Every convex function is locally Lipschitz ($\mathbb{R^n}$)

I know that if $f$ is convex function so $f$ is continuous. And I know too that partial derivatives exists. What can I do?

Solution 1:

You can use this 1D lemma, which gives an estimate for Lipschitz constant:

Lemma: If $f:\mathbb R\to \mathbb R$ is convex, then it is Lipschitz on any interval $[a,b]$ with Lipschitz constant $L$ such that $$L\le 2\max_{[a-1,b+1]}|f|\tag{1}$$

Proof: Let $y=g(x)$ be the equation of some secant line to the graph of $f$, based on two points $x_1,x_2\in [a,b]$. By convexity, $g(a-1)\le f(a-1)$ and $g(b+1)\le f(b+1)$. Hence, the (constant) slope of $g$ satisfies $$ g' = \frac{g(b+1)-g(x_1)}{b+1-x_1}\le \frac{f(b+1)-f(x_1)}{1} \le 2\max_{[a-1,b+1]}|f| $$ and similarly for $g'\ge -2\max_{[a-1,b+1]}|f|$. $\qquad \Box$

With this lemma, the problem reduces to 1D case. Let $M$ be the supremum of $|f|$ on a ball of radius $R+1$ cantered at the origin (denoted $B_{R+1}$). Consider the restriction of $f$ to a line segment contained in $B_R$. By the lemma, this restriction is Lipschitz with constant $2M$. It follows that $f$ is Lipschitz with constant $2M$ on $B_R$, since the Lipschitz condition involves two points, and two points lie on some line. $\qquad \Box$

I posted something of this kind before but within the context of another question.