Relationship between smoothness and Lipschitzness

Question

Given a convex function $f:\Omega\mapsto\mathbb{R}$ has function value bounded $|f(x)|\leq B$, diameter of the convex domain bounded $\|x-y\|_2\leq D, x,y\in \Omega$ and $\beta$ smoothness: $\|\nabla f(x)-\nabla f(y)\|_2\leq \beta\|x-y\|_2$. How do we determine the Lipschitz constant for $f$? I found in this paper that the Lipschitz constant is at most $\frac{2B}{D}+\frac{\beta D}{2}$ but I do not know how to derive it. Do anyone have some idea about this? Thanks!

Answer 1

Because the boundness of $f$, we have $f\in L^{\infty}$. Due to the smoothness, we have $\nabla f\in Lip$, that is $\nabla^2 f\in L^{\infty}$. Therefore, we may use the Gagliardo–Nirenberg interpolation theorem to get an answer... However, I believe that there may be a direct method.

Update: we can use taylor's formular. $$ f(y)=f(x)+\nabla f(x)\cdot(y-x)+\frac{1}{2}\nabla^2f(\xi)(y-x)^2 $$ It implies that

$$ |\nabla f|\le |\frac{f(y)-f(x)}{y-x}|+\frac{\nabla^2 f(\xi)}{2}|y-x|\le \frac{2B}{|y-x|}+\frac{\beta}{2} |y-x| $$

$$ \min_{0\le |x-y|\le L} \frac{2B}{|y-x|}+\frac{\beta}{2} |y-x| =\frac{2B}{L}+\frac{\beta}{2}L $$

the last equality needs the assumption that $L^2\le \frac{4B}{\beta}$.