$ with $w$. For example $y_2 = u_2 + 2u_1 + 3u_0 = 0 + 2 + 3 = 5$.

The state has two entries, one per past input the filter needs. That is minimal for a kernel of length 3 whose last entry is nonzero. An image convolution needs memory in two directions, which is what the 2-D Roesser model provides.

5. Constrained & Safe Deep RL

Constrained RL treats safety as a budget: maximise reward while keeping an expected cost below a limit $d$ (Module 8). The problems below ask what that formulation guarantees and when, and how it scales to deep networks and language models.

R1Guarantees during learning at deep-RL scale

The question
Can safety hold throughout learning, not only at convergence, for neural world models and vision inputs?

In plain words. Most constrained-RL algorithms promise that the policy they converge to satisfies the constraint. That says nothing about the thousands of episodes during training, which on a real robot is exactly when crashes happen. Guarantees during learning exist for small or structured problems. Extending them to neural world models and camera inputs is open.

Symbols in this problem

What is known. SafeMDP (Turchetta et al., NeurIPS 2016), ActSafe (As et al., ICLR 2025) and safe exploration via policy priors (Wendl et al., ICLR 2026) give guarantees during learning. They rely on calibrated uncertainty, known Lipschitz constants or a safe prior policy. Certifying those ingredients for high-dimensional neural models is open.

Why it is hard. To stay safe while learning, you must know that an action is safe before you try it. That needs a trustworthy model, or a trustworthy prior, exactly where the learner has little data. With neural models, the uncertainty estimates that would provide this are not calibrated (see T4).

Worked example — safe at the end versus safe throughout (invented numbers)

Let the cost of an episode be $C = 1$ if it ends in a crash and $C = 0$ otherwise, without discounting. The constraint $\mathbb E[C] \le d = 0.05$ then says: a crash in at most 5% of episodes. (With discounting a late crash would count less, and the budget would no longer be a crash probability.) Two learners train for 1,000 episodes:

  • Learner A explores aggressively. It crashes in about 300 of its first 1,000 episodes, then converges to a policy that crashes in 2% of episodes.
  • Learner B never exceeds the 5% rate, so it crashes in at most about 50 episodes, but it needs twice as many episodes to reach the same final reward.

A convergence guarantee is satisfied by both: their final policies meet $\mathbb E[C] \le d$. Only B is safe during learning. On a real robot, A's 300 crashes are the whole problem.

R2Constraint satisfaction at the last iterate

The question
Do Lagrangian methods with function approximation satisfy the constraint at every iterate, or at least at the final one, rather than only on average?

In plain words. Lagrangian methods turn "maximise reward subject to cost at most $d$" into a game. The policy maximises reward minus a price $\lambda$ times cost; the price rises when the cost is too high and falls when it is low. This game can circle around its solution forever. The policy overshoots the constraint, the price goes up, the policy becomes too careful, the price goes down, and so on. On average the constraint holds, but the policy you deploy, the last one, may violate it.

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What is known. Lagrangian iterates oscillate, and most guarantees are for averaged iterates. Last-iterate results exist, but under restrictive assumptions: regularised or optimistic updates for tabular softmax policies (Ding et al., NeurIPS 2023), or gradient-domination conditions (Montenegro et al., NeurIPS 2024).

Why it is hard. Gradient descent–ascent on a non-convex–concave saddle problem rotates around the saddle point instead of converging to it. The PID view of Module 8 explains the oscillation, but it does not remove it.

Worked example — a policy that circles its constraint forever

One decision per episode: the risky action gives reward 1 and cost 1, the safe action gives 0 and 0. The policy picks the risky action with probability $p$, so $J_r = J_c = p$. The constraint is $J_c \le d = 0.5$. The best feasible policy is $p = 0.5$, with price $\lambda = 1$.

The Lagrangian is $\mathcal L(p, \lambda) = p - \lambda(p - 0.5)$. Take simultaneous projected gradient steps with step size $\eta = 0.5$:

$$p_{k+1} = \mathrm{clip}_{[0,1]}\big(p_k + \eta\,(1 - \lambda_k)\big), \qquad \lambda_{k+1} = \big[\lambda_k + \eta\,(p_k - 0.5)\big]_+ .$$

Starting from $p_0 = 0$, $\lambda_0 = 0$:

$k$01234567891011
$p_k$00.51111110.8750.6250.2810
$\lambda_k$0000.250.50.7511.251.51.6881.751.641

While the price is low, $p$ climbs to 1, which has cost 1, twice the limit. The price then rises until $p$ collapses to 0 (safe but useless), the price falls, and the cycle repeats. After 400 iterations $p$ still swings between 0 and 1, while the average of all iterates is $0.507$, close to the optimum. A guarantee about averages is true here, and it tells you nothing about the policy at iteration 400.

R3Choosing and reporting the constraint

The question
Which constraint type fits which application, and how should results be reported so that methods can be compared? The options are expected cost, chance, CVaR, state-wise or almost-sure constraints.

In plain words. "Safe" can be made precise in several ways. The cost can be small on average (expected cost), a crash can be rare (chance constraint), even the bad cases can be not too bad (CVaR), the cost can be small at every step (state-wise), or a crash can never happen (almost surely). These are different promises, and a policy can satisfy one while badly failing another. Results are also sensitive to the chosen limit $d$, so a single number per method can mislead.

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What is known. Expected-cost constraints allow rare catastrophes. Stricter semantics trade tractability for guarantees. Lagrangian methods are sensitive to the cost limit, and Spoor et al. (arXiv 2025) recommend reporting results over task-specific sets of cost limits.

Why it is hard. Expected cost keeps the CMDP a linear program over occupancy measures (Module 8 §2). Chance constraints, CVaR constraints and almost-sure budgets on the accumulated cost do not, and need extra state or other reformulations (Module 8 §6, Module 8 §7). One useful exception: "never fail", with a nonnegative failure cost, is the linear constraint $J_c(\pi) = 0$, because a nonnegative random variable with mean zero is zero almost surely.

Worked example — one policy, three verdicts

A policy crashes in 5% of episodes. A crash costs 10, and every other episode costs 0, so $C = 10$ with probability $0.05$ and $C = 0$ otherwise.

  • Expected cost. $\mathbb E[C] = 10 \times 0.05 = 0.5$. With limit $d = 1$: satisfied.
  • Chance constraint "crash in at most 1% of episodes": $\mathbb P(C \gt 0) = 0.05 \gt 0.01$: violated.
  • $\mathrm{CVaR}_{0.9}$. The worst 10% of episodes are the 5% crashes (cost 10) and 5% normal ones (cost 0), so $\mathrm{CVaR}_{0.9} = (0.05 \times 10 + 0.05 \times 0)/0.10 = 5$. The formula gives the same: $\mathrm{VaR}_{0.9} = 0$, so $\mathrm{CVaR}_{0.9} = 0 + 0.5/0.1 = 5$. With limit 1: violated.

The same policy is "safe" under the expected-cost definition and unsafe under the other two. This is why papers must state which constraint they enforce. It is also why results reported for one value of $d$ can rank methods differently than for another.

R4Cost estimation under distribution shift, and offline-to-online safety

The question
How can a policy learned from logged data be fine-tuned online without violating constraints, when learned cost critics are unreliable outside the data?

In plain words. Offline RL learns from a fixed log of past experience, without new trials. To respect a constraint it must estimate the cost of actions, including actions the log never contains. Learned critics guess those costs by extrapolation, and the guesses are often too optimistic. The policy then drifts toward exactly the untested actions whose cost was underestimated. Being pessimistic about unseen actions fixes this but makes the policy timid. Moving safely from offline training to online fine-tuning is the open part.

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What is known. Off-policy and offline critics underestimate the cost of out-of-distribution actions, which motivated CPQ (Xu et al., AAAI 2022). Pessimism can make policies over-conservative. Feasibility-based methods such as FISOR (Zheng et al., ICLR 2024) improve offline safety, but safe offline-to-online fine-tuning with guarantees is largely unsolved.

Why it is hard. The data say nothing about unseen actions, so any estimate there is an assumption. Optimistic assumptions break safety, pessimistic ones break performance. And once the policy goes online, the data distribution changes again (see S1).

Worked example — a critic that extrapolates a cost downward

The logged actions are spread evenly over $[0, 0.5]$, and the true cost of action $a$ is $c(a) = a^2$. Fit a straight line to these data by least squares, weighting all of $[0, 0.5]$ equally. The best fit is

$$\hat c(a) = 0.5\,a - \tfrac{1}{24}, \qquad \text{slope } 0.5, \ \text{intercept } -0.042,$$

and on the data it is accurate to within $0.042$.

At the unseen action $a = 2$ the line predicts $\hat c(2) = 0.958$, but the true cost is $4$. A policy that trades reward against predicted cost will happily choose $a = 2$ if it brings more reward, because it believes the cost is below 1. A neural critic extrapolates in less predictable ways, but the problem is the same. CPQ's remedy is to make out-of-distribution actions look costly on purpose, at the price of never trying them.

R5Constrained alignment of foundation models

The question
What can be guaranteed when the constraint itself is a learned, possibly misspecified, cost model, as in RLHF with safety constraints?

In plain words. Safe RLHF fine-tunes a language model to be helpful while keeping a learned "harmfulness" score below a limit. Both the helpfulness reward and the harm cost are themselves learned from human judgements, so they are approximations. If the cost model underestimates harm, the optimiser exploits exactly that error. The true harm can then exceed the limit even though the training constraint looks satisfied.

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What is known. Safe RLHF (Dai et al., ICLR 2024) trains a cost model from human judgements and runs Lagrangian PPO. Theory is starting to appear:

Both rest on realisability and data-coverage assumptions. Misspecified neural reward and cost models, and PPO-style training with many interacting constraints, remain open.

Why it is hard. The optimiser actively searches for responses where the learned cost is low, so cost-model errors are not random noise: they are selected for. Bounds therefore need control of the error everywhere the policy might go, which brings back the coverage problem of R4.

Worked example — how a 0.2 error in the cost model breaks the limit

There are two possible responses: a helpful-but-harmful one (reward $1.0$, true cost $0.6$) and a safe one (reward $0.7$, cost $0.1$). The model gives the harmful response with probability $q$, and the limit is $d = 0.3$.

  • With the true costs. $0.6q + 0.1(1 - q) \le 0.3$ gives $q \le 0.4$, and the best reward is $0.7 + 0.3 \times 0.4 = 0.82$.
  • With a cost model that rates the harmful response at $0.4$ instead of $0.6$. $0.4q + 0.1(1 - q) \le 0.3$ gives $q \le 2/3$. The optimiser picks $q = 2/3$, whose true cost is $0.6 \cdot \tfrac23 + 0.1 \cdot \tfrac13 = 0.43 \gt 0.3$.

The training constraint is met exactly, and the true constraint is violated by 44%. The error was "only" $0.2$ on one response, but the optimiser moved probability onto that response precisely because of the error.

R6Unifying CMDP learning with control certificates

The question
How should CMDP methods be combined with barrier functions, reachability, safe value functions and Lipschitz-bounded policies, and validated on real systems?

In plain words. Constrained RL and control theory express safety in different languages: an expected-cost budget, a failure penalty, a barrier-function filter, a reachable set. Each has strengths, and combining them could give policies that learn well and are provably safe. Today such combinations are built case by case. No common theory says which combination guarantees what, and they are rarely compared on the same control benchmarks.

Symbols in this problem

What is known. Individual combinations exist: HJ feasibility in FISOR, safe value functions (Massiani et al.), CBF filters around RL policies. They are ad hoc, and control-oriented benchmarks such as safe-control-gym (Yuan et al., RA-L 2022) are rarely used for comparison.

Why it is hard. The guarantees live on different objects. A CMDP constrains an average over episodes, a penalty constrains optimal policies, and a filter constrains each action in each state. Composing them requires stating all three in one framework, which is the goal of the unified safety-filter view (Module 10 §8).

Worked example — three ways to say "do not fall off the shelf"

A robot moves on a shelf $x \in [0, 1]$ with dynamics $\dot x = u$. Falling off at $x \lt 0$ is a failure, and the task rewards being near the edge, say $r = 1 - x$.

  • CMDP. Cost $c = 1$ for falling, constraint $J_c(\pi) \le d = 0.01$. Guarantee: the expected discounted number of falls is at most 0.01. Occasional falls are allowed.
  • Failure penalty (T6). Reward $r - p$ at failure, with $p \gt p^\star$. Guarantee: every optimal policy avoids failure whenever avoiding it is possible. Nothing is said about the non-optimal policies met during learning.
  • CBF filter with $h(x) = x$: allow only inputs with $\dot h = u \ge -\alpha x$, for example with $\alpha = 2$. Guarantee: $x(t) \ge 0$ for all time, whatever the policy proposes, provided the model $\dot x = u$ is right.

These promise different things: an average, a property of optimal policies, and a pointwise invariance. That is why combining them needs care.

6. Safety Filters & Learned Certificates

A safety filter sits between a policy and the system, and corrects unsafe actions before they are executed (Module 10). It separates performance (the policy's job) from safety (the filter's job). The problems below are about what happens when the filter itself is learned, when it has no valid action, and how it interacts with learning and with the real world.

F1Filters built on learned models

The question
When the dynamics or the safety certificate are learned, what does the filter's guarantee actually cover?

In plain words. A classical filter relies on a known model of the system. Newer filters learn the model, and sometimes even what "failure" looks like, from data such as camera images. The filter is then only as reliable as what it learned. When the learning errors have verified bounds, robust filters can absorb them into a guarantee (Module 10 §9); for filters that learn from images, such bounds are not available today.

Symbols in this problem

What is known. Latent safety filters (Nakamura, Peters & Bajcsy, RSS 2025) enforce safety in a world model's latent space. The authors state that their practical implementation lacks formal assurances, because errors in the learned world model, failure classifier and HJ value function are not bounded. How such errors propagate to the safety guarantee is open. Calibrated, verifiable uncertainty for ensembles and world models is missing more generally (see also T4).

Why it is hard. Three learned components feed into each other. Errors in the world model change which states the filter thinks are reachable, errors in the failure classifier change what it avoids, and errors in the value function change where it intervenes. Bounding each error is hard; bounding how they combine is harder.

Worked example — a filter that is perfect relative to a classifier that is not

Suppose the failure classifier labels 98% of truly failing states as failures and misses the other 2%. A filter that is perfect with respect to the classifier never enters a state the classifier calls "failed". It can still drive into the 2% of failure states the classifier missed, because from its point of view those states are safe.

Now suppose the task rewards getting close to obstacles. Then the policy is drawn toward exactly the region where such misses are likely, and the filter will not stop it. The statement "the filter never enters a labelled failure" stays true, and it says nothing about those 2%.

F2Verifying learned certificates at scale

The question
Can neural barrier functions, neural Lyapunov functions and learned reachability value functions be formally verified beyond low-dimensional systems?

In plain words. A neural barrier or Lyapunov function is trained so that its defining condition holds at sampled states. A certificate needs the condition at every state, including the infinitely many that were not sampled. Formal verifiers can check this, but their cost explodes with the dimension of the state and the size of the network.

Symbols in this problem

What is known. Training losses on samples do not make a certificate. Policy neural CBFs (So et al., ICRA 2024) and DeepReach (Bansal & Tomlin, ICRA 2021) scale well but are not verified. Formally verified Lyapunov-stable neural controllers exist for modest sizes (Yang et al., ICML 2024); SMT solvers and bound-propagation verifiers still limit the size.

Why it is hard. Exact verification of ReLU networks is NP-hard, so in the worst case its cost grows exponentially. Sampling-based certificates face the same growth in a more visible form, as the example shows.

Worked example — how many samples a sampling-based certificate needs

Suppose the trained condition satisfies $\psi(x_i) \ge m = 0.1$ at every point of a grid with spacing $r$ on $[0, 1]^d$, and $\psi$ is $L_\psi$-Lipschitz in the $\ell_\infty$ norm with $L_\psi = 50$. Every state lies within $\ell_\infty$-distance $r/2$ of a grid point, so

$$\psi(x) \ \ge\ m - L_\psi\,\frac r2 \ \gt\ 0 \qquad\text{whenever}\qquad r \ \lt\ \frac{2m}{L_\psi} = 0.004 .$$

For a concrete strict certificate, take 251 equally spaced cell centres per axis, with spacing $r=1/251\lt0.004$. The cells cover $[0,1]$ and every point is within $r/2$ of a centre. This needs:

  • $d = 2$: $251^2 = 63{,}001$ evaluations, which is easy;
  • $d = 6$: $251^6 \approx 2.5 \times 10^{14}$, which is infeasible;
  • $d = 12$: about $6.3 \times 10^{28}$.

Exact verifiers avoid the grid, but they face the same exponential growth in the worst case. This is why verified neural certificates exist mainly for low-dimensional systems.

F3Feasibility, deadlock and liveness

The question
How can a filter guarantee both safety and task completion when constraints interact, inputs are bounded, or the barrier has high relative degree?

In plain words. At each moment, a CBF filter picks the action closest to what the policy wants among the actions that keep the system safe. Sometimes no such action exists: the motors are not strong enough, or two safety constraints demand opposite things. Then the filter has no answer. And even when it always has an answer, it may keep the system safe by stopping it forever, short of its goal: safe but useless.

Symbols in this problem

What is known. The CBF-QP can become infeasible under input bounds or several interacting constraints. Combined CLF–CBF quadratic programs can introduce spurious asymptotically stable equilibria (Reis, Aguiar & Tabuada, L-CSS 2021). Reach-avoid formulations address liveness for specific settings (Hsu et al., RSS 2021).

Why it is hard. Feasibility depends on the barrier, the input bounds and the dynamics together. Checking that a barrier is feasible everywhere is itself a verification problem (F2). The natural remedy, a barrier that accounts for the input limits, is essentially the viability kernel, which is hard to compute.

Worked example — a barrier condition that no admissible input can satisfy

A cart moves toward a wall at position $0$: $\dot p = v$, $\dot v = u$, $|u| \le 1$. Safety means $h(p) = p \ge 0$. The input appears only in $\ddot h = u$, so $h$ has relative degree 2 and needs a high-order CBF. With $\alpha_1(s) = \alpha_2(s) = s$:

$$\psi_1 = v + p, \qquad \dot\psi_1 + \psi_1 = (u + v) + (v + p) \ \ge\ 0 \iff u \ \ge\ -2v - p .$$

At $p = 1$, $v = -3$ (moving toward the wall at speed 3), the filter needs $u \ge 6 - 1 = 5$, but the motor gives at most $1$. The QP is infeasible.

The physics agrees. Braking at full strength from speed 3 takes a distance of $v^2/2 = 4.5 \gt 1$, so a crash is already unavoidable, even though $h = 1 \gt 0$ says "safe". (Also $\psi_1 = -2 \lt 0$: the state has already left the HOCBF's own safe set.) The barrier ignored the input limit. A barrier that accounts for it, $h(p, v) = p - v^2/2$ when $v \lt 0$, describes exactly the viability kernel of this system, and computing such kernels is hard in general.

F4Filters that do not ruin performance

The question
How do we train a policy that works well with its filter, instead of being corrected by it at deployment?

In plain words. A filter makes a policy safe by overriding it, and every override is a moment where the policy does not do what it was trained to do. If overrides are frequent, performance collapses. If the learner does not account for them, it learns from actions it never actually took. The goal is policies that rarely need correcting, because they learned with the filter in the loop.

Symbols in this problem

What is known. Filters shift the action distribution, can cause chattering, and degrade RL or imitation performance. Training with the filter in the loop helps (Pizarro Bejarano, Brunke & Schoellig, RA-L 2025). Keeping filtered actions consistent with the training distribution of a generative policy is the aim of PACS (Römer et al., ICRA 2026).

Why it is hard. With a fixed filter in the loop, the policy learns in the combined system "world plus filter". That is a valid learning problem, but many proposed actions map to the same executed action, so the learning signal about them is flat. If the filter itself is learned or retuned, the combined system changes during training. Without the filter, the policy is trained for a world it will never face. Generative policies add a twist: their outputs come from a learned distribution, and pushing an action outside that distribution can produce motions the policy has never seen.

Worked example — what the learner records matters

A policy proposes speed $a = 2$. The filter allows at most $1$ and executes $a_{\rm safe} = 1$, and the reward reflects speed 1.

  • Record the proposed action (proposed $a = 2$, reward of speed 1). This is correct for the combined system "world plus filter", in which proposing 2 means driving at 1. It becomes wrong as soon as the filter changes, or if the policy is later run without the filter: the data would still claim that proposing 2 earns the reward of speed 1.
  • Record the executed action (executed $a_{\rm safe} = 1$, reward). This is correct for the physical system. But the data were generated by "policy plus filter", not by the policy being trained, so the update is off-policy and needs corrections.

Neither choice is wrong in itself: the recorded action and the learning objective have to match. At deployment, if the filter changes the proposed action in 30% of the steps, the executed behaviour differs from what the policy was trained to produce in those steps. That can push the system into states the policy rarely saw, and how often this happens has to be measured separately. Training with the filter in the loop reduces this mismatch.

F5Real-world effects

The question
How do guarantees survive sampled-data implementation, delays, partial observability, multiple agents, and generative policies that output long action chunks?

In plain words. Safety proofs are usually written for an idealised controller that measures the exact state, acts instantly and continuously, and is alone. Real controllers sample the state every few milliseconds and act with delay. They see the world through noisy sensors, share space with other agents, and, with generative policies, output whole chunks of motion at once. Each of these gaps can break a guarantee unless a margin accounts for it.

Symbols in this problem

What is known.

Why it is hard. Each effect needs its own margin, margins add up, and too much margin makes the system useless. Bounding the errors of learned perception, which PACS and measurement-robust CBFs both require, is itself an open problem.

Worked example — the margin that sampling costs

A robot obeys $\dot x = u$ with $|u| \le 1$ and must keep $x \ge 0$. Its controller runs every $\Delta t = 0.1$ s and holds $u$ constant in between.

  • Between two samples, $x$ can move by up to $|u|\,\Delta t = 0.1$. A filter that only checks $x \ge 0$ at the sampling instants can therefore reach $x = -0.1$ before it acts again.
  • The fix is a margin: enforce $x \ge 0.1$ at each sample. Then $x \ge 0.1 - 0.1 = 0$ until the next one.
  • With an extra actuation delay of $0.05$ s, the previous input keeps acting for another $0.05$ s, and the margin grows to $0.15$.

Every such effect costs a piece of the safe set. This is why tight, verified bounds on delays and errors matter.

7. Statistical Guarantees & Trustworthy Certificates

Some guarantees are not proofs about every input, but statements that hold with high probability over random data or random noise. They need fewer assumptions about the system and more about the data (Module 15 §7). The last problem in this section asks how far any certificate, deterministic or statistical, can be trusted.

S1Statistical guarantees under feedback and distribution shift

The question
Can conformal-type guarantees hold in closed loop, where the controller's actions change not only which situations it sees next but also how the environment responds?

In plain words. Conformal prediction wraps any predictor in an error bar that is correct with a chosen probability, say 90%, using nothing but a list of past prediction errors. The catch is one assumption: future errors must be statistically interchangeable with past ones. A robot that acts on the error bars changes the world it is predicting, and that breaks the assumption.

Symbols in this problem

What is known.

Why it is hard. Feedback couples the data to the policy. When the policy changes how the environment responds, even the conditional distribution of outcomes shifts, so tomorrow's scores are not exchangeable with today's.

Worked example — a 90% error bar from nine numbers, and how feedback breaks it

Nine calibration errors, in metres and sorted: $0.2, 0.3, 0.3, 0.4, 0.5, 0.6, 0.8, 0.9, 1.2$. For 90% coverage ($\delta = 0.1$), split conformal prediction uses the score at position

$$p = \big\lceil (n + 1)(1 - \delta) \big\rceil = \lceil 10 \times 0.9 \rceil = 9,$$

so the error bar is $C = R^{(9)} = 1.2$ m. If the errors are exchangeable, this procedure covers the next error with probability at least 90%. The 90% is an average over both the calibration data and the new error; for these particular nine numbers the coverage can be higher or lower (Module 15 §5). For 95% the position would be $\lceil 9.5 \rceil = 10 \gt 9$: with only nine scores the bar would be infinite, and you need more data.

Now the robot uses the 1.2 m bars to plan closer to pedestrians than before. The pedestrians react by swerving, and the new errors are $1.0, 1.5, 1.8, 2.0$. They are not exchangeable with the calibration errors, and the bar covers only one of the four. The guarantee did not fail by bad luck; its assumption stopped holding.

S2Beyond randomized smoothing

The question
Are there certificates as strong as randomized smoothing at large radii that are deterministic, cheap, and work for norms other than $\ell_2$?

In plain words. Randomized smoothing turns any classifier into a new one, the smoothed classifier, which returns the class that is most likely when Gaussian noise is added to the input. If one class is likely enough, no small perturbation can change the smoothed classifier's answer. Those probabilities are estimated by classifying many noisy copies of the input. So the certificate holds only with high probability, it needs thousands of network evaluations per input, and it works well only for the $\ell_2$ norm. For other norms, especially $\ell_\infty$ (every pixel may change a little), its guarantees fade in high dimensions.

Symbols in this problem

What is known.

Why it is hard. An $\ell_\infty$ ball of radius $\varepsilon$ in $d$ dimensions has corners at $\ell_2$ distance $\varepsilon\sqrt d$ from its centre, so any method built on $\ell_2$ geometry pays a factor $\sqrt d$. Kumar et al. show this is not just a weakness of one conversion: smoothing with i.i.d. noise cannot avoid it.

Worked example — from a vote to a radius, and what it means for $\ell_\infty$

Take noise level $\sigma = 0.5$, and suppose that after sampling the top class has $p_A \ge 0.9$. The certified $\ell_2$ radius is

$$R = \sigma\,\Phi^{-1}(0.9) = 0.5 \times 1.2816 = 0.641 .$$

Translating to $\ell_\infty$. A CIFAR-10 image has $d = 32 \times 32 \times 3 = 3072$ values. An $\ell_\infty$ perturbation of size $\varepsilon$ can have $\ell_2$ size $\varepsilon\sqrt{3072} = 55.4\,\varepsilon$. So the $\ell_2$ certificate covers only $\varepsilon \lt 0.641/55.4 = 0.0116$, about $2.95/255$, well short of the usual $8/255$.

Sampling also caps the radius. Suppose all $n = 100{,}000$ noise samples vote for $A$. The best one-sided lower bound at confidence $1 - 0.001$ is $p_A = 0.001^{1/n} = 0.99993$ (Primer C), so

$$R \ \le\ \sigma\,\Phi^{-1}(0.99993) \ \approx\ 3.81\,\sigma ,$$

however robust the classifier really is.

S3Certificates you can trust

The question
How do we make sure a published certificate and its computed numbers are actually correct?

In plain words. A certificate is a proof, and proofs can be wrong. Some published certificates had subtle errors that went unnoticed until someone built a counterexample. Others are computed numerically: an SDP solver returns a matrix that it says is positive semidefinite, but only up to a rounding tolerance. A certificate that is "almost" valid is not valid.

Symbols in this problem

What is known. Published proofs have needed corrections, including the original coupled-multiplier LipSDP variant (Pauli et al., L-CSS 2022) and the $\tau \gt 0$ case of chordal LipSDP (Xue et al.). Certificates computed with floating-point SDP or LP solvers are only approximately feasible. The same lesson appeared while writing these notes: independent review caught many statements that did not match their cited sources.

Why it is hard. Floating-point arithmetic cannot represent most numbers exactly, and SDP solvers stop at a tolerance. Exact checks are possible, but they cost more and need the certificate to have some slack.

Worked example — a matrix a solver would accept

Consider

$$M = \begin{bmatrix}1 & 1\\ 1 & 1 - 10^{-9}\end{bmatrix}.$$

Its determinant is $(1 - 10^{-9}) - 1 = -10^{-9} \lt 0$. The determinant is the product of the eigenvalues, so one eigenvalue is negative: about $-5 \times 10^{-10}$, while the other is about 2. $M$ is not positive semidefinite, yet a solver working with tolerance $10^{-8}$ would report it as feasible. If $M \succeq 0$ was the condition certifying a Lipschitz bound, that bound is not proved.

The standard remedy: ask the solver for a margin, $M \succeq 10^{-6} I$, then round the solution to rational numbers and check positive definiteness exactly, for example with an exact rational $LDL^{\top}$ factorisation.

8. Where the Two Lines Meet

Our synthesis, not a published result
The Trimpe and Pauli lines use complementary mathematics. The first gives statistical guarantees from kernel and GP bounds; the second gives deterministic certificates from quadratic constraints and LMIs. The four directions below combine them. They are research ideas suggested by the gaps above, not claims from the literature. No joint Trimpe–Pauli paper exists. The two lines meet only indirectly, for instance through Johannes Köhler, who comes from Allgöwer's Stuttgart group (where Pauli did her PhD) and co-authored Hose et al. (B3) with Trimpe.

B1Certified Lipschitz constants as checkable assumptions for safe exploration

The direction
LoSBO and GoSafeOpt need Lipschitz constants of the unknown performance and constraint functions. When part of the closed loop is a neural network with a certified Lipschitz bound (LipSDP, LipKernel), and the plant has a known Lipschitz bound, composition gives a certified contribution to that constant.

In plain words. LoSBO needs a Lipschitz constant, but nobody knows it (T1). The Pauli line computes certified Lipschitz constants of neural networks. If the controller is such a network, could we assemble LoSBO's constant from certified pieces instead of guessing it? The obstacle is that constants multiply when maps are chained, and a closed loop chains the same map many times.

Symbols in this direction

The catch. Lipschitz constants multiply under composition. A closed loop evaluated over $T$ steps can have a constant as large as $L^T$ when the per-step constant is $L$, so this worst-case bound can become too loose over long horizons unless the loop contracts or the available margins are large. This is a reason to look at contracting architectures such as RENs (Module 13 §5).

Worked example — how a per-step constant becomes a closed-loop constant

By induction (Discussion Question 3), two trajectories of $x_{t+1} = F(x_t)$ satisfy $\|x_t - y_t\| \le L^t\,\|x_0 - y_0\|$. So the cost $J$ has Lipschitz constant at most

$$L_c \sum_{t=0}^{T} L^t .$$

With $L_c = 1$ and $T = 50$:

  • $L = 1.1$ (slightly expanding): $1.1^{50} \approx 117$, and the sum is about $1{,}281$;
  • $L = 0.9$ (contracting): the sum is at most $1/(1 - 0.9) = 10$, here $9.95$.

A 10% expansion per step makes the constant more than a hundred times larger than a 10% contraction does.

From initial states to controller parameters. These numbers measure how $J$ reacts to the initial state $x_0$. LoSBO tunes controller parameters $\theta$, so it needs how $J$ reacts to $\theta$. A parameter change perturbs every step. If $F$ is $L$-Lipschitz in $x$ and $L_\theta$-Lipschitz in $\theta$, the state deviation $e_t = \|x_t(\theta) - x_t(\theta')\|$, from the same initial state, obeys

$$e_t \ \le\ L\,e_{t-1} + L_\theta\,\|\theta - \theta'\|, \quad e_0 = 0 \qquad\Longrightarrow\qquad e_t \ \le\ L_\theta\,\|\theta - \theta'\|\sum_{s=0}^{t-1} L^s .$$

The same geometric sums appear, now multiplied by $L_\theta$. If the tuned parameters enter the network as inputs, $u = \pi(x, \theta)$, one certified Lipschitz bound of $\pi$ with respect to $(x, \theta)$ supplies both factors. Either way, a constant in the thousands would shrink LoSBO's safe intervals, of radius $(y - E - h)/L$, to almost nothing. Certified constants are most informative here for contracting loops, short horizons or large safety margins.

B2Statistical model-error bounds plus IQC analysis of neural loops

The direction
Fiedler, Scherer & Trimpe (CDC 2021) turned GP uniform error bounds into an uncertainty description for robust (LFR/IQC) synthesis of linear controllers. Pauli-style IQC analysis handles neural-network controllers. Together they could give closed-loop guarantees, holding with probability at least $1-\delta$, for a learned model controlled by a learned policy.

In plain words. Learning gives statements like "the model error is at most 0.1, with probability 99%". Robust control gives statements like "the loop is stable for every model error up to 0.1". Chaining the two yields "the loop is stable with probability 99%", a guarantee neither tool provides alone. Fiedler, Scherer and Trimpe did this for linear controllers; doing it for neural-network controllers analysed with Pauli-style IQCs is the proposed direction.

Symbols in this direction
Worked example — from an error bar to an uncertainty block

A GP learns an unknown nonlinearity $f$ on the region where the system operates. Suppose the posterior standard deviation there is at most $\bar\sigma = 0.05$, and that with $\beta = 2$ the uniform bound holds with probability $1 - \delta$. Then the residual $\Delta(x) = f(x) - \mu(x)$ satisfies

$$|\Delta(x)| \ \le\ \beta\,\bar\sigma = 0.1 \qquad\text{for every } x \text{ in the region.}$$

Robust control treats $\Delta$ as an unknown but bounded block. If the loop with the learned model $\mu$ is stable for every such block (an LMI test), then the true loop is stable with probability at least $1 - \delta$, as long as the state stays in the region where the bound holds (which itself must be certified, for example with an invariant set).

The open step is to replace the linear controller by a neural network. Its activations need their own quadratic constraints (Module 14 §2), and both kinds of constraints must fit into one LMI.

B3Certified approximate MPC with Lipschitz-bounded networks

The direction
Both lines have worked on neural approximations of MPC. On the Trimpe side, Hose et al. (TCST 2025) add online safety augmentation. On the Pauli side, Drummond et al. (L4DC 2022) bound the difference between the MPC law and the network. Lipschitz-by-design parameterisations could make both the approximation-error bounds and the runtime certificates cheaper.

In plain words. Model predictive control is powerful but slow: it solves an optimisation problem at every step. A neural network can learn to imitate it and run in microseconds, but MPC's guarantees then carry over only if the imitation is accurate everywhere. Lipschitz-bounded networks make "accurate everywhere" checkable from a finite set of tests.

Symbols in this direction
Worked example — certifying the approximation error everywhere from a grid

Suppose a robust MPC design tolerates input errors up to $\bar\eta = 0.05$. On a grid with spacing $r = 0.005$, the largest measured error is $0.02$. The error function $e(x)$ is Lipschitz with constant at most $L_{\rm NN} + L_{\rm MPC}$; say that is $4$. Every state is within distance $r/2 = 0.0025$ of a grid point, so

$$e(x) \ \le\ 0.02 + 4 \times 0.0025 = 0.03 \ \le\ 0.05 = \bar\eta \qquad\text{for every } x \text{ in the region.}$$

The network inherits the MPC's guarantees. This interpolation argument needs a certified bound on $L_{\rm NN}$ (or a directly certified Lipschitz bound on the approximation error). With a loose constant, say $400$ instead of $4$, the coarsest grid that still works would be a hundred times finer in every dimension.

B4Uncertainty-aware safety filters with certified ensembles

The direction
UPSi's guarantees rest on assumptions about its ensemble (T4). Lipschitz-bounded ensemble members would bound how disagreement, and hence the certain set, can change along a predicted trajectory. That could turn part of the assumption into a certificate.

In plain words. UPSi trusts its ensemble only in a "certain set", where the members agree. Today the certain set is estimated, not certified. If each member is Lipschitz-bounded, their disagreement cannot jump: it can only grow at a known rate as you move away from a point where it was measured. That turns "the members agree here" into a certified statement about a whole neighbourhood.

Symbols in this direction
Worked example — disagreement cannot jump

Each member can change by at most $L\,\|x - x_0\|$ between $x_0$ and $x$, possibly in opposite directions, so $D$ is $2L$-Lipschitz. If $D(x_0) = 0.05$ and $L = 1$, then for every $x$ with $\|x - x_0\| \le 0.1$:

$$D(x) \ \le\ 0.05 + 2 \times 1 \times 0.1 = 0.25 .$$

A certain set defined by $D \le 0.25$ therefore provably contains the whole ball of radius $0.1$ around $x_0$, without evaluating the ensemble anywhere else. What this does not fix is the problem of T4: small disagreement still does not mean small error.

9. Gaps in These Notes

Topics these notes touch only briefly or not at all, which would make good future modules:

TopicWhy it belongs here
Data-driven control (Willems' fundamental lemma, behavioural systems)Controllers with guarantees designed directly from data, without first identifying a model. This is a strong line at Stuttgart (Berberich, Koch, Allgöwer), Pauli's former group.
Contraction theory in depthThe incremental stability notion behind RENs and many Lipschitz-by-design arguments. Module 13 states the contraction conditions it needs, but not the general theory (contraction metrics and their use in nonlinear control design).
Learned dynamics with guarantees (Koopman operators, physics-informed models)Model classes whose error can be bounded; the missing ingredient in T4 and F1.
Multi-agent safetyInteracting agents, game-theoretic filters and learned graph barriers (GCBF+); Module 10 mentions them only briefly.
Robust and distributionally robust MDPsWorst-case models over an uncertainty set: a middle ground between CMDPs and robust control.
Safety under partial observabilityBelief-space safety and perception in the loop (F5).
Formal specifications (temporal logic, shields in depth)Richer safety requirements than "stay in a set"; Module 10 introduces shields only.
Safety for LLM-based agents beyond Safe RLHFConstraints on tool-using and embodied language agents (R5).

Discussion Questions

These have no single answer; each points at the core difficulty of one or more problems above. A suggested angle is hidden under each.

Question 1 — Why can't data bound an RKHS norm? (T1)

You observe $f(x_1),\dots,f(x_n)$ exactly, without noise. Show that for the squared-exponential kernel no finite set of such observations implies an upper bound on $\|f\|_k$. What minimal extra assumption would make a bound possible?

Show a suggested angle
The SE kernel's RKHS is infinite-dimensional, so the orthogonal complement of $V = \operatorname{span}\{k(\cdot,x_i)\}$ is nonzero. Any $g \perp V$ satisfies $g(x_i) = \langle g, k(\cdot,x_i)\rangle = 0$ by the reproducing property. Then $f + c\,g$ matches all observations, and $\|f + c g\|_k^2 = \|f\|_k^2 + 2c\langle f,g\rangle + c^2\|g\|_k^2 \to \infty$ as $c \to \infty$. An upper bound therefore needs information that no data set contains. One option is to assume a norm bound directly, which is the very assumption we wanted to avoid. Another is to put a prior over functions, which yields a probabilistic statement instead of a bound. The alternative is to drop the RKHS-norm assumption altogether and use a regularity notion that is easier to justify, such as a Lipschitz constant, as LoSBO does. That supports a different safety argument, but it does not bound $\|f\|_k$.
Question 2 — Can LoSBO get stuck? (T2)

Describe a situation in which LoSBO stays safe but stops expanding its safe set long before it reaches the optimum. Which ingredient of SafeOpt's proof prevents this there?

Show a suggested angle
LoSBO's GP chooses where to sample. If the kernel's length scale is much too long, the GP can be confidently wrong and pessimistic about points near the safe-set boundary. It may then keep sampling interior "maximisers" whose measurements never push a Lipschitz cone outwards. In SafeOpt, the confidence intervals are assumed valid, so uncertainty sampling of expanders must eventually either certify new points or show they cannot be certified. LoSBO keeps safety without that assumption, and so also loses the lever that forces expansion.
Question 3 — How fast do closed-loop Lipschitz constants grow? (B1)

For $x_{t+1} = F(x_t)$ with $F$ $L$-Lipschitz, bound $\|x_T - y_T\|$ in terms of $\|x_0 - y_0\|$. When is the bound useful for certifying a closed-loop cost over $T$ steps?

Show a suggested angle
By induction, $\|x_t - y_t\| \le L\,\|x_{t-1} - y_{t-1}\| \le \dots \le L^t \|x_0 - y_0\|$. A cost $\sum_{t=0}^{T} c(x_t)$ with $c$ being $L_c$-Lipschitz therefore has constant at most $L_c \sum_{t=0}^{T} L^t$. That is at most $L_c/(1-L)$ if $L \lt 1$ (a contraction), whatever the horizon, and grows like $L^T$ if $L \gt 1$. So certified per-step bounds are most useful for contracting closed loops, or over short horizons.
Question 4 — Incremental vs non-incremental multipliers (P2)

Coupled multipliers are valid in DeepSDP-type (non-incremental) verification but not in LipSDP-type (incremental) certificates. In one or two sentences, why?

Show a suggested angle
In the non-incremental setting a coupling term compares two coordinates of one vector: $\big(v_1 - v_2,\ \varphi(v_1) - \varphi(v_2)\big)$ is a genuine chord of $\varphi$, because the same scalar function acts on both coordinates, so the slope bounds apply. In the incremental setting it compares increments between two inputs. There, $\Delta z_1 - \Delta z_2 = [\varphi(v_1) - \varphi(\bar v_1)] - [\varphi(v_2) - \varphi(\bar v_2)]$ is a difference of two chords taken at different places on the graph, and nothing bounds its ratio to $\Delta v_1 - \Delta v_2$. Module 12, §5 gives a two-neuron ReLU counterexample.
Question 5 — When does feedback break exchangeability? (S1)

A robot uses conformal bounds on a pedestrian predictor to plan. Give a concrete way in which the robot's own plan can make the next calibration scores non-exchangeable with the old ones.

Show a suggested angle
Pedestrians react to the robot. Once the robot drives more aggressively because its bounds are tight, pedestrians swerve more, so prediction errors grow. The new scores come from a different distribution than the calibration scores collected under the old behaviour. Exchangeability, and with it the coverage guarantee, is lost.
Question 6 — Expected cost vs failure probability (R3)

Construct a random cost $Z \ge 0$ with $\mathbb E[Z] \le 0.01$ but $\mathbb P(Z \gt 0) = 0.5$. What does this say about using expected-cost constraints for safety?

Show a suggested angle
Let $Z = 0.02$ with probability $0.5$ and $Z = 0$ otherwise. Then $\mathbb E[Z] = 0.01$ while half of all episodes incur a cost. An expected-cost budget constrains how much cost accumulates on average, not how often cost occurs, so it cannot certify a failure probability without extra structure. That is why chance, CVaR and almost-sure constraints exist (Module 1 §1, Module 8 §6).

Key Papers

PaperVenueProblems
On Safety in Safe Bayesian Optimization (Fiedler, Menn, Kreisköther, Trimpe)TMLR 2024T1, T2
Reliable sampling-based RKHS norm estimation via superconvergence (Wenzel, Tokmak, Fiedler)arXiv 2026T1
Open Problem: Tight Online Confidence Intervals for RKHS Elements (Vakili, Scarlett, Javidi)COLT 2021T3
Tighter Confidence Bounds for Sequential Kernel Regression (Flynn, Reeb)AISTATS 2025T3
Uncertainty-Aware Predictive Safety Filters for Probabilistic Neural Network Dynamics (Frauenknecht et al.)RLC 2026T4, B4
Event-Triggered Time-Varying Bayesian Optimization (Brunzema et al.)TMLR 2025T5
Safe Time-Varying Optimization based on Gaussian Processes with Spatio-Temporal Kernel (Li, Zagorowska, De Pasquale, Rupenyan, Lygeros)NeurIPS 2024T5
Viability of Future Actions: Robust Safety in RL via Entropy Regularization (Massiani et al.)ECML-PKDD 2025T6
A Decade of Bayesian Optimization for Controller Tuning and Robot Learning (Stenger et al.)arXiv 2026T7
Demystifying Lipschitz verification: positive matrices, negative results (Kuang et al.)arXiv 2026P1
Linear systems with neural network nonlinearities: improved stability analysis via acausal Zames–Falb multipliers (Pauli et al.)CDC 2021P2
Synthesizing Neural Network Controllers with Closed-Loop Dissipativity Guarantees (Junnarkar, Arcak, Seiler)Automatica 2026P3
LipNeXt: Scaling up Lipschitz-based Certified Robustness to Billion-parameter Models (Hu, Hu, Fredrikson)ICLR 2026P4
LipKernel: Lipschitz-Bounded Convolutional Neural Networks via Dissipative Layers (Pauli et al.)Automatica 2026P4, P5, B3
Last-Iterate Global Convergence of Policy Gradients for Constrained RL (Montenegro et al.)NeurIPS 2024R2
Towards a Practical Understanding of Lagrangian Methods in Safe RL (Spoor et al.)arXiv 2025R3
Control Barrier Function Based Quadratic Programs Introduce Undesirable Asymptotically Stable Equilibria (Reis, Aguiar, Tabuada)IEEE L-CSS 2021F3
Safety Filtering While Training (Pizarro Bejarano, Brunke, Schoellig)IEEE RA-L 2025F4
The Safety Filter: A Unified View of Safety-Critical Control in Autonomous Systems (Hsu, Hu, Fisac)Annu. Rev. 2024F1–F5
Curse of Dimensionality on Randomized Smoothing for Certifiable Robustness (Kumar, Levine, Goldstein, Feizi)ICML 2020S2
Learning-enhanced robust controller synthesis with rigorous statistical and control-theoretic guarantees (Fiedler, Scherer, Trimpe)CDC 2021B2