# property of convex functions

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.

• @GeraldEdgar I think you missed that $\alpha,\beta,A,B>0$. Aug 21 '17 at 22:15
• The right-hand inequality does not even require convexity of P or f provided P is bounded and f is continuous.: let g=max{f,0} and h=-min{f.0}. Then the right-hand inequality follows from the fact that 2$\int g$ $\leq$ 2(sup g) m(P)=2(sup f) m(P) by changing f to -f. Note that sup f $\geq$ 0 by the hypothesis (unless f is 0). Aug 22 '17 at 7:43
• It is one of those situations where sending the reader to the library wastes more of his time than writing a 5-line proof, so I would advise not to bother about any references here though the choice is yours. Still I'm unable to understand what $A$ and $B$ are doing in a completely homogeneous problem. Aug 22 '17 at 22:32
• Fedja, indeed $A$ and $B$ can be take to be zero. Thanks for pointing that out, I edited the question. I was blinded with the particular application of this inequality that interests me, and that naturally involves the constants $A,B$. Anyway, If you know a 5 line proof for the first inequality please share it with us. My argument is more convoluted, that is why I am (still) interested in a reference. Aug 23 '17 at 1:55
• Probably you want $P$ to be open (or at least has positive measure). Otherwise when $P$ is a segment in $\mathbb R^2$ the left bound cannot hold. Aug 23 '17 at 5:14

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 $\int_P |f|\ge \int_K|f|\ge 4^{-n-1}|P|$.