This is a follow up to this question, where the optimal constant was left open.
Let $P \subset \mathbb{R}^n$ be bounded, convex, and open. Let \begin{equation} \mathcal{H} := \{f : P \rightarrow \mathbb{R} : f\text{ is convex and }\int_P f d\lambda = 0\} \end{equation} Question:
What is the largest possible constant $\alpha > 0$ purely depending on $n$ and $P$ such that \begin{equation} \forall f \in \mathcal{H}: \int_P \left| f \right| d\lambda \geq -\alpha \inf_{P} f \end{equation}
Notes:
The question is the same if one restricts $\mathcal{H}$ to functions with $\inf_P f = -1$.
This answer provides a lower bound of $\alpha \geq 4^{-n-1} |P|$. @fedja already mentions that a similar argument can be used to obtain a sharp constant, but I haven't been able to work it out. Hence this question.
Update: Some examples:
It must hold $\alpha \leq 2^{-1} |P|$, since for the map $f(x_1,...,x_n) = x_1-1/2$ on $P = [0,1]^n$ we have $\int_P |f| d\lambda = 1/4 = - (2^{-1} |P|) \inf_P f$. So if $\alpha > 2^{-1} |P|$, then $\int_P |f| d\lambda < - \alpha \inf_P f$.
On the other hand, $\alpha = 2^{-1} |P|$ is in general false even if one replaces convex functions by restrictions of affine maps. While it would be true in one dimension for these functions, a friend of mine gave the following example for $n=2$: Let $P = \{(x,y) : 0 \leq x \leq 2, 0 \leq y \leq x/2 \}$ (i.e. a triangle) and $f(x,y) = x - 4/3$. Then $\int_P f d\lambda = 0$, but $\int_P |f| d\lambda = \frac{2^5}{3^4} \approx 0.395 < 2/3 = - (2^{-1} |P|) \inf_P f$.
Here you can check if I messed up the integrals :)
