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Suppose I have a continuously differentiable function $f : \mathbb{R}^n \to \mathbb{R}^m$ with a bounded Jacobian $J_f$. Is $f$ Lipschitz continuous? If so, how do I calculate a Lipschitz constant?

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  • $\begingroup$ I searched around math.stackexchange.com for a while without finding a good answer, so I decided to add my question and solution here. $\endgroup$
    – Paul Wintz
    Commented Mar 21, 2023 at 9:00
  • $\begingroup$ That answer is relevant, but is only a special case of this question. $\endgroup$
    – Paul Wintz
    Commented Mar 21, 2023 at 20:43
  • $\begingroup$ Um, how is it not answering your question? That Q is in some sense more general, allowing different convex domains. $\endgroup$ Commented Mar 21, 2023 at 22:01
  • $\begingroup$ Yes, it's more general in that sense, but it only addresses a fixed Lipschitz constant, namely 2. $\endgroup$
    – Paul Wintz
    Commented Mar 21, 2023 at 22:14
  • $\begingroup$ @PaulWintz But there is nothing special about $2$, and the same proof can be easily adapted to any Lipschitz constant. $\endgroup$ Commented Mar 22, 2023 at 16:39

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In Khalil's Nonlinear Systems (Third Edition), Lemma 3.1 (simplified) states:

Let $f:\mathbb{R}^n \to \mathbb{R}^m$ be continuous. Suppose that the Jocobian $J_f$ exists and is continuous. If there is a constant $L \geq 0$ such that $ \left\|J_f(x) \right\| \leq L $ for all $x \in \mathbb{R}^n$, then $\| f(x)-f(y) \| \leq L\|{x-y}\| $ for all $x,y \in \mathbb{R}^n$.

Thus, if $\left\|J_f(x) \right\|$ is bounded, then $f$ is globally Lipschitz. The smallest Lipschitz constant is

$$L = \sup_{x \in \mathbb{R}^n} \left\|J_f(x) \right\|. $$

Note that $\|J_f(x)\|$ is the induced matrix norm of $J_f(x)$. If the $2$-norm is used in the definition of Lipschitz continuity, then the induced matrix norm is the spectral norm (equal to the largest absolute value among the eigenvalues of $J_f(x)$).

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