Cauchy-Schwarz and pigeonhole

My own interpretation (which I guess is pretty similar to the one above):

Suppose you have r pigeons and n holes, and want to minimize the number of pairs of pigeons in the same hole. This can easily be seen as equivalent to minimizing the sum of the squares of the number of pigeons in each hole.

Classical Cauchy Schwartz: $x_1^2+...+x_n^2 \ge\displaystyle\frac1n(x_1+...+x_n)^2$.

Discrete Cauchy Schwartz: If you must place an integer number of pigeons in each hole, the number of pairs of same-hole pigeons is minimized when you distribute the pigeons as close to evenly as possible subject to this constraint.

Pigeonhole: In the case $r=m+1$, the most even split is $(2,1,1,...,1)$, which has a pair of pigeons in the same hole.

Very interesting. Here's an attempt at an answer:

Suppose there are $n$ pigeonholes, with hole $k$ containing $f(k)$ pigeons, for $k = 1, ..., n$. The total number of pigeons is $m=\sum^n_{k=1} f(k)=f\cdot u$ in the standard inner product, where $u(k) = 1$ for all $k$. The Cauchy-Schwarz inequality implies that $\|f\| \|u\| \ge |f\cdot u| = m$, so $\|f\| \ge m/\|u\| = m/\sqrt{n}$. On the other hand, if $r$ is the maximum of $f(k)$ for $k = 1, ..., n$, then $\|f\| = \sqrt{\sum^n_{k=1} f(k)^2} \le \sqrt{\sum^n_{k=1} r^2} =r \cdot \sqrt n$. Putting these together, we have $r \sqrt n \ge \|f\| \ge m/\sqrt n$, so $r \ge m/n$. That's the pigeonhole principle: if there are $m$ pigeons in $n$ holes, then some hole has at least $m/n$ pigeons, which becomes $\lceil m/n\rceil$ if pigeons can't be divided.

That does seem like a very special case of the Cauchy-Schwarz inequality, which makes it somewhat unsatisfying. It's probably not exactly what Terry Tao had in mind, but it's not completely bogus, either.

I recently learned about a nice formulation of this connection from a version of the Cauchy–Schwarz inequality stated in Bateman's and Katz's article [1, page 588 Lemma 2.1].

I imagine this version is probably part of the appropriate combinatorial folklore. Also, it's essentially the interpretation given in Kevin's answer.

Proposition. Let $X$ and $Y$ both be finite sets and let $f \colon X \to Y$ be a function.

• $\#(\ker f) \cdot \#(Y) \ge \#(X)^2$. (Where $\ker f$ is the kernel of a function $f$.)
• Equality holds if and only if every fiber has the same number of elements.

Reference

[1] Bateman, Michael; Katz, Nets Hawk. New bounds on cap sets. J. Amer. Math. Soc. 25 (2012), no. 2., 585–613. MR2869028