Log concavity implies unimodality

I have been to many talks where the speaker says something like “its not hard to see that log concavity (with no internal zeroes) implies unimodality,” but I have never seen the proof of this statement written down. So here it is, inspired by a recent talk by Petter Brändén.

Definitions

A sequence finite sequence \(a_1, a_2, \ldots, a_n\) is said to be “unimodal” if \[ a_1 \leq a_2 \leq \cdots a_m \geq a_{m+1} \geq \cdots \geq a_n.\]

A finite sequence is said to be “log(arithmically) concave” if \[ a_k^2 \geq a_{k-1}a_{k+1} \] for all \(k\).

Proof

Suppose \(\{a_i\}_{i=1}^n\) is a sequence of positive numbers with no internal zeroes which is log concave. Now suppose for the sake of contradiction that \(\{a_i\}\) is not unimodal. Then there exists an index \(k\) such that \(a_{k-1} \geq a_k \leq a_{k+1}\). In particular this means that both \(a_k \leq a_{k+1}\) and \(a_k \leq a_{k-1}\). Then since \(a_k > 0\) (note we are using both positivity and the fact there are no internal zeros to establish this) we have \[ a_k^2 \leq a_{k+1}a_k \leq a_{k+1}a_{k-1}. \] This contradicts the assumption that \(\{a_i\}\) was log concave. QED.

I hope that you would agree it is not too bad of a proof, but in case you (like me) have always been too embarassed to ask to see it, here it is.

Edit (19JAN23): Thanks to Anastasia Nathanson for catching a typo! Its now fixed.