Assume that $ f:\mathbb{R}^n\to[0,+\infty) $ is a convex function and $ f(0)=0 $. It is obvious that $ 0 $ is the minimal point of $ f $. I want to ask if there exists $ \delta>0 $ such thta $ f(x)\leq C|x| $ for some $ C>0 $ if $ |x|<\delta $.
It is easy to show that as $ n=1 $, wuch $ C $ and $ \delta $ exist. However for $ n\geq 2 $, things may be a little bit complicated. Can you give me some hints or references?
Actually such a linear upper bound exists for every $\delta > 0$ and all $x$ with $\Vert x \Vert \le \delta$ if $f: \Bbb R^n \to \Bbb R$ is convex with $f(0) =0$.
Convex functions on $\Bbb R^n$ are continuous (see for example here), therefore $$ M_{\delta} = \max \{ f(x) : \Vert x \Vert = \delta\} $$ exists and is finite. Then we have for all $x$ with $\Vert x \Vert \le \delta$ $$ \begin{align} f(x) &= f\left(\left(1-\frac{\Vert x \Vert}{\delta}\right)\cdot 0 + \frac{\Vert x \Vert}{\delta} \cdot \frac {\delta x}{\Vert x \Vert} \right) \\ &\le \left(1-\frac{\Vert x \Vert}{\delta}\right) \underbrace{f(0)}_{= 0} + \frac{\Vert x \Vert}{\delta} \underbrace{f\left( \frac {\delta x}{\Vert x \Vert}\right)}_{\le M_{\delta}} \\ &\le \frac{M_{\delta}}{\delta} \Vert x \Vert \, . \end{align} $$
