It is probably an easy question, but somehow I am stuck.


**Question** Is the following statement true? If yes, how to prove it?



> 
> Suppose that $f\in C^1(\mathbb{R}^n)$ is convex and
> $$
> \langle\nabla f(x)-\nabla f(y),x-y\rangle \leq L|x-y|^2
> $$
> for some $L>0$ and all $x,y\in\mathbb{R}^n$. Does it follow that
> $$
> |\nabla f(x)-\nabla f(y)|\leq L|x-y|
> $$
> for all $x,y\in\mathbb{R}^n$?
> 
> 
> 



Convexity and Lipschitz continuity
That's a standard result in convex optimization. For example Theorem 2.1.5 in Nesterov's "Introductory Lectures on Convex Optimization" states that the following are equivalent:


* $f$ is $C^1$, convex and the gradient $\nabla f$ is $L$-Lipschitz
* for all $x,y$: $0\leq f(y) - f(x) - \langle\nabla f(x),y-x\rangle \leq \tfrac{L}2
\|x-y\|^2$
* for all $x,y$: $\tfrac1L\|\nabla f(x)-\nabla f(y)\|^2 \leq \langle\nabla f(x)-\nabla f(y),x-y\rangle$
* for all $x,y$: $\langle\nabla f(x)-\nabla f(y),x-y\rangle \leq L\|x-y\|^2$


(In case you are interested: The proof there is directly for $C^1$ functions and no second derivatives are used at intermediate steps.)


