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?
Best Answer
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