# Calculate Lipschitz constant of continuously differentiable function $\mathbb{R}^n \to \mathbb{R}^m$ [duplicate]

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?

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

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 $$J_f(x)$$).