# Volume inequality between projections of a convex symmetric set in $\mathbb R^3$

Let $K$ be a centrally symmetric convex body in $\mathbb R^3$ with volume ${\rm vol}(K)=1$. For any subset $F \subset \lbrace1,2,3\rbrace$, let $K_F$ be the projection of $K$ in $\mathbb R^F$.

Question: What is the best constant $C$, such that

$${\rm vol}(K_{\lbrace 1 \rbrace}) \leq C \cdot {\rm vol}(K_{\lbrace 1,2 \rbrace}) \cdot {\rm vol}(K_{\lbrace 1,3 \rbrace}).$$

Here, the volume of $K_F$ is computed in $\mathbb R^F$. With some work one can prove that $C=2$ is good enough. This is elementary and follows for example from two applications of Lemma 3.1 in

J. Bourgain and V.D. Milman, New volume ratio properties for convex symmetric bodies in $\mathbb R^n$, Invent. Math. 1987 vol. 88 (2) pp. 319-340.

My guess is that maybe $C=1$ works, but I am not sure.

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## 2 Answers

We can exchange the condition $\mathop{\rm vol}K=1$ to ${\rm vol}K_{\lbrace 1 \rbrace}=1$. In this case we need to show that $$\mathop{\rm vol}K\leqslant C\cdot \mathop{\rm vol}K_{\lbrace 1,2 \rbrace} \cdot \mathop{\rm vol}K_{\lbrace 1,3 \rbrace}.$$ The later is equivalent to the following: $$\int\limits_0^1 u{\cdot}v{\cdot}dx\leqslant C\cdot \int\limits_0^1 u{\cdot}dx\cdot\int\limits_0^1 v{\cdot}dx\ \ \ \ \ (*)$$ for two positive convave functions $u,v\colon[0,1]\to\mathbb R$. WLOG we can assume that both functions are PL and both bave $0$ at $0$ andat $1$. Further we can assume that $$\int\limits_0^1 u{\cdot}dx=\int\limits_0^1 v{\cdot}dx=1. \ \ \ \ \ (**)$$ We can move extremal points of $u$ and $v$ one by one keeping the identity $( * * )$ and increasing LHS of $( * )$. This way we may reduce number of extremal points in $u$ and $v$; the process might terminate if you get one or two extremal point in each.

It seems that $$u=v=2-4\cdot|x-\tfrac 12|$$ is the worst case. So $C=\tfrac43$ is the best constant.

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This formalization makes the example $u=v=|x-1/2|$ obvious, which shows $c\geq 4/3$. Or, in $\mathbb R^3$ terms and renormalized, $|x_1|\leq x_2\leq 1$, $|x_1|\leq x_3\leq 1$. $K_{1}$ has volume $2$, $K_{1,2}$ has volume $1$, as does $K_{1,3}$, and $K_{1,2,3}$ has volume $2/3$. Renormalizing we get $4/3 \leq C$ – Will Sawin Jan 21 '12 at 1:58
Yes, it seems $C=\tfrac43$ is the optimal constant. I will update my answer. – Anton Petrunin Jan 21 '12 at 2:19
Anton and Will, thanks a lot for this answer. – Andreas Thom Jan 21 '12 at 10:22

Picking up from anton's answer, notice that by Cauchy-Schwartz:

$\int_0^1 u \cdot v \cdot dx \leq \sqrt{ \int_0^1 u^2 \cdot dx \cdot \int_0^1 v^2 \cdot dx}$

So a bound in the case $u=v$ is sufficient. We wish to show:

$\int_0^1 u^2 dx \leq C \left( \int_0^1 u dx\right)^2$

Fix a triangle with the same average value of $u$, and that attains its maximum over the same $x$-coordinate. If $u$ is equal to that triangle, a trivial calculation gives a value of $C$ of $4/3$. We want to show that it not being a triangle makes $\int u^2$ no bigger. In fact, letting $t$ be the triangle, we have $\int u^2 \leq \int ut\leq \int t^2$.

This is because $(t-u)$ is positive closer to the maximum value, and negative further, so $u$ and $t$ are both higher where $(t-u)$ is positive than where it is negative, so $\int u(t-u)\geq 0$, $\int t(t-u)\geq 0$.

Therefore $C=4/3$ is a bound, obtained by setting $u$ and $v$ to triangles. One can check that this does indeed come from a polytope in $\mathbb R^3$ - an octahedron, for instance.

Edit: This proof isn't quite complete, because the area on one side of the triangle is not necessarily the area on one side of the function. We can remedy this by choosing an unbalanced "triangle" with a discontinuity where the maximum should be, which still satisfies $C=4/3$.

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