Prove a function to be Lipschitz

Question

With the condition $$ 0\le f(y)-f(x)-\langle f'(x),y-x\rangle \le \frac{L}{2} \|x-y\|^2, $$ I want to prove $f'(x)$ is Lipschitz with constant $L$ and is convex.

It is easy to see that $f $ is convex from the first inequality. From the second inequality I derived $$ \langle f'(y)-f'(x),y-x\rangle \le L \|y-x\|^2 $$ by exchanging the role of $x$ and $y$ and add the two inequalities. However I don't know what to do next. Could anybody tell me?

Thanks.

Answer 1

EDIT: As it stands, this answer only holds for the 1-dimensional case

so you have

$$ 0<\langle f'(y)-f'(x),y-x\rangle \leq L \|y-x\|^2\\ \Leftrightarrow\\ 0<[f'(y)-f'(x)](y-x) \leq L \|y-x\|^2 $$ and now you do a case study, case $1):y>x$ and case $2):y<x$.

In case $1)$ we get by dividing $(y-x)$ $$ 0<[f'(y)-f'(x)](y-x) \leq L \|y-x\|^2 \\ \Leftrightarrow \\ 0<[f'(y)-f'(x)] \leq L \|y-x\| $$ which is the same as in case $2)$ (because $0<[f'(y)-f'(x)](y-x)$), so you are done.

bests

Answer 2

If the function $f$ is twice differentiable then your condition implies $\|f''(x)\|\le L$ which easily gives the Lipschitz condition on $f'$. Look, for example, here, p. 2.

Otherwise, for a convex differentiable $f$ the following statements are equivalent:

1. $\langle f'(y)-f'(x),y-x\rangle\le L\|y-x\|^2$, $\forall x,y$.
2. $g(x)=\frac{L}{2}\|x\|^2-f(x)$ is convex.
3. $\langle f'(y)-f'(x),y-x\rangle\ge \frac{1}{L}\|f'(y)-f'(x)\|^2$, $\forall x,y$.
4. $\|f'(y)-f'(x)\|\le L\|y-x\|$, $\forall x,y$.

Proof: left as en exercise ;-)