Continuously differentiable function of a convex set

Question

I have an analysis problem that I am trying to work through.

The problem:

Let $V$ be a convex open set in $R^2$ and let $f: V\to R$ be continuously differentiable in V. Show that if there is a positive number $M$ such that $|\nabla f(x)|\leq M$ for all $x\in V$, then there is a positive number $L$ such that $$|f(x)-f(y)|\leq L|x-y|$$ for all $x,y\in V$.

Is this result still true if $V$ is instead assumed to be open and connected? Prove or disprove with a counterexample.

What I have so far:

Since $V$ is a convex open set in $R^2$, $\lambda x + (1-\lambda)y \in V$ for every $\lambda \in (0,1)$ whenever $x,y\in V$. Then by the given information $$|\nabla f(\lambda x +(1-\lambda)y))|\leq M$$.

Since $f$ is continuously differentiable, $$|f(x)-f(y)|\leq |\nabla f(\lambda x +(1-\lambda)y))||x-y|\leq M|x-y|$$.

Taking $M=L$, we see $$|f(x)-f(y)|\leq L|x-y|$$

I'm not sure if I did this correctly. And I am not sure if this applies when V is assumed to be open and connected.

Answer 1

There's a slight issue with your use of quantifiers. Let's look closer at what's going on.

Fix $x,y\in V$. Define a new function $g:[0,1]\to\mathbb{R}$ by $g(t)=f(ty+(1-t)x)$. This is well defined because the set $V$ is convex, so $ty+(1-t)x\in V$ for every $t$ and thus it makes sense to compute $f(ty+(1-t)x)$. The function is differentiable by the chain rule, so, by the Mean Value Theorem, there must be a $c\in (0,1)$ such that $$ g(1)-g(0)=g'(c)$$ By the chain rule, we can compute $$ g'(c)=\nabla f(cy+(1-c)x)\cdot (y-x) $$ By the Cauchy-Schwarz inequality and the bound we are given, $$ |g'(c)|\le M|y-x|. $$ Note that we need the inequality $|\nabla f(cy+(1-c)x)|\le M$. Since we know nothing about the point $cy+(1-c)x$, other than that it lies in the line segment joining $x$ and $y$, we need the inequality $|\nabla f(z)|\le M$ for all $z\in V$, even though we're using it just once.

Since $g(1)=x$ and $g(0)=y$, putting everything together you get the desired inequality.

Regarding the non-convex case, the answer is no, see mean value property of derivatives in high dimensions or Convex function problem and mean value theorem

A remark on the non-convex case: if $V$ is open and connected, it is connected by polygonal arcs, so you can use this result locally. However, since you have no control on the number of straight lines in the polygonal arc. That's not a proof, but a heuristic for why this will fail.