13. Lipschitz-by-Design Networks & Direct Parameterizations

Orthogonal and Cayley layers, SLL, Sandwich layers, Pauli's Cayley–Gramian CNNs and LipKernel, RENs, and the certified-robustness state of the art

Before you start

This module assumes:

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

Contents
1. Constrain or Parameterize? 2. Spectral Normalization, Parseval, Cayley, SOC, AOL 3. SLL and Sandwich Layers: LMIs Solved by Construction 4. Lipschitz-Bounded CNNs: Cayley–Gramian and LipKernel (Pauli et al.) 5. Recurrent Equilibrium Networks and R2DN 6. Certified Robust Accuracy: State of the Art 2026 Walkthrough: Deriving the Sandwich Layer From the LMI Interactive: Cayley Transform and a 1-Lipschitz Layer Graded practice: Easy / Medium / Hard Original research exercises Application lab & chapter review Exercises and readiness check Key Papers Flashcards
Study route — Start with a scalar layer, then read the construction

On a first pass, follow the links below and solve the Easy set before opening the research exercises. Try each question on paper; open its hint only when stuck, then compare your reasoning with the worked solution. Use Medium questions to connect the algebra and Hard questions to check the assumptions.

§1: certified parameter sets · §2: norm preservation and Cayley · §3: scalar SLL and Sandwich · §4: optional systems extension

Practice: 4 Easy · 4 Medium · 4 Hard · original research exercises. Check readiness or use the course study guide.

A scalar model of “by construction”

Suppose a scalar weight must obey $|w|\le1$. The free parameterization $w=\tanh\theta$ meets the constraint for every real $\theta$: training can move $\theta$ freely. It reaches only $(-1,1)$, so validity does not imply completeness at the boundary. In contrast, adding a penalty $\lambda w^2$ to a loss lets training trade a larger weight against a smaller loss; a finite penalty does not enforce $|w|\le1$.

The matrix constructions replace this interval fact by identities such as $Q^\top Q=I$ or $H=PP^\top\succeq0$. Before reading their formulas, check the dimensions, the domain of each inverse, and the set for which completeness is claimed. Matrix multiplication and transposes and Gram matrices and PSD factors supply the needed algebra.

Module 12 answered the analysis question: given the weights, LipSDP returns a certified Lipschitz bound, and constrained training (ADMM, barriers) can push it down. This module turns the question around and builds the certificate into the architecture: every weight is a function of free parameters (smooth in the Cayley-based constructions), chosen so that the Lipschitz certificate holds for every value of the free parameters in its stated domain. For a direct parameterization with domain $\mathbb R^N$, training is then unconstrained SGD (see Primer B), and every iterate, including the deployed one, is certified in exact arithmetic. The Cayley–Gramian construction below additionally requires nonsingular gain factors; approximate orthogonal updates need their numerical error included.

The algebra is small: a Schur complement turns a Lipschitz LMI into a norm bound, and a norm bound is met by construction with a matrix with orthonormal columns, which the Cayley transform produces from arbitrary matrices. The control-theoretic twist, due to Patricia Pauli and co-authors, is to treat convolutions as dynamical systems (with Dennis Gramlich and Frank Allgöwer), so that the storage matrix of their dissipativity certificate can be chosen as the inverse of a controllability Gramian (see Primer D) in closed form (with Ruigang Wang, Ian Manchester and Frank Allgöwer). The same ideas give recurrent models that are contracting by construction (RENs, R2DN), which Module 14 uses as certified controllers.

Notation for this module (and clashes)
Network $f_\theta:\mathbb R^{n_0}\to\mathbb R^{n_{L+1}}$ with weights $W_k$, biases $b_k$ and an elementwise activation (see Primer E) $\varphi$ slope-restricted in $[0,1]$ (ReLU, tanh; Module 12's $\alpha=0,\ \beta=1$; Wang & Manchester write $\sigma$). Hidden layer $k$ maps $z_k\in\mathbb R^{n_k}$ to $z_{k+1}=\varphi(W_kz_k+b_k)$ with $W_k\in\mathbb R^{n_{k+1}\times n_k}$; the last layer is affine. $\|\cdot\|$ is the Euclidean or spectral norm (see Primer A).
Lipschitz bound: $\operatorname{Lip}(f)$ is the smallest $\gamma$ with $\|f(x)-f(y)\|\le\gamma\|x-y\|$ for all $x,y$; "nonexpansive" means $\operatorname{Lip}\le1$. All constructions below that invert a gain assume $\gamma\gt0$ and $\rho\gt0$. The bound is written $\gamma$ in the Sandwich/LBDN and REN papers, $\rho$ in Pauli et al. (CDC 2023) and LipKernel; both are the bound itself (certificates contain $\gamma I$ or $X_0=\rho^2I$), whereas LipSDP's $\rho=L^2$ is the squared bound. No discount factor or information gain appears here, so $\gamma$ is free for this role.
Multipliers: Module 12's diagonal $T=\operatorname{diag}(\lambda_i)$ is called $\Lambda\succ0$ here (see Primer A), as in the papers; in SLL, $T$ is the diagonal matrix in $W^{\top}W\preceq T$ and the multiplier is $2T^{-1}$.
Clashes: in Section 3 and the walkthrough, $A_k,B_k$ are the blocks of a Cayley-generated matrix with orthonormal columns (Wang & Manchester); in Sections 4–5, $(A,B,C,D)$ are state-space matrices and the Cayley blocks are $U,V$ (Pauli). Layer gain matrices are $Q_i$ in Pauli et al. (CDC 2023) and $X_k$ in LipKernel; the REN IQC uses an unrelated $(Q,S,R)$ with $Q\preceq0$; $L$ counts hidden layers in Section 3 but is a Lipschitz constant in Section 6, and $L_i$ in Section 4 is a matrix square root. $X_k,Y_k$ are free Cayley parameters, but in the one-layer identity of Section 3 and Exercise 13.3, $U$ and $Y$ are the layer's input and output matrices (the paper's notation); $Q$ is also an orthogonal matrix (Section 2) and SLL's diagonal scaling (Section 3); in Section 5, $\alpha$ and $\bar\alpha$ are contraction rates, unrelated to the slope bound $\alpha=0$.

1. Constrain or Parameterize?

The training problem we would like to solve is

$$\min_{\theta}\ \mathcal L(f_\theta)\qquad\text{s.t.}\qquad \operatorname{Lip}(f_\theta)\le\gamma ,$$

where $\mathcal L$ is the training loss, $\theta$ collects the trainable parameters and "s.t." means "subject to": a constrained optimization problem whose feasible points are the $\theta$ satisfying the constraint (see Primer B). Computing $\operatorname{Lip}(f_\theta)$ exactly is NP-hard already for two-layer ReLU networks (Virmaux & Scaman, NeurIPS 2018), so the constraint is always replaced by a certificate set $\Theta_{\rm cert}(\gamma)\subseteq\{\theta:\operatorname{Lip}(f_\theta)\le\gamma\}$, e.g. "there is a diagonal $\Lambda\succ0$ satisfying the LipSDP LMI" (Fazlyab et al., NeurIPS 2019; Module 12). There are three ways to live with that set.

  1. Constrain: optimize over $\Theta_{\rm cert}$ with ADMM (Pauli et al., L-CSS 2022), log-det barriers (Pauli et al., CDC 2022) or projections (Module 12). The set is not jointly convex in $(W,\Lambda)$; the substitution $\hat W=\Lambda W$ makes it an LMI, but every step still touches a semidefinite constraint. Normalized training times on the CDC 2022 toy example: 1 (nominal), 7.35 (projected), 78.72 (ADMM), 1.15–1.30 (barrier); Wang & Manchester report that barriers and projections become the bottleneck around $10^3$ neurons.
    Derivation — why $\hat W=\Lambda W$ gives an LMI

    In the one-hidden-layer LMI of Section 3, $H=\begin{bmatrix}\gamma I&-W_0^{\top}\Lambda&0\\-\Lambda W_0&2\Lambda&-W_1^{\top}\\0&-W_1&\gamma I\end{bmatrix}\succeq0$, the only products of unknowns are $\Lambda W_0$ and $W_0^{\top}\Lambda$. With the new variable $\hat W_0=\Lambda W_0$ the off-diagonal blocks become $-\hat W_0^{\top},-\hat W_0,-W_1^{\top},-W_1$, so $H$ is affine in $(\hat W_0,\Lambda,W_1,\gamma)$: an LMI. The network is recovered as $W_0=\Lambda^{-1}\hat W_0$ ($\Lambda$ is positive diagonal, hence invertible). Only the constraint becomes convex; the training loss, written in the new variables, is still nonconvex.

  2. Regularize: penalize a differentiable bound, e.g. the loop-transformation bound LipLT with a margin regularizer (Fazlyab, Entesari, Roy & Chellappa, NeurIPS 2023). Cheap, but nothing forces the certified bound below a prescribed $\gamma$.
    Going deeper — what LipLT computes, and why a penalty is not a constraint

    A bound routine returns a differentiable $U(\theta)\ge\operatorname{Lip}(f_\theta)$, and training minimizes $\mathcal L(f_\theta)+\lambda U(\theta)$ with $\lambda\ge0$ (or a margin loss built from $U$). For one hidden layer, $f(x)=W_1\varphi(W_0x+b_0)+b_1$ with slopes in $[0,1]$, the loop transformation splits $\varphi(t)=\tfrac t2+r(t)$: $r$ has slopes in $[-\tfrac12,\tfrac12]$, so it is $\tfrac12$-Lipschitz, and $f(x)=\tfrac12W_1W_0x+W_1r(W_0x+b_0)+\text{const}$. Bounding the two terms separately gives

    $$\operatorname{Lip}(f)\le U=\tfrac12\|W_1W_0\|_2+\tfrac12\|W_1\|_2\|W_0\|_2\ \le\ \|W_1\|_2\|W_0\|_2 ,$$

    never worse than the product bound because the first term sees the combined direction of $W_1W_0$; deeper networks repeat the transformation layer by layer (Fazlyab et al., NeurIPS 2023). A finite $\lambda$ lets training trade a larger $U$ for a smaller loss, so nothing enforces $U(\theta)\le\gamma$.

  3. Parameterize: choose a smooth map whose image lies inside $\Theta_{\rm cert}(\gamma)$ and train over its unconstrained argument. This module.
Definition — Convex and direct parameterizations (Wang & Manchester 2023, Def. 2.3), plus certified and complete ones
A parameterization is a differentiable map $\phi=\mathcal M(\theta)$ from free parameters $\theta\in\Theta\subseteq\mathbb R^N$ to weights and biases $\phi=\{(W_k,b_k)\}$; it is convex if $\Theta$ is convex and direct if $\Theta=\mathbb R^N$. We call it certified if $\mathcal M(\Theta)\subseteq\Theta_{\rm cert}(\gamma)$ and complete (for that certificate) if $\mathcal M(\Theta)=\Theta_{\rm cert}(\gamma)$. A certified direct parameterization turns the constrained problem into $\min_{\theta\in\mathbb R^N}\mathcal L(f_{\mathcal M(\theta)})$, solvable by any first-order method without projections, with every iterate satisfying $\operatorname{Lip}\le\gamma$.
Key insight
Every construction on this page uses one move: an LMI says a matrix is positive semidefinite, PSD matrices have square roots, and square roots (or matrices with orthonormal columns) can be generated from arbitrary matrices. The idea goes back to Lipschitz-bounded equilibrium networks (Revay, Wang & Manchester, 2020), in the spirit of Burer–Monteiro factorizations $H=PP^{\top}$ (see Primer A). The art is to choose the factorization so that the sparsity pattern (see Primer A) of the LMI, which encodes the architecture, survives.

Layer-wise versus network-wise certificates

The cheapest certificate multiplies layer bounds, $\operatorname{Lip}(f)\le\prod_k\|W_k\|$ for 1-Lipschitz activations (Module 12); making every layer 1-Lipschitz (Section 2) is the corresponding design principle. It ignores that the directions amplified by one layer need not be those amplified by the next. Network-level certificates such as LipSDP couple the layers through multipliers, and their parameterizations produce networks whose individual layers have norms well above 1 while the network is still $\gamma$-Lipschitz (Wang & Manchester, Remark C.2; see the explorer). LipKernel's Fig. 1 shows it geometrically: propagating ellipsoids through a two-layer network certifies bound 1 where propagating balls certifies almost 2.

Expressivity: why gradient norm has to be preserved

Theorem — Norm-constrained monotone networks cannot preserve gradient norm (Anil, Lucas & Grosse 2019, Thm 1)
Let $f:\mathbb R^n\to\mathbb R$ be a feedforward network with weight matrices $\|W\|_2\le1$ and 1-Lipschitz, elementwise, monotone activations. If $\|\nabla f(x)\|_2=1$ (see Primer B) for almost every $x$ (see Primer 0), then $f$ is affine (a linear map plus a constant).

In words: backpropagation multiplies the gradient by $W_k^{\top}$ (norm at most 1) and by diagonal activation Jacobians with entries in $[0,1]$; to arrive with norm exactly 1, every unit that matters must sit on a slope-1 piece everywhere, so $f$ is affine. The norm bound makes each factor non-expansive; "elementwise and monotone" restricts the Jacobian to a diagonal with entries in $[0,1]$ (a sorting activation escapes because its Jacobian is a permutation). Hence a ReLU network with $\|W_k\|\le1$ cannot represent $|x|$ and can be too restrictive for Wasserstein critics, which optimize over all scalar 1-Lipschitz functions (Anil et al., ICML 2019). The remedy: weights may be taken orthonormal without changing the function (their Thm 2, under the unit-gradient hypothesis; stated below), and activations must be gradient-norm preserving: GroupSort sorts groups of pre-activations, and its group-size-2 case MaxMin maps $(a,b)\mapsto(\max(a,b),\min(a,b))$. With both, scalar-output networks approximate every 1-Lipschitz function (their Thm 3, below).

Background — Wasserstein critics

For probability distributions $\mu,\nu$ on $\mathbb R^n$ with finite mean distance from the origin, the Wasserstein-1 distance $\mathcal W_1(\mu,\nu)$ is the smallest expected travel distance $\mathbb E\|X-Y\|$ over all random pairs $(X,Y)$ with $X\sim\mu$, $Y\sim\nu$ (the cheapest way to move the mass of $\mu$ onto $\nu$). Kantorovich–Rubinstein duality turns this into a maximization over scalar functions, $\mathcal W_1(\mu,\nu)=\sup_{\operatorname{Lip}(f)\le1}\big(\mathbb E_\mu f-\mathbb E_\nu f\big)$, and a maximizing $f$ is called a critic. Example: point masses at $0$ and $2$ have $\mathcal W_1=2$, attained by $f(x)=-x$. A Wasserstein GAN learns the critic with a 1-Lipschitz network; if the network family cannot represent good critics (e.g. $|x|$-shaped ones), the estimated distance is too small. This is the application that motivated gradient-norm-preserving networks (Anil et al., ICML 2019).

Theorem — Orthonormal weights lose nothing (Anil, Lucas & Grosse 2019, Thm 2)
Let $f:\mathbb R^n\to\mathbb R$ be a feedforward network with 1-Lipschitz activations and weights $\|W_k\|_2\le1$. If $\|\nabla f(x)\|_2=1$ for almost every $x$, then each weight $W\in\mathbb R^{m\times k}$ can be replaced, without changing $f$, by a matrix $\tilde W$ whose singular values all equal $1$: orthonormal columns if $m\gt k$, orthonormal rows if $m\lt k$, orthogonal if $m=k$. The scalar output and the unit-gradient hypothesis are essential; the theorem says nothing about Lipschitz networks in general.
Theorem — MaxMin networks are universal 1-Lipschitz approximators (Anil, Lucas & Grosse 2019, Thm 3)
Let $X\subset\mathbb R^n$ be compact (closed and bounded; see Primer 0), $1\le p\le\infty$, and $g:X\to\mathbb R$ with $|g(x)-g(y)|\le\|x-y\|_p$. For every $\varepsilon\gt0$ there is a scalar-output MaxMin network $f$ (any finite depth and width, with biases) whose first weight matrix has $\|W_1\|_{p,\infty}=1$ and whose other weights have $\|W_i\|_\infty=1$, such that $\sup_{x\in X}|f(x)-g(x)|\le\varepsilon$. Here $\|W\|_{p,\infty}=\max_{\|x\|_p\le1}\|Wx\|_\infty$ and $\|W\|_\infty=\max_i\sum_j|W_{ij}|$ (see Primer A).
In words: the family is dense in the 1-Lipschitz functions for the sup norm: one error bound for all inputs at once, reached by letting the network grow as $\varepsilon$ shrinks. It is approximation, not exact representation by a fixed architecture.
Caveat
The universality theorem is for scalar outputs and these mixed $\infty$-norm constraints; it is not a universality result for orthogonal ($\ell_2$) layers or vector-valued maps. MaxMin is not an elementwise slope-restricted map, so plain LipSDP does not apply; on its residual-ReLU rewrite LipSDP only certifies $\sqrt2$, which is why Pauli et al. (ICLR 2024) derived quadratic constraints for GroupSort (Module 12).

The table orders the approaches of this module by what holds for every parameter value; "layer" certificates make each layer 1-Lipschitz, "network" certificates couple layers through multipliers or gain matrices.

ApproachWhat holds for every parameter valueMechanismLevelMain cost / limitation
Spectral normalization$\|W\|\le1$ (only if $\sigma(W)$ is exact)divide by a power-iteration estimate (see Primer A)layerloose product bound
Parseval networksapproximately $WW^{\top}=I$ (orthonormal rows)retraction after each updatelayerno exact certificate
Cayley / Orthogon$W$ orthogonal (no eigenvalue $-1$)Cayley transform per FFT frequency (see Primer D)layerFFTs in every pass, inverses when the weights change; circular padding (see Primer E)
SOCorthogonal up to truncation errorexponential of a skew-symmetric conv (see Primer D)layererror $\|J\|^k/k!$
BCOPorthogonal convolutionblock products of projectorslayerincomplete
AOL$\|WD\|\le1$diagonal rescalinglayertight only near orthogonal $W$
SLL$W^{\top}W\preceq T$ in a residual layer (see Primer E)analytic diagonal $T$ (Gershgorin)layerper-layer certificate, no coupling between layers
Sandwich (LBDN)the full LipSDP LMICayley blocks and $\Psi=\operatorname{diag}(e^d)$networkFourier-domain convolutions
Cayley–Gramian, LipKernellayer-wise dissipativity LMIsGramian, Cayley, Cholesky (see Primer A)networkinverses at training time only
REN, R2DNcontraction and an incremental IQC$H=X^{\top}X+\epsilon I$ read blockwisedynamicequilibrium solve, REN only (see Primer E)
LipNeXt$W^{\top}W=I$ up to the truncation error of each update, reset by a polar retraction every epochoptimization on the orthogonal manifoldlayernot a parameterization

2. Spectral Normalization, Parseval, Cayley, SOC, AOL

The layer-wise route makes every affine layer 1-Lipschitz and uses $\operatorname{Lip}(g\circ h)\le\operatorname{Lip}(g)\operatorname{Lip}(h)$. A linear layer $x\mapsto Wx$ is 1-Lipschitz iff $\|W\|\le1$. It is gradient-norm preserving, i.e. backpropagation $g\mapsto W^{\top}g$ keeps $\|g\|$ for every $g$ (Li et al., NeurIPS 2019), iff $WW^{\top}=I$ (orthonormal rows); the forward map keeps every $\|x\|$ iff $W^{\top}W=I$ (orthonormal columns), and a square orthogonal $W$ does both. A tall $W$ with orthonormal columns has all singular values 1 but is not gradient-norm preserving: $W=(1,0)^{\top}$ sends $g=(0,1)^{\top}$ to $0$. The methods differ in how exactly and cheaply they enforce this, and whether they work for convolutions.

Spectral normalization and Parseval networks

Key equation — Spectral normalization with one power-iteration step (Miyato et al. 2018)
$$\bar W_{\rm SN}=\frac{W}{\sigma(W)},\qquad \tilde v\leftarrow\frac{W^{\top}\tilde u}{\|W^{\top}\tilde u\|},\quad \tilde u\leftarrow\frac{W\tilde v}{\|W\tilde v\|},\quad \sigma(W)\approx\tilde u^{\top}W\tilde v ,$$ with $\tilde u$ kept from one SGD step to the next, so one power iteration per update is commonly used in practice. With exact, positive $\sigma(W)$, every normalized layer has norm $1$, so 1-Lipschitz activations give the network bound $\prod_l\sigma(\bar W^l_{\rm SN})=1$. The one-step estimate only approximates this normalization; a zero matrix needs no normalization (Miyato, Kataoka, Koyama & Yoshida, ICLR 2018).
Pitfall — power iteration underestimates
For unit vectors $\tilde u^{\top}W\tilde v\le\sigma(W)$, so dividing by the estimate $\hat\sigma$ gives $\|W/\hat\sigma\|=\sigma(W)/\hat\sigma$, which exceeds 1 whenever $\hat\sigma\lt\sigma(W)$; an iteration that looks converged proves nothing. For a certificate, divide by a proven upper bound $U\ge\sigma(W)$ (then $\|W/U\|\le1$), e.g. from Gram iteration, which bounds convolution spectral norms from above at every iterate (Delattre, Barthélemy, Araujo & Allauzen, ICML 2023). SN also attenuates gradients, since singular values below 1 are allowed.
Fact — Gram iteration gives upper bounds at every step (Delattre et al. 2023)
For a real matrix $M$ with singular values $\sigma_1\ge\sigma_2\ge\dots$, set $G_0=M$ and $G_{t+1}=G_t^{\top}G_t$. Then $G_t$ has singular values $\sigma_i^{2^t}$, so $$U_t:=\|G_t\|_F^{1/2^t}=\Big(\sum_i\sigma_i^{2^{t+1}}\Big)^{1/2^{t+1}}\ \ge\ \sigma_1=\|M\|_2,$$ and $U_t$ decreases to $\|M\|_2$ as $t$ grows (repeated squaring lets the largest singular value dominate). Every $U_t$ is a valid upper bound, even before convergence. For a circular convolution, apply it to each (complex) frequency block, with $G_t^{*}G_t$ in place of $G_t^{\top}G_t$ (see the background below), and take the maximum over frequencies. In floating point the iterates are rescaled to avoid overflow; the rescaling and rounding must be accounted for in a certificate.

Parseval networks keep a weight $W\in\mathbb R^{d_{\rm out}\times d_{\rm in}}$, $d_{\rm out}\le d_{\rm in}$, close to a Parseval tight frame, $WW^{\top}=I$ (orthonormal rows), with the Frobenius-norm regularizer $R_\beta(W)=\tfrac\beta4\|WW^{\top}-I\|_F^2$, whose gradient is $\beta(WW^{\top}-I)W$ (see Primer B), and take one unit gradient step on it after every main update: $W\leftarrow(1+\beta)W-\beta WW^{\top}W$ (Cisse et al., ICML 2017; the paper prints the penalty as $\tfrac\beta2\|W^{\top}W-I\|_2^2$, which does not have this gradient). In the SVD $W=U\Sigma V^{\top}$ this maps each singular value $s\mapsto(1+\beta)s-\beta s^3$, which has an attracting fixed point at $s=1$ for $0\lt\beta\lt1$ (see Primer D). The result is only approximately orthogonal, so there is no certificate without a post-hoc norm computation.

Derivation — the Parseval gradient and the singular-value map

Gradient. With $E=WW^{\top}-I$ (symmetric), $R_\beta=\tfrac\beta4\operatorname{tr}(E^2)$, so $dR_\beta=\tfrac\beta2\operatorname{tr}(E\,dE)$ with $dE=dW\,W^{\top}+W\,dW^{\top}$ (see Primer B). Both terms equal $\operatorname{tr}\big((EW)^{\top}dW\big)$ by the cyclic property of the trace, hence $dR_\beta=\beta\operatorname{tr}\big((EW)^{\top}dW\big)$ and $\nabla R_\beta=\beta(WW^{\top}-I)W$.

Singular values. With the SVD $W=U\Sigma V^{\top}$, $WW^{\top}W=U\Sigma^3V^{\top}$, so the step $W-\nabla R_\beta=(1+\beta)W-\beta WW^{\top}W=U\big((1+\beta)\Sigma-\beta\Sigma^3\big)V^{\top}$ acts on each singular value by $g(s)=(1+\beta)s-\beta s^3$. Now $g(1)=1$ and $g'(1)=1+\beta-3\beta=1-2\beta$, and $|g'(1)|\lt1$ exactly when $0\lt\beta\lt1$: then iterates that start close to $1$ converge to $1$. This is a local statement, not convergence from arbitrary weights.

The Cayley transform

Theorem — Cayley transform of a skew-symmetric matrix
Let $A^{\top}=-A\in\mathbb R^{n\times n}$ and $Q=(I-A)(I+A)^{-1}$. Then (i) $I+A$ is invertible; (ii) $Q$ is orthogonal; (iii) $-1$ is not an eigenvalue of $Q$ and $\det Q=+1$; (iv) every orthogonal $Q$ without eigenvalue $-1$ is the Cayley transform of exactly one skew-symmetric matrix, $A=(I+Q)^{-1}(I-Q)$. So the Cayley transform is a bijection between skew-symmetric matrices and orthogonal matrices without eigenvalue $-1$ (Trockman & Kolter, ICLR 2021).
Proof sketch — why the Cayley transform is orthogonal

(i) $x^{\top}Ax$ equals its own transpose $x^{\top}A^{\top}x=-x^{\top}Ax$, so it is $0$ and $x^{\top}(I+A)x=\|x\|^2\gt0$ for $x\ne0$: the kernel of $I+A$ is trivial.

(ii) $I\pm A$ are polynomials in $A$, so they commute (see Primer A), and so do $I-A$ and $(I+A)^{-1}$. With $(I+A)^{\top}=I-A$ and $(I-A)^{\top}=I+A$: $$Q^{\top}Q=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}=(I-A)^{-1}(I-A)(I+A)(I+A)^{-1}=I .$$

(iii) If $Qv=-v$, $v\ne0$, put $w=(I+A)^{-1}v\ne0$: then $(I-A)w=-(I+A)w$, i.e. $2w=0$, a contradiction. Also $\det Q=\det\big((I+A)^{\top}\big)/\det(I+A)=1$.

(iv) $Q(I+A)=I-A$ is equivalent to $(I+Q)A=I-Q$, uniquely solvable exactly when $-1\notin\operatorname{spec}(Q)$, with solution $A=(I+Q)^{-1}(I-Q)=(I-Q)(I+Q)^{-1}$ (functions of $Q$ commute, as in (ii)). It is skew-symmetric: using $Q^{\top}=Q^{-1}$, $I-Q^{-1}=Q^{-1}(Q-I)$ and $(I+Q^{-1})^{-1}=(Q+I)^{-1}Q$, $$A^{\top}=(I-Q^{-1})(I+Q^{-1})^{-1}=Q^{-1}(Q-I)(Q+I)^{-1}Q=(Q-I)(Q+I)^{-1}=-A .\qquad\blacksquare$$

Two dimensions. For $A=\begin{bmatrix}0&a\\-a&0\end{bmatrix}$, $Q=\frac{1}{1+a^2}\begin{bmatrix}1-a^2&-2a\\2a&1-a^2\end{bmatrix}$, a rotation by $\theta=2\arctan a$ (put $a=\tan(\theta/2)$ in the double-angle formulas). As $a\to\pm\infty$ the angle tends to $\pm\pi$ without reaching it: $Q=-I$ is missing, and so are all reflections. Trockman & Kolter note that a fixed diagonal $\pm1$ factor recovers them, but that choice is discrete and cannot be learned by gradient descent. Explorer (a) below animates this. Rectangular version: for free $X\in\mathbb R^{n\times n}$, $Y\in\mathbb R^{m\times n}$ and $Z=X-X^{\top}+Y^{\top}Y$, the matrix $\big[(I+Z)^{-1}(I-Z);\,-2Y(I+Z)^{-1}\big]$ has orthonormal columns (walkthrough, Step 3). Pauli et al. write the same map as $U=(I+M)^{-1}(I-M)$, $V=2Z(I+M)^{-1}$ with $M=Y-Y^{\top}+Z^{\top}Z$; the sign of the lower block is irrelevant.

Cayley convolutions. The 2-D DFT block-diagonalizes a circular convolution with $c$ channels on $n\times n$ images, $\mathrm{FFT}(\mathrm{conv}_W(X))[:,i,j]=\tilde W[:,:,i,j]\,\tilde X[:,i,j]$, and transposes or inverts it frequency by frequency. Trockman & Kolter skew-symmetrize per frequency ($\tilde A=\tilde W-\tilde W^{*}$), apply the Cayley transform to each $c\times c$ block and return with an exactly real inverse FFT. Price: $n^2$ inverses of $c\times c$ matrices, $O(n^2c^3)$ operations (see Primer 0), per layer and training forward pass (with frozen weights the transformed blocks can be cached; see "Inference cost" below), circular padding only, strides emulated by invertible downsampling.

Background — complex frequency blocks and invertible downsampling

Complex matrices. For $z=a+\mathrm ib$ ($\mathrm i^2=-1$) the conjugate is $\bar z=a-\mathrm ib$ and $|z|^2=\bar zz=a^2+b^2$. For complex vectors and matrices the adjoint $M^{*}=\bar M^{\top}$ takes the place of the transpose, and $\|v\|_2^2=v^{*}v$. Hermitian ($M^{*}=M$), skew-Hermitian ($M^{*}=-M$) and unitary ($M^{*}M=I$) are the complex versions of symmetric, skew-symmetric and orthogonal; e.g. the $1\times1$ matrix $[\mathrm i]$ is skew-Hermitian and unitary. For skew-Hermitian $A$, the Cayley construction has invertible $I+A$, gives a unitary $Q$ with no eigenvalue $-1$, and is a bijection onto the unitary matrices with that exclusion, with ${}^{\top}$ replaced by ${}^{*}$. Here $|\det Q|=1$; the conclusion $\det Q=+1$ is specific to real skew-symmetric $A$ and need not hold over $\mathbb C$: $A=[\mathrm i]$ gives $Q=[-\mathrm i]$ (see Primer A).

Why frequency blocks. The unitary DFT $F$ ($F^{*}F=I$; Primer D) turns a circular convolution $C$, after grouping the $c$ channels of each frequency together, into $FCF^{*}=\operatorname{blkdiag}_\omega(\tilde W_\omega)$ with one $c\times c$ block per frequency $\omega=(i,j)$; the colons in the formula above select all channels. Hence $\|C\|_2=\max_\omega\|\tilde W_\omega\|_2$, and $C$ is orthogonal iff every block is unitary, which the Cayley transform of the skew-Hermitian $\tilde A_\omega$ guarantees. For a real filter the blocks come in conjugate pairs, $\tilde W_{-\omega}=\overline{\tilde W_\omega}$; skew-symmetrization and the Cayley transform preserve this pairing, so the inverse FFT returns a real layer. (The DFT is the change of coordinates, the FFT a fast algorithm for it; $O(n^2c^3)$ counts $n^2$ dense $c\times c$ solves at cubic cost each, a scaling statement, not a runtime.)

Invertible downsampling rearranges instead of deleting: in 1-D the even-indexed entries go to one channel and the odd-indexed entries to another; in 2-D the four parity classes of pixel positions become four channel groups. This permutes coordinates, so it preserves the Euclidean norm and is invertible; ordinary subsampling discards entries.

SOC and BCOP

For skew-symmetric $J$, $\exp(J)^{\top}\exp(J)=\exp(-J)\exp(J)=I$. Skew Orthogonal Convolutions build a filter whose Jacobian is skew-symmetric ($L=M-\mathrm{conv\_transpose}(M)$ for an arbitrary filter $M$, their Theorem 2) and apply the convolution exponential truncated after $k$ terms (Singla & Feizi, ICML 2021).

Theorem — Truncation error of the exponential (Singla & Feizi 2021, Thm 3)
For skew-symmetric $J$ and $S_k(J)=\sum_{i=0}^{k-1}J^i/i!$: $\ \|\exp(J)-S_k(J)\|_2\le\|J\|_2^{\,k}/k!$. With $k=12$ and $\|J\|_2\le1.8$ the bound is $1.8^{12}/12!\approx2.4\times10^{-6}$.
Caveat — the deployed layer is $S_k(J)$, not $\exp(J)$
$\|S_k(J)\|\le1+\|J\|^k/k!$, so a certificate for an SOC network must carry this factor per layer or keep $\|J\|$ small (the authors bound it with spectral normalization). Cayley layers have no such error: the inverse is exact.

BCOP builds orthogonal convolutions as block convolutions of an orthogonal matrix with symmetric projectors, $W=H\,\Box\,[P_1\ \ I-P_1]\,\Box\cdots\Box\,[P_{K-1}\ \ I-P_{K-1}]$, $P_i=P_i^2=P_i^{\top}$ (1-D case; see the fact below). It also shows why such parameterizations are incomplete: 1-D orthogonal convolutions with kernel size $K$ and $n$ channels form $2(K-1)n+2$ connected components, while a continuous map from $\mathbb R^N$ reaches only one (Li et al., NeurIPS 2019): $\mathbb R^N$ is path-connected and continuous images of path-connected sets are path-connected, so a continuous parameterization cannot jump between components, however many parameters it has (see Primer 0; the simplest case is $O(1)=\{-1,1\}$, which no continuous map from $\mathbb R$ covers).

Fact — Projector filters are orthogonal convolutions (BCOP, Li et al. 2019)
Write $(M\,\Box\,N)_r=\sum_jM_jN_{r-j}$ for the convolution of two matrix-valued kernels $M,N$; it is the kernel of the composition of the two convolutions. Let $H\in\mathbb R^{n\times n}$ be orthogonal and $P_1,\dots,P_{K-1}$ orthogonal projectors ($P_i=P_i^{\top}=P_i^2$). Then the $K$-tap kernel $H\,\Box\,[P_1\ \ I-P_1]\,\Box\cdots\Box\,[P_{K-1}\ \ I-P_{K-1}]$ preserves the energy $\sum_t\|x_t\|^2$ of every two-sided square-summable (or periodic) signal.
Proof: the two-tap filter $[P\ \ I-P]$ has frequency response $\hat P(\omega)=P+e^{-\mathrm i\omega}(I-P)$, and since $P(I-P)=0$, $$\hat P(\omega)^{*}\hat P(\omega)=\big(P+e^{\mathrm i\omega}(I-P)\big)\big(P+e^{-\mathrm i\omega}(I-P)\big)=P^2+(I-P)^2=I .$$ Composition multiplies frequency responses, and products of unitary matrices (and the constant $H$) are unitary, so by Parseval the composed filter preserves energy. In words: each projector splits the channels into two orthogonal subspaces and delays one of them, which cannot change the total energy. With zero padding and cropping the layer is only nonexpansive.

AOL: almost-orthogonal layers

Theorem — AOL rescaling (Prach & Lampert 2022, Thm 1)
For $P\in\mathbb R^{n\times m}$ without zero columns let $D=\operatorname{diag}(d_i)$, $d_i=\big(\sum_j|P^{\top}P|_{ij}\big)^{-1/2}$. Then $\|PD\|_2\le1$, with $D^{\top}P^{\top}PD=I$ when $P$ has orthogonal columns.
Proof sketch — the AOL bound in one line

With $M=P^{\top}P$ and $a_i=d_iv_i$, using $|a_i||a_j|\le\tfrac12(a_i^2+a_j^2)$ and the symmetry of $|M|$: $$v^{\top}DP^{\top}PDv=\sum_{i,j}M_{ij}a_ia_j\le\sum_{i,j}|M_{ij}|\,\frac{a_i^2+a_j^2}{2}=\sum_i\Big(\sum_j|M_{ij}|\Big)d_i^2v_i^2=\|v\|^2 .$$ If the columns of $P$ are orthogonal, $M$ is diagonal and every step is an equality. $\blacksquare$

AOL needs no inverse, FFT or iteration; for convolutions the rescaling is channel-wise, so the trained layer is an ordinary convolution, and the learned $PD$ ends up almost orthogonal (Prach & Lampert, ECCV 2022). Section 3 shows it is one analytic solution of a single matrix inequality.

The cost of orthogonality

Expressivity: 1-Lipschitz layers compose to networks whose true constant is often far below the certified 1. Wang & Manchester's toy regression of a discontinuous function (their Table 1) measures tightness as empirical lower bound over certified bound:

Model$\gamma=1$$\gamma=5$$\gamma=10$
AOL77.2%45.2%47.9%
Orthogonal (Cayley)74.1%72.8%64.5%
SLL99.9%90.5%67.9%
Sandwich99.9%99.3%94.0%

Inference cost: Fourier-domain layers recompute their per-frequency Cayley inverses whenever the weights change, i.e. in training. With frozen weights and a fixed image size the transformed blocks can be computed once and cached (Wang & Manchester's code does this in evaluation mode), so an inference pass needs FFTs, per-frequency matrix products and inverse FFTs but no inverses, and no image-size kernel has to be materialized. LipKernel (Section 4) reports standard-form kernels two to three orders of magnitude faster in its benchmark, which matters inside a feedback loop.

3. SLL and Sandwich Layers: LMIs Solved by Construction

Section 2 bounded weights. The two constructions here take a LipSDP-type LMI, which also involves the activation through its multiplier, and solve it analytically: SLL layer by layer for residual blocks, the Sandwich parameterization for the whole network.

One inequality behind SN, orthogonal layers, AOL and CPL

Theorem — The unifying algebraic condition (Araujo, Havens, Delattre, Allauzen & Hu 2023, Thm 1)
Let $W\in\mathbb R^{m\times n}$ and let $T$ be a nonsingular diagonal matrix with $W^{\top}W\preceq T$. Then (1) $g(x)=WT^{-1/2}x+b$ is 1-Lipschitz, and (2) $h(x)=x-2WT^{-1}\varphi(W^{\top}x+b)$ is 1-Lipschitz if $\varphi$ is ReLU, tanh or sigmoid (slope-restricted in $[0,1]$), applied elementwise (Araujo et al., ICLR 2023). $W^{\top}W\preceq T$ forces $T\succ0$ and is linear in $T$: an SDP condition in a diagonal unknown.

Statement 1 is one line: $\|g(x)-g(y)\|^2=(x-y)^{\top}T^{-1/2}W^{\top}WT^{-1/2}(x-y)\le\|x-y\|^2$. Every known 1-Lipschitz layer is a choice of $T$: SN is $T=\|W\|^2I$ in Statement 1; orthogonal layers are $T=I$ with $W^{\top}W=I$; AOL is $T=\operatorname{diag}\big(\sum_j|W^{\top}W|_{ij}\big)$, for which $T-W^{\top}W$ is diagonally dominant with nonnegative diagonal, hence PSD; CPL (Meunier et al. 2022) is the SN choice in Statement 2; SLL is Statement 2 with $T_{ii}=\sum_j|W^{\top}W|_{ij}\,q_j/q_i$ for learnable $q_i\gt0$ (e.g. $q_i=e^{r_i}$ with free $r_i$). Here $|W^{\top}W|_{ij}$ is the absolute value of the $(i,j)$ entry. Since $T_{ii}\ge(W^{\top}W)_{ii}=\|w_i\|^2$ for the $i$-th column $w_i$ of $W$, this $T$ is nonsingular unless a column of $W$ vanishes; adding a fixed $\epsilon\gt0$ to every $T_{ii}$ removes that exception and keeps $T\succeq W^{\top}W$, because $\epsilon I\succeq0$.

Derivation — the residual layer is 1-Lipschitz, straight from the incremental quadratic constraint

Step 1 (QC). For $h(x)=Hx+G\varphi(W^{\top}x+b)$ and two inputs, let $\Delta\varphi=\varphi(W^{\top}x+b)-\varphi(W^{\top}y+b)$ and $\Delta v=W^{\top}\Delta x$. Slope restriction gives $\Delta\varphi_i(\Delta v_i-\Delta\varphi_i)\ge0$ per channel; a conic combination with a diagonal $\Lambda\succeq0$ gives $2\Delta\varphi^{\top}\Lambda(W^{\top}\Delta x-\Delta\varphi)\ge0$ (Module 2).

Step 2 (S-procedure). Subtracting this nonnegative term, $$\begin{aligned}\|\Delta x\|^2-\|\Delta h\|^2&\ge\|\Delta x\|^2-\|H\Delta x+G\Delta\varphi\|^2-2\Delta\varphi^{\top}\Lambda(W^{\top}\Delta x-\Delta\varphi)\\&=\begin{bmatrix}\Delta x\\\Delta\varphi\end{bmatrix}^{\top}\begin{bmatrix}I-H^{\top}H&-H^{\top}G-W\Lambda\\-G^{\top}H-\Lambda W^{\top}&2\Lambda-G^{\top}G\end{bmatrix}\begin{bmatrix}\Delta x\\\Delta\varphi\end{bmatrix},\end{aligned}$$ so PSD of this matrix makes $h$ 1-Lipschitz (Theorem 4 of Araujo et al.; with $H=0$ it is one-layer LipSDP).

Step 3 (solve it). Choose $H=I$, $G=-2WT^{-1}$, $\Lambda=2T^{-1}$: the $(1,1)$ block is $0$, the off-diagonal block is $2WT^{-1}-2WT^{-1}=0$, and the $(2,2)$ block is $4T^{-1}(T-W^{\top}W)T^{-1}$, PSD iff $W^{\top}W\preceq T$ (congruence with $T^{-1}$). $\blacksquare$ The residual structure "uses up" the input block, so the certificate collapses to one inequality of the size of the layer width, which is why SLL scales.

Theorem — SLL's analytic multiplier via Gershgorin (Araujo et al. 2023, Thm 3)
Let $T$ be nonsingular diagonal. If for some diagonal $Q\succ0$ the matrix $T-QW^{\top}WQ^{-1}$ is diagonally dominant with positive diagonal, then $T\succeq W^{\top}W$, and $h(x)=x-2WT^{-1}\varphi(W^{\top}x+b)$ is 1-Lipschitz for ReLU, tanh or sigmoid. With $Q^{-1}=\operatorname{diag}(q_i)$ the choice $T_{ii}=\sum_j|W^{\top}W|_{ij}\,q_j/q_i$ makes it diagonally dominant with nonnegative diagonal (a diagonal entry is $0$ if column $i$ of $W$ is orthogonal to all others), which is all Gershgorin needs for the conclusion (Exercise 13.2(b)). To use $T^{-1}$, require nonzero columns of $W$, or add $\epsilon I$ with $\epsilon\gt0$ to this choice of $T$.

Why: $T-QW^{\top}WQ^{-1}=Q(T-W^{\top}W)Q^{-1}$ is similar to the symmetric $T-W^{\top}W$ (a similarity $QMQ^{-1}$ keeps the eigenvalues themselves; a congruence $Q^{\top}MQ$ would keep only their signs), so it has the same real eigenvalues. Gershgorin's theorem puts every eigenvalue of a square matrix $M$ in one of the discs $|z-M_{ii}|\le\sum_{j\ne i}|M_{ij}|$; here the discs are centred at nonnegative diagonal entries with radii no larger than those entries, so the eigenvalues are $\ge0$. The $q_i$ are trained with the weights; $q_i\equiv1$ recovers AOL's choice of $T$, here used in the residual layer of Statement 2.

Caveat — only diagonal multipliers
Step 1 needs $\Lambda$ diagonal. The NeurIPS 2019 version of LipSDP also allowed coupled "repeated nonlinearity" multipliers; Pauli et al. (L-CSS 2022) gave counterexamples, the current arXiv version of LipSDP keeps only diagonal $T$, and SLL restricts its Theorem 4 accordingly. Wang & Manchester's one-sentence reason: the nonlinearities $\varphi(v_i)$ may be repeated, but their increments $\varphi(v_i^a+\Delta v_i)-\varphi(v_i^a)$ are not, because they depend on the different operating points $v_i^a$ (Module 12).

The Sandwich layer: a complete parameterization of LipSDP

Wang & Manchester consider $z_0=x$, $z_{k+1}=\varphi(W_kz_k+b_k)$ ($k=0,\dots,L-1$), $y=W_Lz_L+b_L$, with $W_k\in\mathbb R^{n_{k+1}\times n_k}$ and $\varphi$ piecewise differentiable and slope-restricted in $[0,1]$ (their Assumption 2.1), and write LipSDP with the bound $\gamma$ itself.

Theorem — LipSDP in block-tridiagonal $\gamma$-form (Wang & Manchester 2023, Thm A.3 and eq. (5))
The network is $\gamma$-Lipschitz if there are positive diagonal $\Lambda_k\in\mathbb D^{n_{k+1}}_{++}$, $k=0,\dots,L-1$, with $$H=\begin{bmatrix}\gamma I&-W_0^{\top}\Lambda_0&&&\\-\Lambda_0W_0&2\Lambda_0&-W_1^{\top}\Lambda_1&&\\&\ddots&\ddots&\ddots&\\&&-\Lambda_{L-1}W_{L-1}&2\Lambda_{L-1}&-W_L^{\top}\\&&&-W_L&\gamma I\end{bmatrix}\succeq0 .$$ For one hidden layer this is Module 12's $M(\rho,T)\preceq0$ with $\alpha=0$, $\beta=1$, $\rho=\gamma^2$, $T=\gamma\Lambda_0$ (Exercise 13.3).

A block-bidiagonal factor $H=PP^{\top}$ automatically has the tridiagonal pattern; the difficulty is to make the middle diagonal blocks diagonal. Wang & Manchester build the blocks of $P$ as $\Psi_kA_k$ and $\Psi_kB_k$ with $[A_k^{\top};B_k^{\top}]$ from a Cayley transform, so that $A_kA_k^{\top}+B_kB_k^{\top}=I$ collapses each diagonal block to a diagonal matrix. Reading off the weights gives:

Derivation — why a block-bidiagonal factor fits the LMI (one hidden layer)

Multiplying blocks (see Primer A), a lower block-bidiagonal $P$ gives a block-tridiagonal $PP^{\top}$, the pattern of $H$: $$P=\begin{bmatrix}D_0&0&0\\E_1&D_1&0\\0&E_2&D_2\end{bmatrix},\qquad PP^{\top}=\begin{bmatrix}D_0D_0^{\top}&D_0E_1^{\top}&0\\E_1D_0^{\top}&E_1E_1^{\top}+D_1D_1^{\top}&D_1E_2^{\top}\\0&E_2D_1^{\top}&E_2E_2^{\top}+D_2D_2^{\top}\end{bmatrix}.$$ Take $D_0=\sqrt\gamma\,I$, $E_1=-\sqrt2\,\Psi_0B_0$, $D_1=\sqrt2\,\Psi_0A_0$. Then the first block is $\gamma I$, the block below it is $-\sqrt{2\gamma}\,\Psi_0B_0=-\Lambda_0W_0$ for the $W_0=\sqrt{2\gamma}\,\Psi_0^{-1}B_0$ of the theorem below (with $\Lambda_0=\Psi_0^2$), and the middle block is $2\Psi_0\big(B_0B_0^{\top}+A_0A_0^{\top}\big)\Psi_0=2\Psi_0^2=2\Lambda_0$: diagonal, exactly because the Cayley blocks satisfy $A_0A_0^{\top}+B_0B_0^{\top}=I$. The output layer's Cayley blocks fill the last row: $E_2=-\sqrt\gamma\,B_1$ and $D_2=\sqrt\gamma\,A_1$ give $E_2D_1^{\top}=-W_1$ and $E_2E_2^{\top}+D_2D_2^{\top}=\gamma I$. So $H=PP^{\top}\succeq0$ for every value of the free parameters (checked numerically for random draws).

Theorem — Direct parameterization of all LipSDP-certified networks (Wang & Manchester 2023, Thm 3.1)
Free parameters: $d_k\in\mathbb R^{n_{k+1}}$ ($0\le k\lt L$), $X_k\in\mathbb R^{n_{k+1}\times n_{k+1}}$, $Y_k\in\mathbb R^{n_k\times n_{k+1}}$, $b_k$ ($0\le k\le L$). Set $\Psi_k=\operatorname{diag}(e^{d_k})$, $[A_k^{\top};B_k^{\top}]=\operatorname{Cayley}(X_k,Y_k)$ and $$W_k=2\,\Psi_k^{-1}B_kA_{k-1}^{\top}\Psi_{k-1},\qquad k=0,\dots,L,$$ with $A_{-1}=I$, $\Psi_{-1}=\sqrt{\gamma/2}\,I$, $\Psi_L=\sqrt{2/\gamma}\,I$. Then the LMI above holds with $\Lambda_k=\Psi_k^2$ (see the caveat below), hence $\operatorname{Lip}\le\gamma$, for every value of the free parameters; conversely, every network satisfying the LMI can be written this way.

The converse is constructive ($\Psi_k=\Lambda_k^{1/2}$, $B_k=\tfrac12\Psi_kW_k\Psi_{k-1}^{-1}A_{k-1}^{-\top}$, $A_k$ from a Cholesky factor of $I-B_kB_k^{\top}$ times an orthogonal matrix chosen to avoid eigenvalue $-1$, then the inverse Cayley map). This recursion is the nonsingular case: on the boundary of the LMI a block $A_{k-1}$ can be singular (for one hidden layer, $\gamma=\Lambda_0=1$, $W_0=\sqrt2$, $W_1=0$ gives $B_0=1$, $A_0=0$), and the paper's Appendix D.3 then replaces $A_{k-1}^{-\top}$ by the pseudoinverse $(A_{k-1}^{\top})^{+}$, or by $I$ when $A_{k-1}=0$, which forces $W_k=0$; its sufficiency proof uses the matching generalized Schur complements. (For a singular pivot block $C$, $\begin{bmatrix}C&B^{\top}\\B&E\end{bmatrix}\succeq0$ iff $C\succeq0$, $(I-CC^{+})B^{\top}=0$ and $E-BC^{+}B^{\top}\succeq0$; the middle range condition, which says that the coupling only acts in directions where $C$ has energy, is what a pseudoinverse alone would miss; see Primer A.) So completeness is relative to LipSDP: every network LipSDP can certify is reached, not every $\gamma$-Lipschitz network. With $h_k=\sqrt2A_{k-1}^{\top}\Psi_{k-1}z_k$ the network becomes a composition of identical modules with decoupled parameters:

$$h_0=\sqrt\gamma\,x,\qquad h_{k+1}=\sqrt2\,A_k^{\top}\Psi_k\,\varphi\big(\sqrt2\,\Psi_k^{-1}B_kh_k+b_k\big),\qquad y=\sqrt\gamma\,B_Lh_L+b_L .$$
Theorem — The sandwich layer is 1-Lipschitz (Wang & Manchester 2023, Thm 3.2 and Prop. 3.3)
For $\Psi=\operatorname{diag}(e^d)$ and $[A^{\top};B^{\top}]=\operatorname{Cayley}(X,Y)$, the layer $h_{\rm out}=\sqrt2A^{\top}\Psi\varphi(\sqrt2\Psi^{-1}Bh_{\rm in}+b)$ is 1-Lipschitz for every $\varphi$ slope-restricted in $[0,1]$. With $\varphi$ the identity it is the linear layer $h_{\rm out}=2A^{\top}Bh_{\rm in}+\hat b$, and a linear layer is 1-Lipschitz iff its weight can be written $W=2A^{\top}B$ this way.

The proof is one identity (Exercise 13.3): with $U=\sqrt2\Psi^{-1}B$, $Y=\sqrt2A^{\top}\Psi$ and $\Lambda=\Psi^2$, $2\Lambda-Y^{\top}Y-\Lambda UU^{\top}\Lambda=2\Psi(I-AA^{\top}-BB^{\top})\Psi=0$, so the certificate sits exactly on the boundary of the PSD cone. $\Psi$ reshapes each activation channel ($\Psi\varphi(\Psi^{-1}v+b)$ still has slopes in $[0,1]$), while $B$ and $A^{\top}$ are the input and output halves of one matrix with orthonormal columns, "sandwiching" the nonlinearity.

Caveat — a factor of 2 in the paper's Section 3
Section 3 of the paper states the multiplier as $\Lambda_k=\tfrac12\Psi_k^2$; that value belongs to the factorization $PP^{\top}$ of a rescaled matrix. For the LMI exactly as written in its eq. (5)/(19), the certificate is $\Lambda_k=\Psi_k^2$, which is what its appendix proofs of Theorems 3.1 and 3.2 use. We checked on random draws (networks with one to four hidden layers): $\Lambda=\Psi^2$ always gives minimum eigenvalue $\ge0$ up to round-off (exactly $0$ for a single sandwich layer), while $\Lambda=\tfrac12\Psi^2$ gives clearly negative eigenvalues in about half of the draws (Exercise 13.3(b) shows why). The weights, the layer and both theorems are unaffected.
How to read the Sandwich formula

Compute it from the inside out: $B$ changes input coordinates, $\Psi^{-1}$ rescales the activation inputs, $\varphi$ acts coordinate by coordinate, $\Psi$ restores the scale, and $A^\top$ forms the output. The two factors $\sqrt2$ belong to the certificate identity and must be kept. For a scalar channel with zero bias and ReLU, positive $\Psi$ scaling commutes with ReLU, so the expression reduces to $2A\operatorname{ReLU}(Bx)$. If also $B\ge0$, this is $2AB\operatorname{ReLU}(x)$; the Medium practice set computes such an example.

For general activations or nonzero biases, do not cancel the diagonal scalings through the activation as if it were linear. The proof uses slope restrictions and the shared identity $AA^\top+BB^\top=I$, so $A$ and $B$ must come from the same construction.

Network-level coupling. $W_k$ depends on the parameters of layers $k$ and $k-1$ ("interlacing"). The paper's Proposition C.1 shows the network satisfies weighted layer-wise norm bounds whose product is at most $\gamma$, while the plain norms $\|W_k\|$ and their product can exceed 1 even for $\gamma=1$ (Remark C.2); on random draws we found products far above 1, and the explorer lets you find your own.

Algorithm 1: Training a $\gamma$-Lipschitz network with sandwich layers
  1. Initialize free parameters $\theta=\{X_k,Y_k,d_k,b_k\}$ arbitrarily (no feasibility needed).
  2. For each minibatch:
  3. per layer: $Z_k=X_k-X_k^{\top}+Y_k^{\top}Y_k$; $A_k^{\top}=(I+Z_k)^{-1}(I-Z_k)$, $B_k^{\top}=-2Y_k(I+Z_k)^{-1}$; $\Psi_k=\operatorname{diag}(e^{d_k})$ // one linear solve per layer
  4. forward pass $h_0=\sqrt\gamma x$, sandwich layers, $y=\sqrt\gamma B_Lh_L+b_L$; loss $\mathcal L$
  5. backpropagate through the solves and take an SGD/Adam step on $\theta$ // no projection, no SDP
  6. Deploy: for dense layers compute the $W_k$ once; the network is $\gamma$-Lipschitz by Theorem 3.1.

"Backpropagate through the solves" needs only the derivative of the defining equation: if a layer computes $v$ from $Mv=r$, then $(dM)\,v+M\,dv=dr$, so $dv=M^{-1}\big(dr-(dM)\,v\big)$, one more solve with the same matrix (see Primer B). Standard automatic differentiation uses this rule for linear solves rather than differentiating the elimination steps.

Convolutions and results. Circular, unstrided convolutions are evaluated in the Fourier domain ($s(\lfloor s/2\rfloor+1)$ parallel inverses of $q\times q$ complex matrices for $q$ channels and $s\times s$ images). SLL was slightly better on CIFAR-10 with a much larger model (41M versus 3M parameters), Sandwich about 4% better on CIFAR-100 (48M versus 118M) and, with last-layer normalization, 33.4% clean / 24.7% certified (see Primer E) at $\varepsilon=36/255$ on Tiny-ImageNet with 39M parameters versus 32.1% / 23.0% for the 1.1B-parameter SLL X-Large as quoted there (Wang & Manchester, ICML 2023); the SLL paper's own table lists 23.2% certified.

Pitfall — what the certified $\gamma$ is a bound for
The certificate covers the parameterized part only. An input normalization $x\mapsto\operatorname{diag}(1/s_i)(x-\mu)$ (per-channel standard deviations $s_i\gt0$) is $\max_i(1/s_i)$-Lipschitz, so it multiplies the bound by $\max_i1/s_i$ (Wang & Manchester: about $4.1\gamma$ in pixel space), and last-layer normalization can push the composite constant above the nominal bound, so certified accuracy must be computed with the normalized last layer. Always state the input space in which a radius is measured. (Clean accuracy counts correctly classified unperturbed test inputs; certified accuracy counts those that are also certified at the stated radius; see Primer E.)

4. Lipschitz-Bounded CNNs: Cayley–Gramian and LipKernel (Pauli et al.)

Convolutions are where LipSDP-style certificates get expensive: as dense layers they are Toeplitz matrices growing with the signal length, in the Fourier domain they need circular padding and an inverse per frequency. Pauli et al. take a systems-theory route: a convolution is a finite impulse response (FIR) filter, a linear state-space system whose state is a window of past inputs. Its certificate is a dissipation inequality with quadratic storage, and the LMI size depends on kernel size and channels only, not on the signal length (Pauli, Gramlich & Allgöwer, L4DC 2023; Module 12).

Convolutional layers as FIR systems

A 1-D convolutional layer with $c_{i-1}$ input channels, $c_i$ output channels and kernel size $\ell_i$ computes $w^i_k=\varphi\big(b_i+\sum_{j=0}^{\ell_i-1}K^i_j\,w^{i-1}_{k-j}\big)$, $K^i_j\in\mathbb R^{c_i\times c_{i-1}}$. With the state $x^i_k\in\mathbb R^{(\ell_i-1)c_{i-1}}$ holding the last $\ell_i-1$ inputs (Pauli et al. CDC 2023, eqs. (5)–(6)):

$$x^i_{k+1}=A_ix^i_k+B_iw^{i-1}_k,\quad y^i_k=C_ix^i_k+D_iw^{i-1}_k+b_i,\quad w^i_k=\varphi(y^i_k),$$ $$A_i=\begin{bmatrix}0&I&&\\&\ddots&\ddots&\\&&0&I\\&&&0\end{bmatrix},\ B_i=\begin{bmatrix}0\\\vdots\\0\\I\end{bmatrix},\ \hat C_i:=\begin{bmatrix}C_i&D_i\end{bmatrix}=\begin{bmatrix}K^i_{\ell_i-1}&\cdots&K^i_1&K^i_0\end{bmatrix}.$$

$A_i$ is a block shift, hence nilpotent ($A_i^{\ell_i-1}=0$), and all learnable kernel entries sit in $\hat C_i$.

Theorem — Layer-wise LMIs certify the whole 1-D CNN (Pauli et al. CDC 2023, Thm 2, from L4DC 2023)
Let all activations be slope-restricted in $[0,1]$ and $\rho\gt0$. Suppose every convolutional layer $i$ has $Q_i\in\mathbb S^{c_i}$ (diagonal if the layer contains max pooling), $P_i\succeq0$ and a positive diagonal $\Lambda_i$ with $$\begin{bmatrix}P_i-A_i^{\top}P_iA_i&-A_i^{\top}P_iB_i&-C_i^{\top}\Lambda_i\\-B_i^{\top}P_iA_i&Q_{i-1}-B_i^{\top}P_iB_i&-D_i^{\top}\Lambda_i\\-\Lambda_iC_i&-\Lambda_iD_i&2\Lambda_i-Q_i\end{bmatrix}\succeq0,\qquad Q_0=\tilde\rho^{\,2}I,$$ and every fully connected layer satisfies $\begin{bmatrix}Q_{i-1}&-W_i^{\top}\Lambda_i\\-\Lambda_iW_i&2\Lambda_i-Q_i\end{bmatrix}\succeq0$, the last (no activation) $\begin{bmatrix}Q_{l-1}&-W_l^{\top}\\-W_l&I\end{bmatrix}\succeq0$, with $Q_{i-1}$ replaced by $I_N\otimes Q_{i-1}$ at the flattening step. Then the CNN is $\rho$-Lipschitz with $\rho=\tilde\rho\prod_s\mu_s$, where $\mu_s$ are the Lipschitz constants of the average-pooling layers ($\mu=1/\sqrt\ell$ for non-overlapping windows of length $\ell$, L4DC 2023, Prop. 7). Both papers define max and average pooling with non-overlapping windows (stride equal to window length), and the diagonal $Q_i$ covers only that case: overlapping max pooling is not even 1-Lipschitz, since $(a,b,c)\mapsto(\max(a,b),\max(b,c))$ has gain $\sqrt2$ when $b$ moves near $(0,1,0)$. LipKernel's pooling LMI below admits overlapping windows through the pooling constant $\rho_p$.

Here $\mathbb S^{c}$ is the set of real symmetric $c\times c$ matrices, and $\|v\|^2_{Q}:=v^{\top}Qv$; this is a squared norm, and $\{v:\|v\|_Q\le1\}$ an ellipsoid, only when $Q\succ0$, which the construction below guarantees through $Q_i=L_i^{\top}L_i$. At the flattening step, $N$ is the number of spatial positions: stacking the channel vectors $u_1,\dots,u_N$ of the positions one after another, $I_N\otimes Q=\operatorname{blkdiag}(Q,\dots,Q)$ and $\operatorname{vec}(u)^{\top}(I_N\otimes Q)\operatorname{vec}(u)=\sum_{k=1}^Nu_k^{\top}Qu_k$ (see Primer A; another flattening order permutes this matrix).

Derivation — the conv LMI is an incremental dissipation inequality, and the layers telescope

Step 1. Run the layer on two inputs from zero initial state and take increments: $\Delta x_{k+1}=A\Delta x_k+B\Delta u_k$, $\Delta v_k=C\Delta x_k+D\Delta u_k$, $\Delta z_k=\varphi(v^a_k)-\varphi(v^b_k)$ (layer index dropped, $u=w^{i-1}$).

Step 2. Evaluate the LMI's quadratic form at $\xi_k=[\Delta x_k;\Delta u_k;\Delta z_k]$ and regroup: $$0\le\underbrace{\Delta x_k^{\top}P\Delta x_k-\Delta x_{k+1}^{\top}P\Delta x_{k+1}}_{V(\Delta x_k)-V(\Delta x_{k+1})}+\|\Delta u_k\|^2_{Q_{i-1}}-\|\Delta z_k\|^2_{Q_i}-2\Delta z_k^{\top}\Lambda(\Delta v_k-\Delta z_k).$$

Step 3. The last term is $\le0$ by the slope restriction, so $V(\Delta x_{k+1})-V(\Delta x_k)\le\|\Delta u_k\|^2_{Q_{i-1}}-\|\Delta z_k\|^2_{Q_i}$: incremental dissipativity with storage $V(\Delta x)=\Delta x^{\top}P\Delta x$ (Willems 1972; Module 2).

Step 4. Sum over $k$ with $\Delta x_0=0$, $V\ge0$: $\sum_k\|\Delta z_k\|^2_{Q_i}\le\sum_k\|\Delta u_k\|^2_{Q_{i-1}}$. The output weight $Q_i$ of layer $i$ is the input weight of layer $i+1$, so the chain telescopes from $Q_0=\tilde\rho^2I$ to the identity weight of the last layer. $\blacksquare$ $Q_i$ is a directional gain: the next layer only has to tolerate an ellipsoid of increments, not a ball, which is where network-level tightness comes from.

Fully connected layers use $W_i=\sqrt2\,\Gamma_i^{-1}V_i^{\top}L_{i-1}$, $L_i=\sqrt2\,U_i\Gamma_i$, $Q_i=L_i^{\top}L_i$, $\Lambda_i=\Gamma_i^{\top}\Gamma_i$, with $\Gamma_i=\operatorname{diag}(\gamma_i)$ and $[U_i;V_i]=\operatorname{Cayley}(Y_i,Z_i)$ (Thm 4): $U_i^{\top}U_i+V_i^{\top}V_i=I$, multiplied by $\sqrt2\,\Gamma_i$ on both sides, becomes $Q_i+\Lambda_iW_iQ_{i-1}^{-1}W_i^{\top}\Lambda_i=2\Lambda_i$, and a Schur complement gives the FC LMI. This is equivalent to the Sandwich parameterization (their Remark 6); what is new is the gain factor $L_i$, passed on to the next layer. Convolutions need one more idea because of the dynamics in the upper-left block.

Step 1: the storage matrix from a controllability Gramian

With $F_i:=\begin{bmatrix}P_i-A_i^{\top}P_iA_i&-A_i^{\top}P_iB_i\\-B_i^{\top}P_iA_i&Q_{i-1}-B_i^{\top}P_iB_i\end{bmatrix}$ the conv LMI reads $\begin{bmatrix}F_i&-\hat C_i^{\top}\Lambda_i\\-\Lambda_i\hat C_i&2\Lambda_i-Q_i\end{bmatrix}\succeq0$, the FC shape with $F_i$ in place of $Q_{i-1}$. We need $F_i\succ0$ for every parameter value.

Lemma — Gramian parameterization of the storage (Pauli, Wang, Manchester & Allgöwer, CDC 2023, Lemma 7)
Let $Q_{i-1}\succ0$. For some $\varepsilon\gt0$ (in fact any) and all $H_i\in\mathbb R^{n_{x_i}\times n_{x_i}}$, $P_i=X_i^{-1}$ with $$X_i=\sum_{k\ge0}A_i^k\big(B_iQ_{i-1}^{-1}B_i^{\top}+H_i^{\top}H_i+\varepsilon I\big)(A_i^{\top})^k$$ renders $F_i\succ0$. The sum is finite because $A_i$ is nilpotent (the paper writes the upper limit as $n_{x_i}-c_{i-1}$; all powers from $\ell_i-1$ on vanish), and $X_i$ is the unique solution of $X_i-A_iX_iA_i^{\top}-B_iQ_{i-1}^{-1}B_i^{\top}=H_i^{\top}H_i+\varepsilon I\succ0$: the controllability Gramian of $\big(A_i,\ [\,B_iQ_{i-1}^{-1/2}\ \ H_i^{\top}\ \ \sqrt\varepsilon\,I\,]\big)$.
Proof sketch — two Schur complements turn the Lyapunov equation into $F_i\succ0$

Step 1. $X_i\succ0$ (the $k=0$ term contains $\varepsilon I$). For the Lyapunov identity write $R_i=B_iQ_{i-1}^{-1}B_i^{\top}+H_i^{\top}H_i+\varepsilon I$ and let $A_i^r=0$. Then $X_i=\sum_{k=0}^{r-1}A_i^kR_i(A_i^{\top})^k$ and $A_iX_iA_i^{\top}=\sum_{k=1}^{r}A_i^kR_i(A_i^{\top})^k$; subtracting, all terms cancel except $R_i$ (the $k=r$ term is zero), so $X_i-A_iX_iA_i^{\top}=R_i$. It is the only solution: the difference $D$ of two solutions satisfies $D=A_iDA_i^{\top}=\dots=A_i^rD(A_i^{\top})^r=0$. Hence $X_i-A_iX_iA_i^{\top}-B_iQ_{i-1}^{-1}B_i^{\top}=H_i^{\top}H_i+\varepsilon I\succ0$.

Step 2. By the Schur complement with respect to $\operatorname{diag}(X_i^{-1},Q_{i-1})\succ0$ (recall $Q_{i-1}=L_{i-1}^{\top}L_{i-1}\succ0$), this is equivalent to $\begin{bmatrix}X_i^{-1}&0&A_i^{\top}\\0&Q_{i-1}&B_i^{\top}\\A_i&B_i&X_i\end{bmatrix}\succ0$.

Step 3. Now take the Schur complement with respect to $X_i\succ0$ and put $P_i=X_i^{-1}$: $\ \operatorname{diag}(P_i,Q_{i-1})-[A_i\ \ B_i]^{\top}P_i[A_i\ \ B_i]=F_i\succ0$. $\blacksquare$

Why the Gramian. Solving a Lyapunov equation in every forward pass would be expensive and awkward to differentiate; for a nilpotent shift the solution is a finite sum of products, and the free $H_i$ lets training move the storage around.

Step 2: the kernel from a Cayley transform

Theorem — Cayley–Gramian convolutional layers (Pauli et al. CDC 2023, Thm 8)
For a convolutional layer with average pooling or no pooling set $$\hat C_i=\sqrt2\,\Gamma_i^{-1}V_i^{\top}L^F_i,\qquad \Gamma_i=\operatorname{diag}(\gamma_i),\quad \begin{bmatrix}U_i\\V_i\end{bmatrix}=\operatorname{Cayley}\begin{bmatrix}Y_i\\Z_i\end{bmatrix},\quad L^F_i=\operatorname{chol}(F_i),$$ with $F_i=L_i^{F\top}L^F_i$ built from $P_i$ of the Lemma, $Q_i=L_i^{\top}L_i$, $L_0=\rho I$, $L_i=\sqrt2\,U_i\Gamma_i$. Assume, as the paper does throughout (its Remark 5), that every $\Gamma_i$ and $L_i$ is nonsingular, so that $Q_{i-1}\succ0$ and the Gramian of the Lemma is defined; positive exponential diagonal entries ensure nonsingular $\Gamma_i$ but not nonsingular $L_i$ (scalar case: $Y_i=0$, $Z_i=1$ give $M_i=Y_i-Y_i^{\top}+Z_i^{\top}Z_i=1$, $U_i=(1-M_i)/(1+M_i)=0$ and $L_i=0$). The exceptional free variables form a set of measure zero, but they exist, so the guarantee is for all admissible free variables, those with every $L_i$ invertible: the conv LMI then holds for all admissible free variables $Y_i\in\mathbb R^{c_i\times c_i}$, $Z_i\in\mathbb R^{\ell_ic_{i-1}\times c_i}$, $H_i$, $\gamma_i$, $b_i$, and the kernel is read off $\hat C_i$. With average pooling $\rho$ is first divided by the pooling constants; max pooling needs diagonal $Q_i$ and a modified formula (Corollary 9).

Proof in one identity: with the multiplier $\Lambda_i=\Gamma_i^2$, the two terms of the Schur complement are $Q_i=L_i^{\top}L_i=2\Gamma_iU_i^{\top}U_i\Gamma_i$ and, because $L^F_iF_i^{-1}L_i^{F\top}=I$ for $F_i=L_i^{F\top}L^F_i$, $\Lambda_i\hat C_iF_i^{-1}\hat C_i^{\top}\Lambda_i=2\Gamma_iV_i^{\top}V_i\Gamma_i$. Multiplying $U_i^{\top}U_i+V_i^{\top}V_i=I$ by $\sqrt2\,\Gamma_i$ on both sides therefore gives $Q_i+\Lambda_i\hat C_iF_i^{-1}\hat C_i^{\top}\Lambda_i=2\Gamma_i^2=2\Lambda_i$, so $2\Lambda_i-Q_i-\Lambda_i\hat C_iF_i^{-1}\hat C_i^{\top}\Lambda_i=0$ and a Schur complement with respect to $F_i\succ0$ gives the LMI. Compare the sandwich layer: $\Psi$ became $\Gamma$, and the Cholesky factor of the Gramian-based $F_i$ plays the role of the input-side scaling.

Algorithm 2: One Cayley–Gramian 1-D convolutional layer (Pauli et al., CDC 2023)
  1. Input: gain factor $L_{i-1}$ of the previous layer ($L_0=\rho I$); free variables $Y_i,Z_i,H_i,\gamma_i,b_i$.
  2. $Q_{i-1}=L_{i-1}^{\top}L_{i-1}$; $\ X_i=\sum_{k=0}^{\ell_i-2}A_i^k(B_iQ_{i-1}^{-1}B_i^{\top}+H_i^{\top}H_i+\varepsilon I)(A_i^{\top})^k$; $\ P_i=X_i^{-1}$ // finite Gramian sum
  3. form $F_i$ and its Cholesky factor $L^F_i$; $\ [U_i;V_i]=\operatorname{Cayley}(Y_i,Z_i)$; $\ \hat C_i=\sqrt2\,\Gamma_i^{-1}V_i^{\top}L^F_i=[K^i_{\ell_i-1}\cdots K^i_1\ \ K^i_0]$
  4. $L_i=\sqrt2\,U_i\Gamma_i$ // directional gain handed to layer $i+1$
  5. apply an ordinary convolution with kernel $K^i$, bias $b_i$, activation and pooling

All sizes are set by $c_{i-1},c_i,\ell_i$; the signal length never enters. On the MIT-BIH arrhythmia ECG data (5 classes, mean of 5 CNNs) the Lipschitz-bounded CNNs reached 84.8%, 90.0% and 94.7% test accuracy for $\rho=5,10,50$ (certified bounds 4.99, 9.85, 45.3), against 94.9% for a vanilla CNN with bound 147 and 92.4% / 90.6% for L2-regularized CNNs (training loss plus a weight penalty $\lambda\sum_k\|W_k\|_F^2$) with bounds 45.7 / 33.9; they degraded more slowly under $\ell_2$ PGD attacks (projected gradient ascent on the loss over an $\ell_2$ ball around the input: a successful attack disproves robustness, a failed one certifies nothing; see Primer E) and could be trained at bounds where L2-regularized training failed (Pauli, Wang, Manchester & Allgöwer, CDC 2023).

LipKernel: 2-D convolutions parameterized in standard form

LipKernel extends the construction to 2-D and 1-D CNNs with pooling, strides, dilation and zero padding, with the goal $\|\mathrm{NN}_\theta(u_a)-\mathrm{NN}_\theta(u_b)\|_Q\le\|u_a-u_b\|_R$ ($(Q,R)$-Lipschitz; $Q=I$, $R=\rho^2I$ is $\rho$-Lipschitz, other choices weight input directions and output classes) (Pauli, Wang, Manchester & Allgöwer, Automatica 2026).

Theorem — Incrementally dissipative layers compose (LipKernel, Def. 4 and Thm 7)
A layer $u_k\mapsto y_k$ is incrementally dissipative with respect to the supply $\sum_{i}\big(\|\Delta u_k[i]\|^2_{X_{k-1}}-\|\Delta y_k[i]\|^2_{X_k}\big)\ge0$ if this holds for all input pairs and horizons. If every layer $k=1,\dots,l$ is, with $X_0=R$ and $X_l=Q$, the network is $(Q,R)$-Lipschitz; $X_0=\rho^2I$, $X_l=I$ gives $\rho$-Lipschitz. Proof: since $u_{k+1}=y_k$ the sum of the layer inequalities telescopes (stated as a theorem in Module 2).

Each layer type has its own LMI: $\varphi\circ$conv (the $3\times3$ block LMI above with $X_{k-1},X_k$ for $Q_{i-1},Q_i$), pooling$\circ\varphi\circ$conv (the $(3,3)$ block becomes $2\Lambda-\rho_p^2X$ with pooling constant $\rho_p$; $X$ diagonal for max pooling), $\varphi\circ$FC, and the last layer $X_{l-1}-W_l^{\top}QW_l\succeq0$. A 2-D convolution is realized as a Roesser model with a horizontal and a vertical state, $$\begin{bmatrix}x_1[i+1,j]\\x_2[i,j+1]\end{bmatrix}=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix}\begin{bmatrix}x_1[i,j]\\x_2[i,j]\end{bmatrix}+\begin{bmatrix}B_1\\B_2\end{bmatrix}u[i,j],$$ $$y[i,j]=C_1x_1[i,j]+C_2x_2[i,j]+Du[i,j]+b,$$ where $A_{21}=0$, $A_{11},C_1,A_{22},B_2$ are fixed shift structures and the kernel blocks fill $A_{12},B_1,C_2,D$ (Pauli, Gramlich & Allgöwer, IFAC-PapersOnLine 2024; the 2-D Lur'e view of CNNs is Gramlich et al., Automatica 2026), with block-diagonal storage $P=\operatorname{blkdiag}(P_1,P_2)$. 2-D Lyapunov equations have no general solution, so LipKernel constructs one for this FIR structure: since $A_{21}=0$ the $x_2$ dynamics are decoupled, $P_2^{-1}$ is a nilpotent Gramian-type sum, and $P_1^{-1}$ a second sum absorbing the coupling through the now free $A_{12},B_1$ (Lemma 16). The remaining blocks $[C_2\ D]$ come from a Cayley transform (Theorem 18), where the needed $2\Gamma-C_1F_1^{-1}C_1^{\top}\succ0$ is secured by $\gamma_i=\epsilon+\delta_i^2+\tfrac12\sum_j|C_1F_1^{-1}C_1^{\top}|_{ij}\,q_j/q_i$: SLL's Gershgorin trick. After training, the bijection $(A,B,C,D)\leftrightarrow K$ returns an ordinary kernel.

Theorem — LipKernel's 2-D layer parameterization (Pauli et al. 2026, Lemma 16 and Thm 18)

Take the Roesser realization above (zero initial states) with $n_1$ horizontal and $n_2$ vertical state coordinates, $m$ input and $c$ output channels, and an elementwise activation slope-restricted in $[0,1]$. The shift blocks $A_{11}\in\mathbb R^{n_1\times n_1}$ and $A_{22}\in\mathbb R^{n_2\times n_2}$ (both nilpotent), $B_2$ and $C_1\in\mathbb R^{c\times n_1}$ are fixed by the realization; $A_{12}$ and $B_1$ are free kernel parameters, and $X_-\succ0$ is the gain matrix handed over by the previous layer. Put

$$A=\begin{bmatrix}A_{11}&A_{12}\\0&A_{22}\end{bmatrix},\quad B=\begin{bmatrix}B_1\\B_2\end{bmatrix},\quad \widetilde X=BX_-^{-1}B^\top=\begin{bmatrix}\widetilde X_{11}&\widetilde X_{12}\\\widetilde X_{12}^\top&\widetilde X_{22}\end{bmatrix}.$$

For arbitrary square $H_i$ of sizes $n_i$ and $\epsilon>0$, compute the finite sums in this order:

$$\begin{aligned}T_2&=\sum_{k\ge0}A_{22}^k(\widetilde X_{22}+H_2^\top H_2+\epsilon I)(A_{22}^\top)^k,\\R_2&=T_2-A_{22}T_2A_{22}^\top-\widetilde X_{22}=H_2^\top H_2+\epsilon I,\\J&=\widetilde X_{12}+A_{12}T_2A_{22}^\top,\\\widehat X_{11}&=A_{12}T_2A_{12}^\top+\widetilde X_{11}+JR_2^{-1}J^\top,\\T_1&=\sum_{k\ge0}A_{11}^k(\widehat X_{11}+H_1^\top H_1+\epsilon I)(A_{11}^\top)^k,\\P&=\operatorname{diag}(T_1^{-1},T_2^{-1}),\\F&=\operatorname{diag}(P,X_-)-[A\ B]^\top P[A\ B]=\begin{bmatrix}F_1&F_{12}\\F_{12}^\top&F_2\end{bmatrix}\succ0.\end{aligned}$$

Partition $F$ after its first $n_1$ coordinates, so $F_2$ has size $d=n_2+m$. With free $r,\delta\in\mathbb R^c$ and upper Cholesky factors ($L^{\top}L=$ the factored matrix), define

$$\begin{aligned}M&=C_1F_1^{-1}C_1^\top,\quad q_i=e^{r_i},\quad \Gamma=\operatorname{diag}\left(\epsilon+\delta_i^2+\tfrac12\sum_j|M_{ij}|q_j/q_i\right),\\L_\Gamma^\top L_\Gamma&=2\Gamma-M,\qquad L_F^\top L_F=F_2-F_{12}^\top F_1^{-1}F_{12},\\ {}[U;V]&=\operatorname{Cayley}(Y,Z),\quad Y\in\mathbb R^{c\times c},\quad Z\in\mathbb R^{d\times c},\\ {}[C_2\ D]&=C_1F_1^{-1}F_{12}-L_\Gamma^\top V^\top L_F,\\L&=UL_\Gamma\Gamma^{-1},\quad X=L^\top L,\quad\Lambda=\Gamma^{-1}.\end{aligned}$$

Then the following dissipation certificate holds:

$$\begin{bmatrix}F&-[C_1\ C_2\ D]^\top\Lambda\\-\Lambda[C_1\ C_2\ D]&2\Lambda-X\end{bmatrix}\succeq0.$$

Why it works: the two storage sums solve the two Lyapunov equations of the Roesser model one after the other ($x_2$ first, because $A_{21}=0$), and Schur complements turn them into $F\succ0$ (Lemma 16). The choice of $\Gamma$ makes $2\Gamma-M\succ0$ by SLL's scaled Gershgorin argument: with $D=\operatorname{diag}(1/q_i)$, row $i$ of the similar matrix $D(2\Gamma-M)D^{-1}$ has a diagonal entry that exceeds the sum of its absolute off-diagonal entries by at least $2\epsilon+2\delta_i^2\gt0$. Finally, $U^{\top}U+V^{\top}V=I$ makes the Schur complement of the certificate with respect to $F$ exactly zero, as in the 1-D proof (we checked the whole construction numerically on random data). Note $\Lambda=\Gamma^{-1}$ here, whereas the 1-D layer used $\Lambda=\Gamma^2$. The structural zeros of the realization must be kept, the bias is free, and $L$ must be invertible to hand $X$ to the next layer (Pauli et al., Automatica 2026).

Why it matters — inference in a control loop
Sandwich and Orthogon convolutions still need an FFT round trip and per-frequency matrix products in every pass after training; their Cayley inverses depend only on the frozen weights and the image size, so they can be cached, as the official code does in evaluation mode. LipKernel layers are standard convolutions. On random weights LipKernel took under 1 ms to about 100 ms, Sandwich and Orthogon about 10 ms to over 10 s, two to three orders of magnitude slower and growing with channels and image size (kernel size barely mattered): about 100 Hz versus 0.1 Hz in a feedback loop, in the authors' words. These are the paper's measurements; it attributes the gap to FFTs and inverses at inference, which caching reduces to FFTs and matrix products. LipKernel also handles zero padding and arbitrary image sizes. The Sandwich implementation uses circular convolutions and caches its factors for one fixed square image size; its layers accept any even size (e.g. $6\times6$), and the network's emulated stride-2 downsampling adds divisibility requirements, so powers of two, which LipKernel states as a requirement, are not intrinsically needed.

Accuracy as reported on MNIST (32×32; 20 epochs; mean of three initializations; certified if the margin exceeds $\sqrt2\rho\varepsilon$):

ArchitectureMethodCert. boundEmp. lower boundCleanCert. 36/25572/255108/255
2C2FVanilla–221.799.0%0.0%0.0%0.0%
2C2FOrthogon10.96094.6%92.9%91.0%88.3%
2C2FSandwich10.91497.3%96.3%95.2%93.8%
2C2FLipKernel10.95296.6%95.6%94.3%92.6%
2CP2F (avg. pool)AOL10.92688.7%85.5%81.7%77.2%
2CP2F (avg. pool)LipKernel10.75991.7%88.0%83.1%77.3%

LipKernel is slightly less expressive than Sandwich but more flexible and much faster at inference, and it beats Orthogon and AOL on clean accuracy.

Caveats
Notation: $\rho$ is the bound itself here ($X_0=\rho^2I$), not LipSDP's squared bound. Exact scalings matter: the layer LMIs hold by construction only with the specific combination of Gramian, Cholesky factor, $\Gamma$ and gain factor; a generic Cayley map applied to a reshaped kernel certifies nothing. Realizations (causal versus centred kernels, state dimensions, strides) differ between arXiv and journal versions of the 2-D papers, so reproduce one paper's realization consistently. Datasets: LipKernel's published experiments are on MNIST, not CIFAR; its case is speed and flexibility, not state-of-the-art certified accuracy (Section 6).

5. Recurrent Equilibrium Networks and R2DN

For dynamic models (system identification, i.e. fitting a dynamical model to input–output data; observers, which estimate an unmeasured state; feedback policies) the model itself must be stable for every parameter value, including those SGD visits on the way. A recurrent equilibrium network (REN) is an LTI system in feedback with an implicit (equilibrium) layer (Revay, Wang & Manchester, IEEE TAC 2024; recurrent and equilibrium layers: Primer E):

$$x_{t+1}=Ax_t+B_1w_t+B_2u_t+b_x,\qquad y_t=C_2x_t+D_{21}w_t+D_{22}u_t+b_y,$$ $$w_t=\sigma\big(D_{11}w_t+C_1x_t+D_{12}u_t+b_v\big),$$

with $\sigma$ elementwise, piecewise differentiable and slope-restricted in $[0,1]$ (Assumption 1). With $D_{11}$ strictly lower triangular the equilibrium is computed row by row (acyclic REN); a deep feedforward network is the case of block sub-diagonal $D_{11}$, so RENs contain the networks of Section 3, all stable LTI systems and previously known contracting RNNs.

Fact — Well-posed equilibrium layers (Revay, Wang & Manchester 2020; condition (12) of the TAC paper)
Write $r=C_1x_t+D_{12}u_t+b_v$, so the layer must solve $w=\sigma(D_{11}w+r)$ for $w\in\mathbb R^q$. It is well-posed if this equation has exactly one solution for every $r\in\mathbb R^q$, so that the model defines an output for every state and input. Under Assumption 1 this holds whenever some positive diagonal $\Lambda$ satisfies $2\Lambda-\Lambda D_{11}-D_{11}^{\top}\Lambda\succ0$ (Revay, Wang & Manchester, 2020). If $D_{11}$ is strictly lower triangular no condition is needed: $w_1=\sigma(r_1)$, $w_2=\sigma\big((D_{11})_{21}w_1+r_2\big)$, and each later coordinate uses only coordinates already computed.
Well-posedness is a property of the equation, not of a solver. Naive iteration $w^{k+1}=\sigma(D_{11}w^k+r)$ converges if, e.g., $\|D_{11}\|_2\le\kappa\lt1$: the map is then a $\kappa$-contraction ($\sigma$ is 1-Lipschitz), so $\|w^k-w^\ast\|\le\kappa^k\|w^0-w^\ast\|$ by Banach's fixed-point theorem (see Primer 0). The well-posedness condition does not imply this (scalar example: $D_{11}=-3$ is well-posed, yet for $\sigma(v)=v$ the iteration $w^{k+1}=-3w^k+r$ diverges), so for full $D_{11}$ the REN paper computes the equilibrium with a monotone operator-splitting method (Peaceman–Rachford) instead.
Definition — Contraction and incremental IQCs (Revay et al. 2024, Defs. 2–3)
The model is contracting with rate $\alpha\in(0,1)$ if for any initial states $a,b$ and the same input $|x^a_t-x^b_t|\le K\alpha^t|a-b|$ for a fixed $K\gt0$ independent of the initial states and common input. It satisfies the incremental IQC $(Q,S,R)$, $Q\preceq0$, if for all pairs of solutions (solution $a$ starts at $x_0=a$ with input $u$ and output $y^a$, solution $b$ at $x_0=b$ with input $v$ and output $y^b$) $$\sum_{t=0}^{T}\begin{bmatrix}y^a_t-y^b_t\\u_t-v_t\end{bmatrix}^{\top}\begin{bmatrix}Q&S^{\top}\\S&R\end{bmatrix}\begin{bmatrix}y^a_t-y^b_t\\u_t-v_t\end{bmatrix}\ \ge\ -d(a,b)\qquad\forall T,$$ for some $d(a,b)\ge0$ with $d(a,a)=0$. $Q=-\tfrac1\gamma I$, $R=\gamma I$, $S=0$ is an incremental $\ell_2$-gain (Lipschitz) bound $\gamma$: multiplying the IQC by $\gamma$ gives $\sum_{t=0}^T|y^a_t-y^b_t|^2\le\gamma^2\sum_{t=0}^T|u_t-v_t|^2+\gamma\,d(a,b)$, a pure Lipschitz bound when $a=b$ because $d(a,a)=0$. With equal input and output dimensions, $Q=0$, $R=-2\nu I$, $S=I$ with $\nu\ge0$ gives the supply $2(u_t-v_t)^{\top}(y^a_t-y^b_t)-2\nu|u_t-v_t|^2$: incremental passivity (strict input passivity if $\nu\gt0$).
Theorem — Contracting RENs (Revay, Wang & Manchester 2024, Thm 1, part 1)
Let Assumption 1 hold and $\bar\alpha\in(0,1]$. If there are $P=P^{\top}\succ0$ and a positive diagonal $\Lambda$ with $$\begin{bmatrix}\bar\alpha^2P&-C_1^{\top}\Lambda\\-\Lambda C_1&\mathcal W\end{bmatrix}-\begin{bmatrix}A^{\top}\\B_1^{\top}\end{bmatrix}P\begin{bmatrix}A&B_1\end{bmatrix}\succ0,\qquad \mathcal W=2\Lambda-\Lambda D_{11}-D_{11}^{\top}\Lambda,$$ then the REN is well-posed ($\mathcal W\succ0$ is implied, and it is precisely the sufficient well-posedness condition for equilibrium networks, their Remark 3) and contracting with some rate $\alpha\lt\bar\alpha$. An analogous LMI (Thm 1, part 2) additionally certifies a given incremental IQC $(Q,S,R)$ with $Q\preceq0$.

Why: take two trajectories with the same input, so $\Delta x_{t+1}=A\Delta x_t+B_1\Delta w_t$ and $\Delta v_t=C_1\Delta x_t+D_{11}\Delta w_t$. The quadratic form of the LMI matrix at $[\Delta x_t;\Delta w_t]$ equals $\bar\alpha^2|\Delta x_t|_P^2-|\Delta x_{t+1}|_P^2-2\Delta w_t^{\top}\Lambda(\Delta v_t-\Delta w_t)$, and the last term is $\le0$ by the diagonal incremental QC. The matrix is $\succeq\eta I$ for some $\eta\gt0$ (its smallest eigenvalue), so $|\Delta x_{t+1}|_P^2\le\bar\alpha^2|\Delta x_t|_P^2-\eta|\Delta x_t|^2\le\alpha^2|\Delta x_t|_P^2$ with $\alpha^2=\bar\alpha^2-\eta/\lambda_{\max}(P)$ (shrink $\eta$ if needed so that $\alpha\gt0$): an incremental Lyapunov function decreasing at a rate $\alpha\lt\bar\alpha$. Iterating and using $\lambda_{\min}(P)|z|^2\le|z|_P^2\le\lambda_{\max}(P)|z|^2$ gives $|\Delta x_t|\le\sqrt{\lambda_{\max}(P)/\lambda_{\min}(P)}\,\alpha^t|\Delta x_0|$: contraction with a constant $K$ that depends only on $P$, not on the initial states or the input.

Caveat — incremental certificates need diagonal multipliers (Remark 5)
Richer multipliers for repeated nonlinearities can certify boundedness, but they are not valid for incremental IQCs and contraction, the same issue as the LipSDP correction of Section 3. And every contracting REN has some finite Lipschitz bound (Thm 2), but only the robust version certifies a prescribed $\gamma$.

From LMI to free parameters. The LMI is convex in $(P,\Lambda)$ for a fixed model but not jointly in the weights, so RENs use an implicit form with extra invertible $E$ and $\Lambda$, $Ex_{t+1}=Fx_t+\mathcal B_1w_t+\mathcal B_2u_t+\dots$ and $\Lambda v_t=\mathcal C_1x_t+\mathcal D_{11}w_t+\mathcal D_{12}u_t+\dots$ (so $A=E^{-1}F$, $C_1=\Lambda^{-1}\mathcal C_1$, and so on), in which contraction is an LMI $\mathcal H(\theta_{\rm cvx})\succ0$ jointly convex in weights, certificate and multiplier. The map $\theta_{\rm cvx}\mapsto\mathcal H$ is onto the positive definite cone and has an explicit right inverse, so writing $\mathcal H=X^{\top}X+\epsilon I$ with a free square $X$ and reading the model off its blocks gives a direct parameterization; for a fixed $\epsilon$ it reaches exactly the matrices $\mathcal H\succeq\epsilon I$, not every $\mathcal H\succ0$. Random free parameters give random contracting systems, a strictly larger family of stable echo state networks than previously known.

Theorem — Direct parameterization of contracting RENs (Revay, Wang & Manchester 2024, Sec. V-A and Thm 3)

Choose the state size $n$, the equilibrium size $q$, $0\lt\bar\alpha\le1$, $\epsilon\gt0$ and free $X\in\mathbb R^{(2n+q)\times(2n+q)}$, $Y\in\mathbb R^{n\times n}$. Partition $\mathcal H=X^{\top}X+\epsilon I$ into blocks of sizes $(n,q,n)$, set

$$\mathcal P=\mathcal H_{33},\quad F=\mathcal H_{31},\quad \mathcal B_1=\mathcal H_{32},\quad \mathcal C_1=-\mathcal H_{21},$$ $$E=\tfrac12\big(\mathcal H_{11}+\mathcal P/\bar\alpha^2+Y-Y^{\top}\big),$$

write $\mathcal H_{22}=\Phi-J-J^{\top}$ with $\Phi$ diagonal and $J$ strictly lower triangular (so $J_{ij}=-(\mathcal H_{22})_{ij}$ for $i\gt j$), and set $\Lambda=\tfrac12\Phi$, $\mathcal D_{11}=J$. By construction

$$\mathcal H=\begin{bmatrix}E+E^{\top}-\mathcal P/\bar\alpha^2&-\mathcal C_1^{\top}&F^{\top}\\-\mathcal C_1&2\Lambda-\mathcal D_{11}-\mathcal D_{11}^{\top}&\mathcal B_1^{\top}\\F&\mathcal B_1&\mathcal P\end{bmatrix}\succ0 .$$

The explicit REN has $A=E^{-1}F$, $B_1=E^{-1}\mathcal B_1$, $C_1=\Lambda^{-1}\mathcal C_1$ and the strictly lower triangular $D_{11}=\Lambda^{-1}\mathcal D_{11}$ (an acyclic REN); $B_2,C_2,D_{12},D_{21},D_{22}$ and the biases are free. Under Assumption 1 every such model is well-posed and contracting with some rate $\alpha\lt\bar\alpha$.

Proof sketch: $E+E^{\top}=\mathcal H_{11}+\mathcal P/\bar\alpha^2\succ0$, so $Ev=0$ forces $v^{\top}(E+E^{\top})v=0$, i.e. $v=0$: $E$ is invertible. Expanding $(\bar\alpha E-\mathcal P/\bar\alpha)^{\top}\mathcal P^{-1}(\bar\alpha E-\mathcal P/\bar\alpha)\succeq0$ gives $\bar\alpha^2E^{\top}\mathcal P^{-1}E\succeq E+E^{\top}-\mathcal P/\bar\alpha^2$, so the $(1,1)$ block may be replaced by $\bar\alpha^2E^{\top}\mathcal P^{-1}E$. A Schur complement with respect to $\mathcal P$ and the substitutions $F=EA$, $\mathcal B_1=EB_1$, $\mathcal C_1=\Lambda C_1$, $\mathcal D_{11}=\Lambda D_{11}$ turn the result into exactly the LMI of the contraction theorem above, with $P=E^{\top}\mathcal P^{-1}E$ (we also checked this numerically). For a full $D_{11}$ the paper takes $\Lambda=e^{\operatorname{diag}(g)}$ and $\mathcal D_{11}=\Lambda-\tfrac12(\mathcal H_{22}+Y_2-Y_2^{\top})$ with free $g,Y_2$ (Revay et al., TAC 2024).

Going deeper — robust RENs: a prescribed incremental IQC

For a prescribed $(Q,S,R)$ with $Q\preceq0$, the paper first chooses $D_{22}$ with $\mathcal R:=R+SD_{22}+D_{22}^{\top}S^{\top}+D_{22}^{\top}QD_{22}\succ0$ and then adds to $\mathcal H$ two positive semidefinite terms that pay for the input–output coupling (their (28)–(29)): with free $\mathcal B_2,C_2,\mathcal D_{12},D_{21}$, $\mathcal C_2=(D_{22}^{\top}Q+S)C_2$ and $\mathcal D_{21}=(D_{22}^{\top}Q+S)D_{21}-\mathcal D_{12}^{\top}$,

$$\mathcal H=X^{\top}X+\epsilon I+K\mathcal R^{-1}K^{\top}-OQO^{\top},\qquad K=\begin{bmatrix}\mathcal C_2^{\top}\\\mathcal D_{21}^{\top}\\\mathcal B_2\end{bmatrix},\quad O=\begin{bmatrix}C_2^{\top}\\D_{21}^{\top}\\0\end{bmatrix}.$$

The blocks are recovered as in the theorem, plus $B_2=E^{-1}\mathcal B_2$ and $D_{12}=\Lambda^{-1}\mathcal D_{12}$; every such model is well-posed, contracting and satisfies the IQC (their Thm 3). For a Lipschitz bound $\gamma$ ($Q=-\tfrac1\gamma I$, $R=\gamma I$, $S=0$) the condition on $D_{22}$ reads $\gamma I-\tfrac1\gamma D_{22}^{\top}D_{22}\succ0$, i.e. $\|D_{22}\|\lt\gamma$, and the paper takes $D_{22}=\gamma N$ with a Cayley-type $N$ whose extra $\epsilon I$ makes it a strict contraction (their (31)–(32)). We checked numerically that this construction satisfies the robust LMI of Thm 1, part 2.

R2DN: removing the equilibrium solve

The implicit layer must be solved at every time step (e.g. by operator splitting), which is slow on GPUs for long sequences. R2DN (Barbara, Wang & Manchester, accepted to CDC 2026) replaces it by an explicit network, $w_t=\phi_g(C_1x_t+D_{12}u_t+b_v)$ with the same linear equations for $x_{t+1}$ and $y_t$, where $\phi_g$ is 1-Lipschitz for every value of its own parameters $g$, e.g. a stack of sandwich layers. The network enters the certificate only through $|\Delta w_t|\le|\Delta v_t|$, and the linear part is directly parameterized so that the interconnection is contracting and, if desired, $\gamma$-Lipschitz from $u$ to $y$ (their Prop. 1 covers general incremental IQCs; the direct parameterization of the Lipschitz case keeps $D_{22}=0$). Training and inference become up to an order of magnitude faster than for RENs with similar performance on system identification, observer design, feedback control and sequential image classification, scaling better with model size.

Fact — The contraction test behind R2DN (derived here; explicit form of Barbara, Wang & Manchester's condition (19))

Let $\phi_g$ be 1-Lipschitz for every value of its parameters, $w_t=\phi_g(C_1x_t+D_{12}u_t+b_v)$ and $x_{t+1}=Ax_t+B_1w_t+B_2u_t+b_x$. If some $P\succ0$ and $0\lt\alpha\lt1$ satisfy

$$\begin{bmatrix}\alpha^2P-A^{\top}PA-C_1^{\top}C_1&-A^{\top}PB_1\\-B_1^{\top}PA&I-B_1^{\top}PB_1\end{bmatrix}\succeq0,$$

then two trajectories with the same input satisfy $|\Delta x_t|_P\le\alpha^t|\Delta x_0|_P$: the R2DN is contracting, and no implicit equation has to be solved.

Proof: with equal inputs, $\Delta x_{t+1}=A\Delta x_t+B_1\Delta w_t$ and $\Delta v_t=C_1\Delta x_t$, and the quadratic form of the matrix at $[\Delta x_t;\Delta w_t]$ equals $\alpha^2|\Delta x_t|_P^2-|\Delta x_{t+1}|_P^2-\big(|\Delta v_t|^2-|\Delta w_t|^2\big)$. The bracket is $\ge0$ because $|\Delta w_t|\le|\Delta v_t|$, so $|\Delta x_{t+1}|_P\le\alpha|\Delta x_t|_P$. $\blacksquare$ The condition says that the linear part, seen from $w$ to $v$, has incremental gain at most one, which is what an incremental small-gain argument with a 1-Lipschitz network needs (their Prop. 1); a 1-Lipschitz network alone guarantees nothing. R2DN parameterizes it like the REN above: $\mathcal H=X^{\top}X+\epsilon I+\operatorname{diag}(C_1^{\top}C_1,\mathcal B_1\mathcal B_1^{\top})$ with free $X,Y,\mathcal B_1,C_1$, $E=\tfrac12(\mathcal H_{11}+\mathcal H_{22}+Y-Y^{\top})$, $A=E^{-1}\mathcal H_{21}$, $B_1=E^{-1}\mathcal B_1$ (their (20)–(21)). A Lipschitz bound from $u$ to $y$ needs the extra blocks of their Section V-C (Barbara, Wang & Manchester, R2DN).

Connection to Module 14 (optional preview)
A contracting REN can be the free parameter of a nonlinear Youla parameterization. In the setting of Wang & Manchester, ACC 2022 (Youla-REN) the plant is linear with an uncertain parameter, $x_{t+1}=A(\rho)x_t+Bu_t+d_t$ with measured state and $\rho$ in a compact set, and a linear state feedback $u=-Kx+v$ is assumed to give a finite $\ell_2$-gain from $v$ to $x$ for every $\rho$. The policy runs a copy of the nominal closed loop, $\hat x_{t+1}=(A(\hat\rho)-BK)\hat x_t+Bv_t$, and applies $u_t=-Kx_t+v_t$ with $v=\mathcal Q_\theta(x-\hat x)$ for a REN $\mathcal Q_\theta$. If the mismatch system from $v$ to $x-\hat x$ has incremental $\ell_2$-gain $\mu$ and $\mathcal Q_\theta$ is a robust REN with Lipschitz bound $\gamma\lt1/\mu$, the closed loop is contracting for every $\rho$ and every $\theta$ (their Thm 1 and Remark 1, an incremental small-gain argument), so policy learning, including RL, is unconstrained and stable at every iterate. The model copy and the base controller are essential: a contracting REN inserted into an arbitrary feedback loop guarantees nothing. Nothing in this module depends on this application; Module 14 develops the partially observed version. The survey by Manchester, Wang & Barbara (Annu. Rev. Control Robot. Auton. Syst. 2026) collects the toolbox: IQC certificates, direct parameterizations, and their use as observers, stabilizing policy classes and learned certificates.

6. Certified Robust Accuracy: State of the Art 2026

For logits $f(x)\in\mathbb R^N$ and prediction $y=\arg\max_if_i(x)$, a Lipschitz certificate turns the margin into a radius: the deterministic guarantee type of Module 1.

Theorem — Margin certificate (Tsuzuku, Sato & Sugiyama 2018, Props. 1–2)
Assume $N\ge2$. The displayed quotients use positive certificate bounds $L$ and $L_{yj}$; a positive upper bound can always be chosen. If a pairwise bound is zero, the corresponding gap is constant and, when positive at $x$, imposes no finite radius restriction. If the logit map is $L$-Lipschitz in $\ell_2$ and $M_f(x)=f_y(x)-\max_{j\ne y}f_j(x)\gt0$, the prediction cannot change for $\|\delta\|_2\lt M_f(x)/(\sqrt2L)$ (Prop. 1). More generally, if $L_{yj}$ bounds the Lipschitz constant of $f_y-f_j$, the radius is $\min_{j\ne y}\big(f_y(x)-f_j(x)\big)/L_{yj}$; $L_{yj}=\sqrt2L$ is always valid and recovers the first rule, and a last linear layer with rows $w_i$ on top of an $L_{\rm sub}$-Lipschitz feature map gives $L_{yj}=L_{\rm sub}\|w_y-w_j\|$ (Prop. 2).

Why $\sqrt2$: $f_y-f_j=(e_y-e_j)^{\top}f$, so $|\Delta(f_y-f_j)|\le\|e_y-e_j\|\,\|\Delta f\|\le\sqrt2L\|\delta\|$, and a label flip needs some margin to reach zero (Tsuzuku et al., NeurIPS 2018). Three conventions are in use: the global rule $M_f/(\sqrt2L)$ (BRONet, LipKernel); a constant $L_{yj}$ per logit difference (GloRo nets, Leino, Wang & Fredrikson, ICML 2021; LiResNet, Hu, Zou, Wang, Leino & Fredrikson, NeurIPS 2023), which for a 1-Lipschitz feature map and a linear last layer with rows $w_i$ can be $\|w_y-w_j\|$; and LipNeXt's decomposition into $N-1$ binary classifiers ("one-vs-rest" in the paper's words, which with $N-1$ classifiers we read as $f_y-f_j$), each certified with radius $|f_y-f_j|/K_j$ for a Lipschitz bound $K_j$, the overall radius being the minimum. The last two coincide when $K_j=L_{yj}$, and with a linear last layer $\|w_y-w_j\|$ is never worse than the $\sqrt2L$ rule based on the product certificate (Exercise 13.5).

Standard $\ell_2$ radii for images in $[0,1]$: $36/255\approx0.141$, $72/255\approx0.282$, $108/255\approx0.424$. CIFAR-10 certified robust accuracy (CRA) as tabulated by LipNeXt (Tables 1–2), no extra data unless marked:

ModelParamsCleanCRA 36/25572/255108/255
Cayley Large (Trockman & Kolter 2021)21M74.661.446.432.1
SOC-20 (Singla & Feizi 2021)27M76.362.648.736.0
AOL Large (Prach & Lampert 2022)136M71.664.056.449.0
SLL X-Large (Araujo et al. 2023)236M73.364.855.747.1
LiResNet (Hu et al.)83M81.069.856.342.9
BRONet (Lai et al. 2025)68M81.670.657.242.5
LipNeXt L32W2048 (Hu, Hu & Fredrikson 2026)256M85.073.258.843.3
BRO + diffusion-generated data68M87.278.367.454.5
LipNeXt L32W2896 + generated data512M92.781.768.655.8

BRONet uses block-reflector orthogonal layers $W=I-2V(V^{\top}V)^{-1}V^{\top}$ for full-column-rank $V\in\mathbb R^{m\times n}$: symmetric, orthogonal, with $n$ eigenvalues $-1$ and $m-n$ eigenvalues $+1$ (precisely the eigenvalue the Cayley transform cannot produce), without iterative approximation (their Prop. 1). The reason: $P_V=V(V^{\top}V)^{-1}V^{\top}$ is the orthogonal projector onto the column space of $V$, $P_V^{\top}=P_V=P_V^2$, so $W^{\top}W=I-4P_V+4P_V^2=I$, $Wx=-x$ on the $n$-dimensional column space and $Wx=x$ on its orthogonal complement. BRONet also trains with a logit-annealing loss: for logits $z$, true class $t$ and $p=\operatorname{softmax}\big((z-\xi e_t)/T\big)$ (temperature $T$, offset $\xi$), $\mathcal L_{\rm LA}=-T(1-p_t)^{\beta}\log p_t$ (their eq. (8); this $\beta$ is unrelated to $\beta$-Abs below). The factor $(1-p_t)^{\beta}$ shrinks the loss of examples that are already classified with a large margin ($0.01$ for $p_t=0.9$, $\beta=2$), so capacity goes to the others; the loss only changes which network training finds, the certificate still comes from the orthogonal layers (Lai, Huang, Kung & Chen, ICML 2025).

LipNeXt drops parameterizations: orthogonal weights are updated directly on the orthogonal manifold by a manifold Adam whose exponential map is replaced by a norm-adaptive truncated Taylor series ("FastExp"); the small loss of orthogonality this causes is removed by a polar retraction (SVD) once per epoch, plus a Lookahead step taken in the tangent space. Convolutions are replaced by a spatial-shift module (each selected circular shift is a norm-preserving permutation), the activation $\beta$-Abs (absolute value on a fraction $\beta$ of the coordinates, identity on the rest) contains MinMax up to an orthogonal change of basis, and scale does the rest (Hu, Hu & Fredrikson, ICLR 2026). On ImageNet at $36/255$ its 2B-parameter model reaches 57.0% clean / 41.2% CRA without generated data, against 49.3% / 37.6% for BRONet (86M) and 45.6% / 35.0% for LiResNet (51M), and 41.0% / 22.4% at $\varepsilon=1$; BRONet with 2M generated images reaches 40.7% CRA, so this comparison mixes data regimes.

Background — optimization on the orthogonal manifold

The square orthogonal matrices $\{W\in\mathbb R^{d\times d}:W^{\top}W=I\}$ form a curved surface in matrix space, a manifold. Differentiating $W(t)^{\top}W(t)=I$ along a curve on it gives $W^{\top}Z+Z^{\top}W=0$ for its velocity $Z=W'(0)$; these admissible directions form the tangent space at $W$ (at $W=I$ in two dimensions, $Z=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ starts a rotation). They are exactly the matrices $Z=WK$ with $K$ skew-symmetric, and the curve $W\exp(tK)$ never leaves the manifold, because $\exp(tK)^{\top}\exp(tK)=\exp(-tK)\exp(tK)=I$.

Manifold gradient descent keeps only the tangent part of the Euclidean gradient $G=\nabla f(W)$, which is $W\operatorname{skew}(W^{\top}G)$ with $\operatorname{skew}(M)=\tfrac12(M-M^{\top})$, and moves along $W\exp(-\eta\operatorname{skew}(W^{\top}G))$ for a step size $\eta$; manifold Adam uses Adam's rescaled skew-symmetric step instead. An update that leaves the manifold (e.g. through a truncated exponential) is pulled back by the polar retraction: if $\widetilde W=U\Sigma V^{\top}$ is an SVD, $UV^{\top}$ is the orthogonal matrix closest to $\widetilde W$ in the Frobenius norm. Lookahead returns every $k$ steps to the weight of $k$ steps earlier and moves half-way along the sum of the skew-symmetric steps taken since, which keeps the averaged weight on the manifold (Hu, Hu & Fredrikson, eqs. (2)–(5)). None of these names proves that the stored weights are orthogonal; the next fact says what the certificate can rely on.

Fact — What truncated updates do to the certified norm (derived here from the SOC bound; FastExp from Hu, Hu & Fredrikson 2026, eq. (4))
FastExp replaces $\exp(K)$ by the partial sum $S_m(K)=\sum_{j=0}^{m}K^j/j!$ with $m=2,3,4$ when $\|K\|_F$ lies below $0.05$, in $[0.05,0.25)$ or in $[0.25,1)$, and uses the exact exponential beyond. For skew-symmetric $K$ the truncation theorem of Section 2 gives $\|\exp(K)-S_m(K)\|_2\le e_m:=\|K\|_2^{m+1}/(m+1)!$, so an update of an orthogonal $W$ satisfies $\|WS_m(K)\|_2\le1+e_m$, and a run of such updates that starts from an orthogonal matrix has norm at most $\prod_t(1+e_{m_t})$. Example: $\|K\|_2\le\|K\|_F\lt0.05$ with $m=2$ gives $e_2\le0.05^3/6\approx2.1\times10^{-5}$ per step, which compounds to about $1.23$ over $10^4$ steps. The polar retraction at the end of each epoch resets the norm to one, so a deterministic certificate should use retracted weights (or carry the product factor), with floating-point error accounted for.
Fact — Why a LipNeXt block is 1-Lipschitz (derived here; definitions from Hu, Hu & Fredrikson 2026, eqs. (6)–(10))
The core block maps $X\mapsto\beta\text{-Abs}\big(MR^{\top}S(R(X+p))+b\big)$, with orthogonal channel mixers $R,M$, a learned positional embedding $p$, the spatial shift $S$, and $[\beta\text{-Abs}(x)]_i=|x_i|$ for $i\le\beta d$, $x_i$ otherwise, applied to the channel vector $x\in\mathbb R^d$ of every pixel. Every factor is 1-Lipschitz: adding $p$ or $b$ does not change differences; orthogonal matrices preserve norms; $S$ shifts groups of channels circularly by one pixel, a permutation of coordinates (a selected single-tap $\pm1$ circular kernel is a norm-preserving signed permutation), while the zero-padded variant drops entries and is only nonexpansive; and $\beta$-Abs is 1-Lipschitz because $\big||a|-|b|\big|\le|a-b|$ in each coordinate. Single-tap shifts are sufficient for this argument; a converse classification of all circular norm-preserving kernels would require extra grid/support conditions. For example, on three pixels the circular kernel $(-1/3,2/3,2/3)$ gives $C=\tfrac23\mathbf1\mathbf1^{\top}-I$, with $C^{\top}C=I$ despite three nonzero taps. MinMax is the case $\beta=\tfrac12$ in rotated coordinates: $R_2=\tfrac1{\sqrt2}\begin{bmatrix}1&-1\\1&1\end{bmatrix}$ maps $(a,b)$ to $\big(\tfrac{a-b}{\sqrt2},\tfrac{a+b}{\sqrt2}\big)$, and taking the absolute value of the first coordinate and applying $R_2^{\top}$ gives $\big(\tfrac{a+b+|a-b|}2,\tfrac{a+b-|a-b|}2\big)=\big(\max(a,b),\min(a,b)\big)$.

Deterministic versus randomized-smoothing certificates

Randomized smoothing certifies a different object. With the base label classifier $C(x)=\arg\max_i f_i(x)$ (fixed tie-breaking) and a noise level $\sigma\gt0$, the smoothed classifier is $g(x)=\arg\max_c\mathbb P_{\varepsilon\sim\mathcal N(0,\sigma^2I)}(C(x+\varepsilon)=c)$, and its certificate is the theorem below (Cohen, Rosenfeld & Kolter, ICML 2019; see Module 15). With an off-the-shelf diffusion denoiser and a large pretrained classifier, Carlini et al. (ICLR 2023) report certified top-1 accuracy of 71.1% / 54.3% / 38.1% / 29.5% on ImageNet at $\ell_2$ radius 0.5 / 1.0 / 1.5 / 2.0 (best noise level per radius).

Theorem — Gaussian smoothing certificate (Cohen, Rosenfeld & Kolter 2019, Thm 1 and Prop. 2)

Let $C:\mathbb R^{n_0}\to\{1,\dots,N\}$ be any deterministic or random base classifier, $\sigma\gt0$, $\varepsilon\sim\mathcal N(0,\sigma^2I)$ and $g$ as above. If a class $A$ and a number $\underline{p_A}\gt\tfrac12$ satisfy $\mathbb P(C(x+\varepsilon)=A)\ge\underline{p_A}$, then

$$g(x+\delta)=A\qquad\text{for all }\|\delta\|_2\lt\sigma\,\Phi^{-1}(\underline{p_A}),$$

where $\Phi$ is the standard normal cumulative distribution function and $\Phi^{-1}$ its quantile function (see Primer C). The probability is unknown, so it is estimated: CERTIFY picks $A$ from a few pilot samples of $C(x+\varepsilon)$, computes $\underline{p_A}$ as a one-sided $(1-\alpha)$ Clopper–Pearson lower bound (see Primer C) from fresh samples ($10^4$ to $10^5$ in the table below), and abstains if $\underline{p_A}\le\tfrac12$. With probability at least $1-\alpha$ over this sampling, a returned class and radius are correct (Prop. 2).

In words: the only randomness in the guarantee is the Monte-Carlo estimate; on the good event it holds for every perturbation in the ball at once. The displayed radius is Cohen's $\tfrac\sigma2\big(\Phi^{-1}(\underline{p_A})-\Phi^{-1}(\overline{p_B})\big)$ with the runner-up bound $\overline{p_B}=1-\underline{p_A}$, since $\Phi^{-1}(1-p)=-\Phi^{-1}(p)$. Nothing is assumed about $C$, so a fixed denoiser $D$ applied before the classifier is covered by taking $C\circ D$ as the base classifier, with no Lipschitz bound on $D$: this is Carlini et al.'s diffusion denoised smoothing.

Lipschitz by design (LipNeXt)Diffusion denoised smoothing (Carlini et al.)
Certified objectthe deployed network $f$the smoothed classifier $g$, not $f$
Guaranteedeterministicwith probability $\ge1-\alpha$ over the Monte-Carlo sampling
Cost per certified inputone forward pass$N=10^4$ (ImageNet) or $10^5$ (CIFAR-10) noisy passes
ImageNet, radius 1.022.4% (2B model, no extra data)54.3% (BEiT-L, 552M diffusion model, pretraining data)
Where it winscost and certainty: one deterministic pass, real time, closed-loop analysiscertified accuracy: on ImageNet even its radius-0.5 number (71.1%) beats LipNeXt at $36/255\approx0.14$ (41.2%); large radii; offline certification
Caveats — comparing certified numbers
Radius conventions: check the margin rule ($\sqrt2L$ versus pairwise) and the input space (pixels versus normalized inputs, Section 3); Lipschitz papers use $36/255$–$108/255$, smoothing papers $0.25$–$3.0$. Data regimes: generated data and pretraining move numbers more than architectures do. Provenance: the SLL paper's main table still prints the original ICLR numbers for SLL X-Large (65.8% / 58.4% / 51.3% at $36/72/108$ over 255), but its erratum (Appendix D, Table 7, October 2023) declares them inaccurate: the implementation added $\varepsilon$ to a denominator, which broke the 1-Lipschitz guarantee. The corrected (exponential-scaling) values are 64.8% / 55.7% / 47.1%, the numbers LipNeXt tabulates. Always check for errata before quoting a certified number. Guarantee type: deterministic and probabilistic certificates are different objects (see Module 15).
Open problems
  • Completeness is relative: Sandwich is complete for LipSDP, not for $\gamma$-Lipschitz networks; the true constant lies between the empirical lower bound and the certified bound, so their ratio (0.76–0.96 in the MNIST table) bounds the remaining conservatism: the certified bound overestimates the true constant by at most a factor $1/0.76\approx1.3$ there.
  • Clean accuracy: even LipNeXt trails unconstrained ImageNet models by a wide margin; expressive 1-Lipschitz architectures remain open.
  • Beyond $\ell_2$ and classification: $\ell_\infty$ certificates from $\ell_2$ bounds lose a $\sqrt n$ factor (see Primer A); control needs Lipschitz bounds of dynamic, closed-loop maps (Modules 11 and 14).
  • Speed versus accuracy: LipKernel gives standard-form convolutions with network-level certificates and zero padding, pooling and strides; LipNeXt shows that FFT-free orthogonal architectures scale. A certified CNN with both LipKernel's architectural flexibility and state-of-the-art accuracy has not been reported.

Walkthrough: Deriving the Sandwich Layer From the LMI

The central derivation of this module, for one hidden layer so that every block is visible: start from the LipSDP condition, turn it into a norm bound, satisfy the norm bound with Cayley factors, and read off the weights. The result is exactly Wang & Manchester's formula (8) for a single hidden layer, and the same three moves reappear in the Cayley–Gramian layers of Section 4. The Schur complement and congruence steps use only Module 2.

Interactive: Cayley Transform and a 1-Lipschitz Layer

Panel (a): the Cayley transform of $A=\begin{bmatrix}s&a\\-a&-s\end{bmatrix}$; for $s=0$, $A$ is skew-symmetric and $W=(I-A)(I+A)^{-1}$ is a rotation by $2\arctan a$, while any $s\ne0$ turns the unit circle into an ellipse. Panel (b): the 2-2-1 network $f(x)=W_1\,\mathrm{ReLU}(W_0x+b_0)$ of the walkthrough built from arbitrary free parameters (for $X_0$ only the skew part $X_0-X_0^{\top}$ enters the Cayley map, so the slider sets its entry $(X_0-X_0^{\top})_{12}$). On each region where the ReLU activation pattern $D=\operatorname{diag}(d_1,d_2)$, $d_i\in\{0,1\}$, is fixed, $f$ is affine with gradient $W_0^{\top}DW_1^{\top}$, so its exact Lipschitz constant is the largest of these gradient norms over the patterns that occur on open sets of inputs: all four when $W_0$ is invertible ($x\mapsto W_0x+b_0$ is then onto), those met along a line when $W_0$ has rank one (the code sorts the two kinks), one when $W_0=0$. The panel compares it with the largest ratio $|f(x)-f(x')|/\|x-x'\|$ over random input pairs, which can only underestimate it, and reports the smallest eigenvalue of the certificate $H(\gamma,\Lambda=\Psi^2)$: a floating-point check of the identity proved in the walkthrough, so values of order $-10^{-16}$ are round-off; the guarantee for all parameters is the proof, not this number. The comparison network skips the Cayley step: $W_0=\sqrt{2\gamma}\,\Psi^{-1}Y_0^{\top}$, $W_1=\sqrt{2\gamma}\,Y_1^{\top}\Psi$.

(a) Cayley transform in two dimensions
(b) A 2-2-1 sandwich network versus the same parameters without the Cayley step

Things to try: push $d_1$ and $d_2$ apart. ReLU is positively homogeneous, so $\Psi\,\mathrm{ReLU}(\Psi^{-1}v+b_0)=\mathrm{ReLU}(v+\Psi b_0)$ and $f(x)=\sqrt{2\gamma}\,B_1A_0^{\top}\mathrm{ReLU}(\sqrt{2\gamma}\,B_0x+\Psi b_0)$: $\Psi$ only rescales the effective bias, and the pattern gradients $2\gamma B_1A_0^{\top}DB_0$ do not contain it. While $W_0$ is invertible all four patterns occur, so the exact constant does not move while $\|W_0\|\,\|W_1\|$ grows far beyond $\gamma$ (for singular $W_0$ the bias can change which patterns occur; for tanh, $\Psi$ would also reshape the activation). "Random free parameters" never breaks the certificate; the comparison network is much steeper.

From the mathematics to a real decision

Learning objectives

A commissioning decision

The robot now uses a learned velocity correction based on enclosure temperature and wall distance. Normalize readings as $z_1=(T-40)/10$ and $z_2=(d-0.5)/0.1$, where temperature is in degrees Celsius and distance in metres. The deployed correction $u$ has units metres per second. The architect chooses

$$u(z)=c\,q^\top\operatorname{ReLU}(Qz+b),\qquad c=0.3\,\mathrm{m/s},\quad \|q\|_2=1.$$

Here $Q$ is a two-by-two orthogonal matrix, $b$ is a dimensionless bias, and ReLU is coordinatewise. Assume exact orthogonality and exact real arithmetic for the first calculation. A later exercise revises that implementation assumption. Training may change $Q$, $q$, and $b$ while preserving these constraints. The structural promise concerns increments, so arbitrary bias does not change it.

Hardware calibration bounds temperature error by 0.2 degrees and distance error by 0.003 metres. The robot's safety layer allows at most $0.012\,\mathrm{m/s}$ change in this correction from sensor error. Assume both errors can occur simultaneously and in either direction. No statistical averaging of their magnitudes is allowed in this worst-case component requirement.

Worked decision, with its limits

Inspect one architecture instance. A Cayley construction with scalar parameter 0.5 can produce $Q=\left[\begin{smallmatrix}0.6&-0.8\\0.8&0.6\end{smallmatrix}\right]$. Its columns have unit norm and inner product zero, so $Q^\top Q=I$. Choosing $q=(1,0)$ is one admissible readout, though the following bound holds for every unit $q$ and every bias.

Compose the gain. Orthogonality preserves Euclidean norm. Coordinatewise ReLU is 1-Lipschitz, and the unit readout has operator norm 1. Therefore $|u(z)-u(\bar z)|\le0.3\|z-\bar z\|_2$ in metres per second. This argument applies to every pair of normalized inputs, without retraining or solving a new SDP after each admissible weight update.

Map the hardware box. The normalized errors satisfy $|\Delta z_1|\le0.02$ and $|\Delta z_2|\le0.03$. Their largest Euclidean norm is $r=\sqrt{0.02^2+0.03^2}\approx0.036056$. The resulting velocity change is at most $0.3r\approx0.010817\,\mathrm{m/s}$. This passes the $0.012\,\mathrm{m/s}$ budget with about $0.001183\,\mathrm{m/s}$ remaining.

Keep the promise narrow enough to be true. The gain bound says nothing about the absolute velocity correction: biases can make it large. A separate output limit or safety filter may be necessary. Likewise, the robot's plant can amplify a small input change over time. The architecture supplies one component inequality for that subsequent dynamical analysis, rather than a complete collision-avoidance proof.

The benefit of the design is predictable sensitivity throughout training. Its cost is a restriction on representable responses. An optimizer cannot fit a desired map whose increments exceed the imposed gain; treating that failure as merely poor optimization overlooks a mathematical incompatibility. B1 makes this tradeoff visible in physical units.

A tempting wrong approach

Common mistake — Normalize by an underestimate

If a matrix has true norm 2 but a power-iteration estimate is 1.9, dividing by that estimate gives norm $20/19\gt1$. Four such layers can have product bound $(20/19)^4\approx1.227738$. The preceding sensor-error bound would then inflate to about $0.013280\,\mathrm{m/s}$ and fail. A numerical estimate close to the norm is useful for training, but a by-construction certificate must bound the implemented operation in the correct direction.

Transfer the argument

Exercise 13.B1 — Medium: Check whether the target response is expressible

On an open operating region the desired correction is $u_*(T,d)=0.04(T-40)+0.02(d-0.5)$, with coefficients carrying the units needed to produce metres per second. Can an exactly matching map on that region have normalized-input Lipschitz constant at most 0.3?

Review: Sensitivity constraints and expressivity.

Show hint

Rewrite the affine target in terms of $z_1,z_2$ and compute its gradient norm.

Show worked solution

Because $T-40=10z_1$ and $d-0.5=0.1z_2$, the target is $u_*(z)=0.4z_1+0.002z_2$. Its gradient norm is $\sqrt{0.4^2+0.002^2}\approx0.400005\,\mathrm{m/s}$ per normalized input unit. On an open region, increments in that gradient direction attain this slope, exceeding 0.3. Exact matching is impossible under the chosen budget. Options include reducing the target response or revisiting the downstream sensitivity allowance; changing the training loss cannot remove the contradiction.

Exercise 13.B2 — Hard: Reserve gain for approximate orthogonality

Replace the exact orthogonal stage by six stages. For each implemented matrix $\widehat Q_i$, an exact orthogonal $Q_i$ exists with $\|\widehat Q_i-Q_i\|_2\le0.002$; activations between stages remain 1-Lipschitz. Find an external scale sufficient to retain overall gain at most 0.3.

Review: Implemented orthogonal-layer bounds.

Show hint

Use the triangle inequality for each matrix norm and multiply the six stage bounds.

Show worked solution

Every implemented norm is at most $\|Q_i\|_2+0.002=1.002$. The stack gain is therefore at most $1.002^6\approx1.012060$. Choose $c\le0.3/(1.002^6)\approx0.296425\,\mathrm{m/s}$. Then $c\prod_i\|\widehat Q_i\|_2\le0.3$, restoring the earlier sensor-error budget. This is a sufficient worst-case allowance; errors need not align to attain it. It relies on the stated operator-error bounds, not merely small entrywise rounding errors or an informal claim that the matrices remain almost orthogonal.

Synthesis and bridge

Architectural constraints turn a certificate into a maintained design invariant, but only for the mathematical architecture and the implemented operations actually covered. Preprocessing and output scaling remain part of the end-to-end map. The target-gradient calculation provides a useful feasibility check before committing substantial training effort.

The next chapter places the neural module inside feedback. There the central question changes from how far a function output can move to whether repeated interaction with the physical plant converges, remains inside an operating set, or stays bounded under disturbances.

Exercises

Readiness check

Check one Cayley rotation and one scalar Sandwich layer without software. Explain the difference between a certificate valid for every parameter and a penalty that merely encourages a small norm.

If this is difficult, revisit the earlier prerequisite, work the Easy questions, and return to the linked section. You can postpone the advanced extensions while building confidence with the core certificate.

Graded practice: build the calculation, then audit the claim

These twelve new questions each include an independent hint and a fully worked solution. The difficulty measures the amount of reasoning, not the amount of notation. The original research exercises follow below.

Easy: read definitions and compute

Easy 1 — Normalize with the correct side of an estimate

$W=\operatorname{diag}(3,1)$ and a power method estimates its norm as $2.5$. Compute $\|W/2.5\|_2$. Instead divide by the Frobenius upper bound $\sqrt{10}$; what norm results?

Review if needed: Primer A: spectral norms and products; this module's relevant section.

Hint

The largest singular value of a positive diagonal matrix is its largest entry.

Worked solution

Step 1. The exact spectral norm is $3$. Dividing by $2.5$ gives norm $3/2.5=1.2$, so the stored layer is expansive.

Step 2. The Frobenius norm is $\sqrt{3^2+1^2}=\sqrt{10}\ge3$. Dividing by it gives spectral norm $3/\sqrt{10}\approx0.948683\le1$.

Step 3. A lower estimate is useful for monitoring, but a certified normalization denominator must be a proven upper bound (or the exact norm).

Easy 2 — A Cayley rotation by hand

For $A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$ compute $Q=(I-A)(I+A)^{-1}$ and $Q(3,-4)^\top$. Check the input and output norms.

Review if needed: Primer A: matrix arithmetic; this module's relevant section.

Hint

The inverse of $I+A$ is $\tfrac12\begin{bmatrix}1&-1\\1&1\end{bmatrix}$.

Worked solution

Step 1. Multiplying gives $Q=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$. Its columns are orthonormal, so $Q^\top Q=I$.

Step 2. $Q(3,-4)^\top=(4,3)^\top$. Both norms are $\sqrt{3^2+4^2}=5$.

Step 3. This is a right-angle rotation. The skew-symmetric free matrix generated a norm-preserving weight by construction, without a norm penalty.

Easy 3 — Forward preservation and backward preservation

Let $W=(1,0)^\top$. Check $W^\top W$ and $WW^\top$. Does it preserve the norm of scalar inputs? Does it preserve every output-gradient norm under $g\mapsto W^\top g$?

Review if needed: Primer A: singular values and rectangular maps; this module's relevant section.

Hint

Test the backward map on $g=(0,2)^\top$.

Worked solution

Step 1. $W^\top W=[1]$, but $WW^\top=\operatorname{diag}(1,0)\ne I_2$. The matrix has orthonormal columns and lacks orthonormal rows.

Step 2. For a scalar $x$, $Wx=(x,0)^\top$, so $\|Wx\|=|x|$. Forward norms are preserved.

Step 3. For $g=(0,2)^\top$, $W^\top g=0$ while $\|g\|=2$. Backward preservation for every $g$ requires $WW^\top=I$, a different condition for rectangular matrices.

Easy 4 — Include input preprocessing in the radius

A logit network has certified bound $\gamma=0.8$ in normalized coordinates. Raw inputs are normalized with coordinate standard deviations $0.5$ and $0.25$. Find a raw-input bound and the strict Euclidean radius when the raw input has margin $0.4$.

Review if needed: Primer A: spectral norms and products; this module's relevant section.

Hint

The normalization matrix is $\operatorname{diag}(2,4)$; constants subtracted from inputs cancel in differences.

Worked solution

Step 1. The normalization map has spectral norm $4$. The composite raw-input logit map is therefore at most $4(0.8)=3.2$-Lipschitz.

Step 2. The standard logit-margin rule gives $r=0.4/(\sqrt2\cdot3.2)=1/(8\sqrt2)\approx0.0883883$.

Step 3. This certifies $\|\delta\|_2\lt r$ measured in raw input units. Using $0.8$ directly would claim a radius four times too large.

Medium: connect two or three steps

Medium 1 — An analytic AOL scaling

For $P=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ form $T_{ii}=\sum_j|(P^\top P)_{ij}|$ and $D=T^{-1/2}$. Verify $\|PD\|_2\le1$ by computing the eigenvalues of $DP^\top PD$.

Review if needed: Primer A: quadratic forms and PSD order; this module's relevant section.

Hint

The Gram matrix is $\begin{bmatrix}1&1\\1&2\end{bmatrix}$, so the row sums are $2$ and $3$.

Worked solution

Step 1. $T=\operatorname{diag}(2,3)$ and $D=\operatorname{diag}(1/\sqrt2,1/\sqrt3)$. Thus $DP^\top PD=\begin{bmatrix}1/2&1/\sqrt6\\1/\sqrt6&2/3\end{bmatrix}$.

Step 2. This symmetric matrix has trace $7/6$ and determinant $1/6$. Its eigenvalues are $1$ and $1/6$, since these have the stated sum and product.

Step 3. The squared singular values of $PD$ are those eigenvalues, so $\|PD\|_2=1$. Equivalently $T-P^\top P=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\succeq0$ gives the certificate directly.

Medium 2 — A residual layer with a sharp condition

In scalar SLL let $W=2$, $b=0$, and $h(x)=x-2WT^{-1}\operatorname{ReLU}(W^\top x)$. Compare $T=4$ and $T=2$. Compute the piecewise slopes in each case.

Review if needed: Primer B: derivatives and directional slopes; this module's relevant section.

Hint

The sufficient condition is $T\ge W^\top W=4$.

Worked solution

Step 1. With $T=4$, the coefficient $2W/T=1$ gives $h(x)=x-\operatorname{ReLU}(2x)$. It equals $x$ on $x\le0$ and $-x$ on $x\ge0$, or $-|x|$.

Step 2. The slopes are $1$ and $-1$, so the map is exactly 1-Lipschitz. The certificate condition holds at equality.

Step 3. With $T=2$, $h(x)=x-2\operatorname{ReLU}(2x)$ has slopes $1$ and $-3$. Its Lipschitz constant is $3$. Invertibility and positivity of $T$ alone are insufficient; $T\ge W^\top W$ matters.

Medium 3 — A scalar sandwich and its multiplier

Use $A=3/5$, $B=4/5$, $\Psi=2$, $b=0$, and ReLU. Find the input coefficient $U=\sqrt2\Psi^{-1}B$, output coefficient $Y=\sqrt2A\Psi$, and exact layer gain. Check $2\Lambda-Y^2-\Lambda^2U^2$ for $\Lambda=4$ and for the mistaken $\Lambda=2$.

Review if needed: Primer A: quadratic forms and PSD order; this module's relevant section.

Hint

$A^2+B^2=1$. The positive scalar coefficients let you move them through ReLU.

Worked solution

Step 1. $U=2\sqrt2/5$ and $Y=6\sqrt2/5$. Thus $Y\operatorname{ReLU}(Ux)=(24/25)\operatorname{ReLU}(x)$, with exact gain $24/25=0.96$.

Step 2. The correct multiplier is $\Lambda=\Psi^2=4$. Here $U^2=8/25$, $Y^2=72/25$, and $8-72/25-16(8/25)=0$. This matches the boundary identity.

Step 3. For $\Lambda=2$, the same expression is $4-72/25-4(8/25)=-4/25$. The wrong multiplier fails this certificate although the layer itself remains 1-Lipschitz.

Medium 4 — A finite Gramian without an infinite sum

Let $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $B=(0,1)^\top$. Compute $X=\sum_{k=0}^\infty A^kBB^\top(A^\top)^k$. Check $X=AXA^\top+BB^\top$ and whether $X^{-1}$ exists.

Review if needed: Primer D: realizations and controllability Gramians; this module's relevant section.

Hint

$A^2=0$, so only $k=0,1$ can contribute.

Worked solution

Step 1. $BB^\top=\operatorname{diag}(0,1)$ and $AB=(1,0)^\top$, so $ABB^\top A^\top=\operatorname{diag}(1,0)$. All later terms vanish and $X=I_2$.

Step 2. $AXA^\top=AA^\top=\operatorname{diag}(1,0)$; adding $BB^\top$ returns $I_2=X$. This is the discrete controllability-Gramian equation.

Step 3. $X$ is positive definite and invertible, with $X^{-1}=I_2$. Nilpotence makes the sum finite; controllability, visible from the independent columns $B,AB$, supplies positive definiteness.

Hard: combine calculations with assumptions

Hard 1 — Completeness depends on the certificate set

A direct parameterization reaches every network certified by a particular LMI at bound $1$. A network has true constant $1$ but the minimum bound from that LMI is $2$. Must the parameterization represent that network at bound $1$? Use $\operatorname{ReLU}(x)-\operatorname{ReLU}(-x)$ as the concrete case for diagonal LipSDP.

Review if needed: Module 12: the LipSDP certificate; this module's relevant section.

Hint

Distinguish the set of truly Lipschitz networks from the subset that the LMI can certify.

Worked solution

Step 1. Completeness says that the image equals the LMI certificate set. It does not say that the certificate set equals all networks with true constant at most $1$.

Step 2. The example computes $x$ and has exact constant $1$. Its weights also admit the independent all-active slope pattern, with slope $2$; diagonal LipSDP must cover that pattern and cannot return a bound below $2$.

Step 3. Therefore this weight representation is outside the LipSDP certificate set at bound $1$, and completeness does not require reaching it there. The same function may have another, simpler certified representation; completeness concerns the stated weight/certificate set.

Hard 2 — Coupled layers can have large ordinary norms

Take identity activations and weights $W_0=\operatorname{diag}(4,1/4)$, $W_1=\operatorname{diag}(1/4,4)$. Compute their norm product and the composite gain. Find $X_1$ so $W_0^\top X_1W_0=I$ and $W_1^\top W_1=X_1$.

Review if needed: Primer A: quadratic forms and PSD order; this module's relevant section.

Hint

The second layer reverses the first layer's directional scaling.

Worked solution

Step 1. Both spectral norms are $4$, so their product is $16$. Yet $W_1W_0=I$, and the composite gain is exactly $1$. Identity activation is slope-restricted in $[0,1]$.

Step 2. Set $X_1=\operatorname{diag}(1/16,16)$. Direct multiplication gives $W_0^\top X_1W_0=I$ and $W_1^\top W_1=X_1$.

Step 3. The intermediate weighted energy is therefore equal to input energy and output energy. This example explains why coupling directions through a certificate can prove gain $1$ while enforcing ordinary norm at most $1$ for each stored layer would exclude these weights.

Hard 3 — Carry a truncation factor through depth

For $K=\begin{bmatrix}0&-0.2\\0.2&0\end{bmatrix}$ deploy $S_2=I+K+K^2/2$ instead of $e^K$. Compute its exact norm and compare it with $1+0.2^3/6$. What bound follows for a composition of ten such layers and 1-Lipschitz activations?

Review if needed: Primer A: spectral norms and products; this module's relevant section.

Hint

$K^2=-0.04I$, so $S_2=0.98I+K$ and $S_2^\top S_2$ is diagonal.

Worked solution

Step 1. $S_2^\top S_2=(0.98^2+0.2^2)I=1.0004I$, so $\|S_2\|_2=\sqrt{1.0004}\approx1.00019998$, slightly above one.

Step 2. The skew-exponential remainder bound is $\|e^K-S_2\|_2\le0.2^3/6\approx0.00133333$. Since $e^K$ is orthogonal, $\|S_2\|_2\le1+0.2^3/6$. The rounded bound $\|S_2\|_2\le1.00133333$ is also valid here by the exact norm in Step 1.

Step 3. For ten layers the remainder-based product bound is $(1+0.2^3/6)^{10}\approx1.01341362$. Exact layer norms give $(\sqrt{1.0004})^{10}\approx1.00200160$. Intermediate activations can reduce the true gain, but calling the truncated network exactly 1-Lipschitz would be unjustified.

Hard 4 — Contraction time and a nonzero initial storage

A recurrent model satisfies $\|\Delta x_t\|_P\le0.8^t\|\Delta x_0\|_P$ for equal inputs. If the initial distance is $3$, find the first integer time at which this bound is at most $0.1$. Separately, a dissipativity certificate gives $\sum_t\|\Delta y_t\|^2\le V_0+4\sum_t\|\Delta u_t\|^2$. If $V_0=4$ and input energy is $9$, what output-energy bound follows?

Review if needed: Primer D: Lyapunov stability and invariant sets; this module's relevant section.

Hint

Take logarithms carefully: $\log0.8\lt0$. For the second question keep the initial storage term.

Worked solution

Step 1. $3(0.8)^t\le0.1$ is equivalent to $t\ge\log(1/30)/\log0.8\approx15.24$, so the first integer is $16$. Indeed the bounds at $15$ and $16$ are about $0.105553$ and $0.0844425$.

Step 2. The output energy is at most $4+4\cdot9=40$, giving output sequence norm at most $\sqrt{40}\approx6.32456$.

Step 3. The pure incremental gain rule $\|\Delta y\|_{\ell_2}\le2\|\Delta u\|_{\ell_2}=6$ requires zero initial storage (for example equal initial states). Contraction under equal inputs and a gain statement under differing inputs answer different questions.

Original research exercises

Continue here when the core calculations and the certificate assumptions are clear. Use the graded questions above as a warm-up; the original derivations below remain available in full.

Exercise 13.1 — The Cayley transform produces orthogonal matrices (proof)

(a) Let $A^{\top}=-A$. Prove that $I+A$ is invertible, that $Q=(I-A)(I+A)^{-1}$ is orthogonal with $\det Q=1$, and that $-1$ is not an eigenvalue of $Q$. (b) Compute $Q$ for $A=\begin{bmatrix}0&a\\-a&0\end{bmatrix}$, identify the rotation angle, evaluate it for $a=0.5$, and name the orthogonal $2\times2$ matrices that are never produced.

Show answer

(a) $x^{\top}Ax$ is a scalar equal to its transpose $x^{\top}A^{\top}x=-x^{\top}Ax$, so it is $0$ and $x^{\top}(I+A)x=\|x\|^2\gt0$ for $x\ne0$: $I+A$ has trivial kernel. $I\pm A$ commute (polynomials in $A$) and $(I+A)^{\top}=I-A$, so $Q^{\top}Q=(I-A)^{-1}(I+A)(I-A)(I+A)^{-1}=(I-A)^{-1}(I-A)(I+A)(I+A)^{-1}=I$. Next, $\det Q=\det(I-A)/\det(I+A)=\det\big((I+A)^{\top}\big)/\det(I+A)=1$. If $Qv=-v$, put $w=(I+A)^{-1}v\ne0$: $(I-A)w=-(I+A)w$ gives $2w=0$, impossible.

(b) $(I+A)^{-1}=\frac{1}{1+a^2}\begin{bmatrix}1&-a\\a&1\end{bmatrix}$ and $I-A=\begin{bmatrix}1&-a\\a&1\end{bmatrix}$, so $Q=\frac{1}{1+a^2}\begin{bmatrix}1-a^2&-2a\\2a&1-a^2\end{bmatrix}$. With $a=\tan(\theta/2)$: $\cos\theta=\frac{1-a^2}{1+a^2}$, $\sin\theta=\frac{2a}{1+a^2}$, a rotation by $\theta=2\arctan a\in(-\pi,\pi)$. For $a=0.5$: $Q=\begin{bmatrix}0.6&-0.8\\0.8&0.6\end{bmatrix}$, $\theta\approx53.13^\circ$. Never produced: the rotation by $\pi$ ($Q=-I$) and all reflections ($\det Q=-1$; they always have eigenvalue $-1$).

Exercise 13.2 — SLL satisfies the LipSDP-type condition (derivation)

Let $W\in\mathbb R^{m\times n}$, $T=\operatorname{diag}(t_i)\succ0$ with $W^{\top}W\preceq T$, and $h(x)=x-2WT^{-1}\varphi(W^{\top}x+b)$ with $\varphi$ slope-restricted in $[0,1]$. (a) Using the incremental QC with multiplier $\Lambda=2T^{-1}$, prove $\|\Delta x\|^2-\|\Delta h\|^2\ge4\Delta\varphi^{\top}T^{-1}(T-W^{\top}W)T^{-1}\Delta\varphi\ge0$. (b) Prove that $T_{ii}=\sum_j|W^{\top}W|_{ij}\,q_j/q_i$ ($q_i\gt0$) satisfies $W^{\top}W\preceq T$. (c) For $W=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ compute $T$ for $q=(1,1)$ and $q=(1,2)$ and check $T-W^{\top}W\succeq0$.

Show answer

(a) $\|\Delta h\|^2=\|\Delta x\|^2-4\Delta x^{\top}WT^{-1}\Delta\varphi+4\Delta\varphi^{\top}T^{-1}W^{\top}WT^{-1}\Delta\varphi$, and the QC with $\Lambda=2T^{-1}$ reads $q:=4\Delta\varphi^{\top}T^{-1}(W^{\top}\Delta x-\Delta\varphi)\ge0$. Hence $$\begin{aligned}\|\Delta x\|^2-\|\Delta h\|^2&=4\Delta\varphi^{\top}T^{-1}W^{\top}\Delta x-4\Delta\varphi^{\top}T^{-1}W^{\top}WT^{-1}\Delta\varphi\\&=q+4\Delta\varphi^{\top}T^{-1}\big(T-W^{\top}W\big)T^{-1}\Delta\varphi\ \ge\ 0 .\end{aligned}$$

(b) With $Q=\operatorname{diag}(1/q_i)$, $T-QW^{\top}WQ^{-1}=Q(T-W^{\top}W)Q^{-1}$ is similar to the symmetric $T-W^{\top}W$. Its entries are $T_{ii}\delta_{ij}-(W^{\top}W)_{ij}q_j/q_i$; since $(W^{\top}W)_{ii}\ge0$, the diagonal entry of row $i$ is $\sum_{j\ne i}|W^{\top}W|_{ij}q_j/q_i$, exactly the sum of the absolute off-diagonal entries of that row. By Gershgorin every eigenvalue $z$ of a matrix $M$ satisfies $|z-M_{ii}|\le\sum_{j\ne i}|M_{ij}|$ for some row $i$; for $M=Q(T-W^{\top}W)Q^{-1}$ every disc is centred at a nonnegative number with radius equal to it, so its real points lie in $[0,2M_{ii}]$. Similar matrices have the same eigenvalues, and those of the symmetric $T-W^{\top}W$ are real, so they are all $\ge0$.

(c) $W^{\top}W=\begin{bmatrix}1&1\\1&2\end{bmatrix}$. For $q=(1,1)$ (AOL): $T=\operatorname{diag}(2,3)$, $T-W^{\top}W=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$, eigenvalues $0,2$. For $q=(1,2)$: $T=\operatorname{diag}(3,2.5)$, $T-W^{\top}W=\begin{bmatrix}2&-1\\-1&0.5\end{bmatrix}$, determinant $0$, trace $2.5$, eigenvalues $0,2.5$. Both are PSD and tight in one direction; neither dominates ($T_{11}$ is smaller for $q=(1,1)$, $T_{22}$ for $q=(1,2)$), which is why SLL learns $q$. SN would use $T=\|W\|^2I=\tfrac{3+\sqrt5}{2}I\approx2.618\,I$.

Exercise 13.3 — The sandwich LMI identity and the factor of 2 (derivation)

Let $AA^{\top}+BB^{\top}=I$, $\Psi\succ0$ diagonal, and write the sandwich layer as $v=Uh_{\rm in}+b$, $z=\varphi(v)$, $h_{\rm out}=Yz$ with $U=\sqrt2\Psi^{-1}B$, $Y=\sqrt2A^{\top}\Psi$. (a) Show $2\Lambda-Y^{\top}Y-\Lambda UU^{\top}\Lambda=0$ for $\Lambda=\Psi^2$ and conclude that $\begin{bmatrix}I&-U^{\top}\Lambda&0\\-\Lambda U&2\Lambda&-Y^{\top}\\0&-Y&I\end{bmatrix}\succeq0$, so the layer is 1-Lipschitz. (b) Evaluate the same expression for $\Lambda=\tfrac12\Psi^2$. (c) Show that for one hidden layer the $\gamma$-form LMI with multiplier $\Lambda$ is equivalent to LipSDP's $M(\rho,T)\preceq0$ ($\alpha=0$, $\beta=1$) with $\rho=\gamma^2$, $T=\gamma\Lambda$.

Show answer

(a) $Y^{\top}Y=2\Psi AA^{\top}\Psi$ and $\Lambda UU^{\top}\Lambda=\Psi^2\cdot2\Psi^{-1}BB^{\top}\Psi^{-1}\cdot\Psi^2=2\Psi BB^{\top}\Psi$, so the expression is $2\Psi(I-AA^{\top}-BB^{\top})\Psi=0$. It is the Schur complement of the $3\times3$ matrix with respect to its two uncoupled identity blocks, so that matrix is PSD. It is Wang & Manchester's LMI with $\gamma=1$, no hidden-to-hidden weights, input matrix $U$ and output matrix $Y$, and the incremental QC gives $\|\Delta x\|^2-\|\Delta h_{\rm out}\|^2\ge2\Delta z^{\top}\Lambda(\Delta v-\Delta z)\ge0$.

(b) $\Psi^2-2\Psi AA^{\top}\Psi-\tfrac12\Psi BB^{\top}\Psi=\Psi\big(\tfrac12I-\tfrac32AA^{\top}\big)\Psi$ using $BB^{\top}=I-AA^{\top}$. This is negative definite whenever $AA^{\top}\succ\tfrac13I$; for $B=0$ and orthogonal $A$ it is $-\Psi^2$. So $\tfrac12\Psi^2$ is not a certificate for the LMI as written; $\Psi^2$ is.

(c) For $\gamma\gt0$, the Schur complement on the last block gives $H\succeq0\iff\begin{bmatrix}\gamma I&-W_0^{\top}\Lambda\\-\Lambda W_0&2\Lambda-\tfrac1\gamma W_1^{\top}W_1\end{bmatrix}\succeq0$. Multiplying by $\gamma$ and negating gives $\begin{bmatrix}-\gamma^2I&W_0^{\top}(\gamma\Lambda)\\(\gamma\Lambda)W_0&-2\gamma\Lambda+W_1^{\top}W_1\end{bmatrix}\preceq0$, which is $M(\rho,T)=\begin{bmatrix}-\rho I&W_0^{\top}T\\TW_0&-2T+W_1^{\top}W_1\end{bmatrix}\preceq0$ with $\rho=\gamma^2$, $T=\gamma\Lambda$; both certify $\sqrt\rho=\gamma$.

Exercise 13.4 — Controllability Gramian of a 2-tap FIR filter (computation)

A single-channel convolution $y_k=K_0w_k+K_1w_{k-1}$ (no activation). (a) Give a state-space realization and check that $A$ is nilpotent. (b) With input gain $Q_0=\rho^2$, free $H=h$ and $\varepsilon\ge0$, compute $X$ of Lemma 7, $P=X^{-1}$ and $F$. (c) For $\rho=h=1$, $\varepsilon=0$ and the unit vector $\tilde u=(\cos\theta,\sin\theta)^\top$ with $\theta=\pi/3$ (as produced by a Cayley transform), set $[C\ \ D]=[K_1\ \ K_0]=\tilde u^{\top}\operatorname{chol}(F)$ and compute the kernel (this is the linear-layer version of Theorem 8: without an activation there is no multiplier, the certificate is just $F-\hat C^{\top}\hat C\succeq0$, and no $\Gamma$ is needed). (d) Verify the gain bound via the dissipation LMI $F-\hat C^{\top}\hat C\succeq0$ and via $\sup_\omega|K_0+K_1e^{-j\omega}|$. (e) Show $|K_0|+|K_1|\le\rho$ for every $\theta$ and admissible $P$; when is it tight?

Show answer

(a) State $x_k=w_{k-1}$: $A=0$, $B=1$, $C=K_1$, $D=K_0$ ($\ell=2$, one state). $A^1=0$: nilpotent of index $\ell-1=1$.

(b) Only the $k=0$ term survives: $X=1/\rho^2+h^2+\varepsilon$, $P=\rho^2/\big(1+\rho^2(h^2+\varepsilon)\big)$, and $F=\operatorname{diag}(P,\ \rho^2-P)$. Whenever $h^2+\varepsilon\gt0$ (always under Lemma 7's $\varepsilon\gt0$, and here also for $\varepsilon=0$, $h\ne0$) we get $P\in(0,\rho^2)$ and $F\succ0$, as Lemma 7 promises; for $h=\varepsilon=0$, $P=\rho^2$ and $F$ is only semidefinite, which is why the lemma needs $\varepsilon\gt0$.

(c) $X=2$, $P=0.5$, $F=\operatorname{diag}(0.5,0.5)$, $\operatorname{chol}(F)=\tfrac1{\sqrt2}I\approx0.7071\,I$, so $K_1=\tfrac{\cos(\pi/3)}{\sqrt2}=\tfrac{\sqrt2}{4}\approx0.3536$ and $K_0=\tfrac{\sin(\pi/3)}{\sqrt2}=\tfrac{\sqrt6}{4}\approx0.6124$.

(d) $F-\hat C^{\top}\hat C=\begin{bmatrix}3/8&-\sqrt3/8\\-\sqrt3/8&1/8\end{bmatrix}\approx\begin{bmatrix}0.375&-0.2165\\-0.2165&0.125\end{bmatrix}$. The exact matrix has determinant $0$ and trace $0.5$, eigenvalues $0$ and $0.5$: PSD. This holds for every unit $\tilde u$: with $F=L^{\top}L$ and $\hat C=\tilde u^{\top}L$, $F-\hat C^{\top}\hat C=L^{\top}(I-\tilde u\tilde u^{\top})L\succeq0$. This is the dissipation inequality $V(x_{k+1})-V(x_k)\le\rho^2w_k^2-y_k^2$ with $V(x)=Px^2$ (the bounded-real lemma), and summing from $x_0=0$ gives $\sum y_k^2\le\rho^2\sum w_k^2$. In frequency, $|K_0+K_1e^{-j\omega}|\le|K_0|+|K_1|=\tfrac{\sqrt2+\sqrt6}{4}\approx0.9659\le\rho=1$, attained at $\omega=0$.

(e) $|K_0|+|K_1|=|\sin\theta|\sqrt{\rho^2-P}+|\cos\theta|\sqrt P\le\sqrt{\sin^2\theta+\cos^2\theta}\,\sqrt{\rho^2-P+P}=\rho$ by Cauchy–Schwarz: the parameterization cannot violate the bound. Equality iff $\tan^2\theta=(\rho^2-P)/P$; for $P=\rho^2/2$ (here) that is $\theta=\pi/4$, giving $K_0=K_1=0.5$ and gain exactly 1.

Exercise 13.5 — Converting certified radii between conventions (computation)

A 3-class network has logits $f(x)=(4.0,\,3.1,\,0.6)$ and predicts class 1. It is a 1-Lipschitz feature map followed by a linear layer with $\|W\|\le L=1$ and $\|w_1-w_2\|=1.2$, $\|w_1-w_3\|=1.35$. (a) Compute $M_f/(\sqrt2L)$. (b) Compute LipNeXt's one-versus-rest radius $\min_j|f_1-f_j|/K_j$ with $K_j=\sqrt2L$ and with $K_j=\|w_1-w_j\|$. (c) Prove $\|w_1-w_j\|\le\sqrt2\|W\|$. (d) Which radii certify $\varepsilon=36/255$ and $72/255$ for inputs in $[0,1]$? Repeat if the certificate was computed for normalized inputs $\tilde x=(x-\mu)/0.25$.

Show answer

(a) $M_f=4.0-3.1=0.9$, radius $0.9/\sqrt2\approx0.6364$.

(b) With $K_j=\sqrt2$: $\min(0.9,3.4)/\sqrt2\approx0.6364$, identical to (a). With pairwise constants: $\min(0.9/1.2,\ 3.4/1.35)=\min(3/4,\ 68/27)=3/4=0.750$ (with $68/27\approx2.519$), about 18% larger.

(c) $w_1-w_j=W^{\top}(e_1-e_j)$, so $\|w_1-w_j\|\le\|W\|\,\|e_1-e_j\|=\sqrt2\|W\|\le\sqrt2L$: pairwise radii are never smaller than the global one.

(d) $36/255\approx0.1412$ and $72/255\approx0.2824$, so both radii certify both levels. With normalization $\|\Delta\tilde x\|=4\|\Delta x\|$, so a radius $r$ in $\tilde x$ is $r/4$ in pixel units: $0.9/(4\sqrt2)\approx0.159$ and $3/16=0.1875$, which certify $36/255$ but not $72/255$. This illustrates the same normalization effect as the "about $4.1\gamma$" reported for Wang & Manchester's normalization layer; the factor in this example is exactly $4$.

Key Papers

PaperVenueContributionWhy read it
Wang & Manchester — Direct Parameterization of Lipschitz-Bounded Deep NetworksICML 2023Sandwich layer; smooth map onto all LipSDP-certified feedforward networks; FFT convolutions.The walkthrough; read the appendix proofs for the exact multiplier.
Araujo, Havens, Delattre, Allauzen & Hu — A Unified Algebraic Perspective on Lipschitz Neural NetworksICLR 2023$W^{\top}W\preceq T$ unifies SN, orthogonal, AOL, CPL; SLL via Gershgorin.Shortest bridge from LipSDP to layer design.
Pauli, Wang, Manchester & Allgöwer — Lipschitz-Bounded 1D Convolutional Neural Networks using the Cayley Transform and the Controllability GramianIEEE CDC 2023Direct parameterization of 1-D CNNs: Gramian storage, Cayley kernels.Every step a Schur complement; the model for Section 4.
Pauli, Wang, Manchester & Allgöwer — LipKernel: Lipschitz-Bounded Convolutional Neural Networks via Dissipative LayersAutomatica 188, 2026Incrementally dissipative 2-D layers via Roesser models; kernels in standard form.Reported inference 2–3 orders of magnitude faster than Fourier layers.
Revay, Wang & Manchester — Recurrent Equilibrium Networks: Flexible Dynamic Models With Guaranteed Stability and RobustnessIEEE TAC 69(5), 2024Contracting, IQC-robust recurrent models for every parameter value.The dynamic counterpart; basis of certified policies.
Barbara, Wang & Manchester — R2DN: Scalable Parameterization of Contracting and Lipschitz Recurrent Deep NetworksCDC 2026 (accepted)LTI system in feedback with a 1-Lipschitz DNN; no equilibrium solve.Up to 10× faster than RENs, same guarantees.
Manchester, Wang & Barbara — Neural Networks in the Loop: Learning with Stability and Robustness GuaranteesAnnu. Rev. Control Robot. Auton. Syst. 9, 2026Survey: IQC certificates plus direct parameterizations.Best overview of this module and the next.
Trockman & Kolter — Orthogonalizing Convolutional Layers with the Cayley TransformICLR 2021Cayley transform per FFT frequency: orthogonal convolutions.The Orthogon baseline and the FFT trick.
Anil, Lucas & Grosse — Sorting out Lipschitz function approximationICML 2019Gradient-norm preservation; GroupSort/MaxMin; universality.Why norm-constrained ReLU nets are weak.
Prach & Lampert — Almost-Orthogonal Layers for Efficient General-Purpose Lipschitz NetworksECCV 2022Diagonal rescaling makes any layer 1-Lipschitz.Simplest certified layer.
Singla & Feizi — Skew Orthogonal ConvolutionsICML 2021Exponential of skew-symmetric convolutions with an error bound.The exponential-map alternative.
Miyato, Kataoka, Koyama & Yoshida — Spectral Normalization for Generative Adversarial NetworksICLR 2018Per-layer normalization by a power-iteration estimate.The cheap baseline and its pitfalls.
Hu, Hu & Fredrikson — LipNeXt: Scaling up Lipschitz-based Certified Robustness to Billion-parameter ModelsICLR 2026Manifold-optimized orthogonal weights, spatial shifts, 1–2B parameters.2026 state of the art, deterministic $\ell_2$.
Lai, Huang, Kung & Chen — Enhancing Certified Robustness via Block Reflector Orthogonal Layers and Logit Annealing LossICML 2025Block-reflector orthogonal layers; logit annealing loss.Previous state of the art.
Carlini, Tramèr, Dvijotham, Rice, Sun & Kolter — (Certified!!) Adversarial Robustness for Free!ICLR 2023Diffusion denoised smoothing with off-the-shelf models.The probabilistic competitor at large radii.

Flashcards