--- date: '2025-08-21' description: global slope bounds, convex subgradients, and logistic loss id: Lipschitzness modified: 2026-09-07 13:37:25 GMT-04:00 tags: - ml - math title: Lipschitzness created: '2025-08-21' published: '2025-08-21' pageLayout: default slug: thoughts/Lipschitzness permalink: https://aarnphm.xyz/thoughts/Lipschitzness.md generator: quartz: v4.6.0 hostedProvider: Cloudflare baseUrl: aarnphm.xyz full: https://aarnphm.xyz/llms-full.txt --- Let $(\mathcal{X},\|\cdot\|)$ be a normed space, let $D\subseteq\mathcal{X}$, and let $f:D\to\mathbb{R}$. The function is $L$-Lipschitz on $D$ when $$ |f(x)-f(y)|\leq L\|x-y\|\qquad\text{for every }x,y\in D. $$ The constant bounds every secant slope at once. It is global on the stated domain: $e^x$ is not Lipschitz on $\mathbb{R}$, but it is Lipschitz on every bounded interval. > \[!note\] Useful rules > > | Construction | Bound | > | -------------------------------------------- | ------------------------ | > | $f+g$ | $L_f+L_g$ | > | $x\mapsto f(Ax)$ | $L_f\|A\|_{\mathrm{op}}$ | > | $x\mapsto\max_i\{\langle a_i,x\rangle+b_i\}$ | $\max_i\|a_i\|_*$ | > > For $u\in\mathcal{X}^*$, the dual norm is $\|u\|_*=\sup_{\|x\|\leq 1}|\langle u,x\rangle|$. These bounds follow from the triangle inequality and $|\langle u,x\rangle|\leq\|u\|_*\|x\|$. Lipschitz continuity implies uniform continuity. In one dimension, a differentiable function with $|f'(x)|\leq L$ is $L$-Lipschitz by the mean value theorem. The converse has to be stated with weaker derivatives: Lipschitz functions on $\mathbb{R}^n$ can have corners, but Rademacher’s theorem says they are differentiable almost everywhere. \[@cobzas2019lipschitzfunctions\] ## secant picture

## convex functions Assume now that $f:\mathbb{R}^n\to\mathbb{R}$ is closed and convex. Because $f$ is finite everywhere, each point has a subgradient. The following statements are equivalent: 1. $f$ is $L$-Lipschitz. 2. For every $x$, every $g\in\partial f(x)$ satisfies $\|g\|_*\leq L$. 3. The effective domain of the Fenchel conjugate is contained in the closed dual ball: $$ \operatorname{dom}f^*\subseteq\{u:\|u\|_*\leq L\}. $$ One direction follows from the subgradient inequality $$ f(y)\geq f(x)+\langle g,y-x\rangle. $$ If $\|g\|_*\leq L$, this gives $f(x)-f(y)\leq L\|x-y\|$; swap $x$ and $y$ for the absolute-value bound. For the conjugate condition, use $f(x)=\sup_{u\in\operatorname{dom}f^*}\{\langle u,x\rangle-f^*(u)\}$. When that domain lies in the closed dual ball, $f(x)-f(y)\leq\sup_{\|u\|_*\leq L}\langle u,x-y\rangle\leq L\|x-y\|$. \[@rockafellar1970convexanalysis; @bubeck2015convexoptimization\] > \[!warning\] Domain matters > > A constrained convex objective often takes the value $+\infty$ outside its feasible set. It is not a real-valued Lipschitz function on $\mathbb{R}^n$. Restricting the first inequality to its effective domain does not recover the equivalences above. For example, the indicator of a closed convex set is constant on its domain, while its boundary subgradients include an unbounded normal cone. ## function Lipschitzness and smoothness Function Lipschitzness bounds changes in $f$; smoothness bounds changes in $\nabla f$. In Euclidean space, $f$ is $L$-smooth when $$ \|\nabla f(x)-\nabla f(y)\|_2\leq L\|x-y\|_2. $$ For a convex differentiable function this implies $$ f(y)\leq f(x)+\langle\nabla f(x),y-x\rangle+\frac{L}{2}\|y-x\|_2^2 $$ and the Baillon-Haddad inequality $$ \langle\nabla f(x)-\nabla f(y),x-y\rangle \geq \frac{1}{L}\|\nabla f(x)-\nabla f(y)\|_2^2. $$ The inner-product structure matters for this co-coercivity statement; writing it for an arbitrary norm and its dual is not generally valid. \[@bubeck2015convexoptimization\] A $\mu$-strongly convex function satisfies $$ f(y)\geq f(x)+\langle g,y-x\rangle+\frac{\mu}{2}\|y-x\|_2^2, \qquad g\in\partial f(x). $$ When $\mu>0$, such a function cannot also be globally Lipschitz on all of $\mathbb{R}^n$: the quadratic lower bound eventually outruns every linear Lipschitz bound. The two properties can coexist on a bounded domain. ## examples - $f(x)=\|x\|$ is $1$-Lipschitz with respect to the same norm. - $f(x)=\langle a,x\rangle$ is $\|a\|_*$-Lipschitz. - $f(t)=\max(0,1-t)$ is $1$-Lipschitz on $\mathbb{R}$. - $f(z)=\log\sum_i e^{z_i}$ is $1$-Lipschitz with respect to $\|\cdot\|_\infty$, since $\nabla f(z)=\operatorname{softmax}(z)$ has $\ell_1$ norm $1$. - If $Q=Q^\top\succeq0$, then $f(x)=\tfrac12x^\top Qx$ is convex and has a $\|Q\|_2$-Lipschitz gradient. When $Q\neq0$, the function itself is not globally Lipschitz on $\mathbb{R}^n$. ## logistic loss For a logit $t$, binary logistic loss has two equivalent label conventions: $$ \ell_{01}(t;y) =-y\log\sigma(t)-(1-y)\log(1-\sigma(t)), \qquad y\in\{0,1\}, $$ $$ \ell_{\pm}(t;y)=\log(1+e^{-yt}), \qquad y\in\{-1,+1\}. $$ Their first derivatives obey $$ |\partial_t\ell(t;y)|\leq 1. $$ For the $0/1$ form, $\partial_t^2\ell_{01}(t;y)=\sigma(t)(1-\sigma(t))$; for signed labels, $\partial_t^2\ell_{\pm}(t;y)=\sigma(yt)\sigma(-yt)$. Both lie in $[0,1/4]$, so the scalar loss is $1$-Lipschitz in $t$ and its derivative is $1/4$-Lipschitz. For parameters $\theta=(w,b)$, let $\widetilde X$ have rows $(x_i^\top,1)$ and define the mean empirical risk $$ J(\theta)=\frac1n\sum_{i=1}^n\ell(\widetilde x_i^\top\theta;y_i). $$ Then $$ \nabla^2J(\theta)=\frac1n\widetilde X^\top S\widetilde X, \qquad 0\preceq S\preceq\frac14I, $$ so $J$ is $L_J$-smooth with $$ L_J\leq\frac{\|\widetilde X\|_2^2}{4n}. $$ For a summed loss, remove the factor $1/n$. \[@freund2018conditionnumberlogisticregression\] See [[thoughts/Logistic regression#MLE derivation and gradients]], [[thoughts/cross entropy]], [[thoughts/norm]], and [[thoughts/linear map#Operator norm and Lipschitzness]].