If $f:\mathbb R^n\to\mathbb R$ is twice continuously differentiable, then $\nabla f$ is Lipschitz continuous

Question

Let $f\in C^2(\mathbb R^n)$. I've read that since $f$ is twice continuously differentiable, $\nabla f$ is Lipschitz continuous. Is that really true?

By the mean-value theorem, $$\left\|\nabla f(x)-\nabla f(y)\right\|\le\left\|\int_0^1\nabla^2f(y+t(x-y))\;dt\right\|\left\|x-y\right\|\le\underbrace{\sup_{t\in[0,1]}\left\|\nabla^2f(y+t(x-y))\right\|}_{=:L(x,y)}\left\|x-y\right\|\tag 1$$ for any operator norm $\left\|\;\cdot\;\right\|$ and $x,y\in\mathbb R^n$. However, $L$ doesn't seem to be uniformly bounded. So, I assume that we cannot prove Lipschitz continuity without making further assumptions.

In the book that I read, they further assume that there is a $x^*\in\mathbb R^n$ with $\nabla f(x^*)=0$ such that $\nabla^2f(x^*)$ is invertible. This implies that there is a $\varepsilon>0$ such that $\nabla^2f(x)$ is invertible, too, and $$\left\|\nabla^2f(x)^{-1}\right\|\le C\;,$$ for some $C>0$, for all $x$ in the (open) ball $B_\varepsilon(x^*)$. Is this somehow helpful?

Answer 1

You just have to note that as $t \mapsto\nabla^2f(y+t(x-y))$ is a continuous function on a compact set $[0,1]$

$$\sup_{t\in[0,1]}\left\|\nabla^2f(y+t(x-y))\right\| \leq C(x,y) $$

In fact since $$\sup_{z : |z| \leq K}\left\|\nabla^2f(z)\right\| \leq C_K$$

You obtain that $f$ is Lipschitz continuous on that compact set which means that $f$ is locally Lipschitz continuous.

For a counter example take $f(x) = x^3$, then $f'(x) = 3x^2$ wich is not globally Lipschitz continuous

Answer 2

Firstly, note @user251257's comment; I'll assume we're on a compact set $X$. I'll also just do it for one dimension, but the result is analogous in two (it's just easier for typesetting in one!).

For a differentiable function $g$, we have $\frac{g(x) - g(0)}{x} \to G$ as $x \to 0$; in particular, there exists $L$ such that

$$g(x) - g(0) \le L'x \;\; \forall x \in X',$$

for some $X' \subseteq X$ with $0 \in X'$, compact by assumption (just take some $X'$, then take a compact set inside it if it isn't compact).

Now, away from $0$ in $X \setminus X'$, we can certainly find some $L''$ such that the above inequality holds $\forall x \in X \setminus X'$, since it is a compact set and there exists $\epsilon > 0$ such that $|x| > \epsilon \; \forall x \in X \setminus X'$.

Now just take $L = \max\{L',L''\}$ and you have that the equality holds for all $x \in X$.

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To directly apply this to your question, we just have to define $g$ correctly. In essence, "$g = f'$", but not quite: you'll need to apply this principle at each point. Notice that I've just done it about $0$; you can just change this to the ordinary definition of differentiable everywhere, instead of just at $0$.

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Hopefully this answer is of help! (=