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

Let DC
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1 Answers1

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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.

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  • Very practical this Lemma. Thanks. – Let DC Jun 23 '15 at 13:42
  • For anyone that tries to better understand this answer in the future, my question has some answers that clarify a bit the proof of the lemma. – Caio Lins Jul 21 '22 at 18:40
  • @LetDC Is there a book that contains the Lemma of this answer? This appears to be stronger, the resulting function is more than locally continuous Lipschitz, it is Lipschitz on bounded intervals. – Ilovemath Jul 02 '24 at 00:46
  • Existence of the supremum should be justified. It is enough to show a convex function is bounded in any $n$-dimensional cube. I don't see how the 1D argument extends to $n$ dimensions, though. There are many such lines, each having its own bounds, how do we know these combine to finite bounds on the entire ball? – AlvinL May 18 '25 at 10:20