Why is Volume^2 at most product of the 3 projections?

Question

Is there a simple proof for $$ \text{Vol}^2(P)\le \prod_{i=x,y,z} \text{Area}(\text{Proj}_i(P)), $$ where $P\subset \mathbb R^3$ and $\text{Proj}_z(P)$ denotes the projection of $P$ to the $z=0$ plane?

I have a complicated proof for this (in any dimension) using shifts but I wonder if this inequality is something well-known/simple.

Answer 1

As I have learned from Gabor Tardos (and Peter Csikvari) this is well-known and can be proved easily using the submodularity of entropy. First we reduce the problem to the finite case by approximating $P$ by a collection of cubes. Then let us select a cube uniformly random and denote its $i$-th coordinate by $X_i$. We have $$\log Vol^2(P)=H(X_1,X_2,X_3)+H(X_1,X_2,X_3)\le H(X_1,X_2)+H(X_1,X_3)-H(X_1)+H(X_1,X_2,X_3)\le H(X_1,X_2)+H(X_1,X_3)+H(X_2,X_3)\le \Pi_{i=x,y,z} Area(Proj_i(P)).$$

Answer 2

This is a special case of the Loomis-Whitney inequality. First, some biographical information:

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• Lynn Harold Loomis was an analyst. He got his PhD under Salomon Bochner at Harvard in 1942. Later he would teach at the same school. He co-aurthored Advanced Calculus with Shlomo Sternberg.
• Hassler Whitney was a reknown topologist. He came from a rather learned family in New York City, studied at Yale where he got degrees in Physics and Music.

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The Loomis Inequality appears in An Inequality Related to the Isoperimetric Inequality. The result states:

Let $m$ be the measure of an open subset of Euclidean $n$-space, $O \in \mathbb{R}^n$. Let $m_1, \dots, m_n$ be the $(n-1)$-dimensional measures of the projections on the coordinate hyperplanes. Then $m^{n-1} \leq m_1\dots m_n$

We can take $n=3$. The Loomis-Whitney paper is 2 pages long and we will summarize it here:

• Step 1 Reduce to a counting problem:

  Let $S$ be a set of cubes from the cubical subdivision of $n$-space and let $S_i$ be the set of $n$-cubes by projecting the cubes of $S$ onto the $i$th coordinate hyperplane. Let $N$ be the number of cubes in $S$ and $N_i$ be the number of cubes in $N_i$ then $N^{n-1} \leq N_1 \dots N_n$

• This cubic apprxoximation works since the number of cubes is a good approximation to the volume or "measure" of the set $S$. We just set the cubes of size $\delta$ and count.

• Step 2 How do we prove the above lemma? The proof is by induction on the number of dimensions:

  • The cubes project into intervals $I_1, \dots, I_k$ if we project to the first axis. Here $k$ might be very large set of intervals.
  • For each interval $I_i$, let $T_i$ be the set of cubes projecting into $T_i$. Let $T_{i \to j}$ be the projection of $T_i$ onto the $j$-th hyperplane.
  • Let $a_i$ be the number of cubes in $T_i$. We can count the cubes so that $$ \begin{align} \sum_{i=1}^k a_i = N \\ \sum_{i=1}^k a_{i \to j} = N_j \end{align}$$ and by induction hypothesis $a_i^{n-2} \leq a_{i \to 2} \dots a_{i \to n}$ which implies $a_i^{n-1} \leq N_i a_{i \to 2} \dots a_{i \to n}$
  • The last step follows from Hölder's inequality ...