I am able to give a proof to the following inequality for convex functions. Most likely this is well known, but I am unable to find a reference. I would appreciate if someone more knowledgeable in the literature of convex analysis could help.


Suppose $P$ is an open bounded convex subset of $\Bbb R^n$ and $f: P \to \Bbb R$ is a convex function such that $\int\_P f =0$. Then there exists positive constants $\alpha,\beta>0$ (not dependent on $f$) such that
$$-\alpha\inf\_P f \leq \int\_P |f(x)|dx^n \leq -\beta \inf\_P f .$$


EDIT: per Fedja's observation, I removed the unnecessary constants from the ineqilities. Also, it is specified that $P$ has to be open and bounded. Also, in lack of a good reference, if someone can furnish a short proof of the first estimate, that would work for me as well.



property of convex functions
*Anyway, If you know a 5 line proof for the first inequality please share it with us*


OK, here goes. 


Assume that $\inf\_P f=-1$ and that it is (nearly) attained at the origin. Let $K=\{x\in P: f(x)\le -\frac 12\}$. Then $E=\{x\in P: f(x)<1\}\subset 4K$ by convexity. Also, $|E|\ge |P|/2$ because otherwise $\int\_P f=\int\_{P\setminus E}f+\int\_E f\ge |P\setminus E|-|E|>0$. Thus, $|K|\ge\frac 12 4^{-n}|P|$ whence $\int\_P |f|\ge \int\_K|f|\ge 4^{-n-1}|P|$. 


If you allow 10 lines instead of 5, you can get the sharp constant via a similar argument, but, since it hasn't been requested, I'll stop here :-)


