Proving lipschitz continuity based on bounded gradient

Question

I would like to prove that if $S$ is a non-empty convex subset, and $f:S \xrightarrow{} R $ is a convex and differentiable function, that the following holds true: $ |f(x)-f(y)| \le L\|x-y \|_{2} \Longleftrightarrow \| \nabla f(x) \|_{2} \le L $

I already have $\rightarrow$, but I am unfinished with $\leftarrow$. The following is what I currently have: From convexity: $f(x) - f(y) \le \nabla f(x)^T(x-y) $

According to Cauchy-Schwartz, the following also holds true:

$ |\nabla f(x)^T(x-y)| \le \|\nabla f(x)\|_{2} \|x-y\|_{2}$

It is given that $\| \nabla f(x) \|_{2} \le L$, so the following is true:

$ |\nabla f(x)^T(x-y)| \le L \|x-y\|_{2} $

Hence, because $\forall a: |a| \ge a$, we have:

$ \nabla f(x)^T(x-y) \le |\nabla f(x)^T(x-y)| \le L \|x-y\|_{2}$

So far I have proven the following: $ f(x)-f(y) \le L\|x-y \|_{2} $

But I am missing the absolute value, I have made some attempts but they all involve making fallacies with inequality signs... Does anyone have an idea?

Answer 1

Hint

Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}\in (x,y)=\{x+t(y-x)\mid t\in (0,1)\}$ s.t. $$f(x)-f(y)=\nabla f(c_{x,y})\cdot (y-x).$$

Answer 2

I can't really tell if these answers other people gave me are very useful or not.

However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) \le \nabla f(y)^T(y-x)$.

You will then end up at $f(x)-f(y) \ge -L\|x-y \|_{2}$, then you will have proven the theorem...