# Log-concavity of repeated convolution

Let $$A = (a_0,a_1,\ldots,a_k)$$ be a sequence of strictly positive numbers, and let $$A^{\ast k}$$ denote the $$k$$-fold repeated convolution (defined by $$A^{\ast 1} = A$$ and $$A^{\ast k+1} = A^{\ast k} \ast A$$).

Is it true that for any such sequence $$A$$, there exists $$n$$ such that $$A^{\ast n}$$ is log-concave?

As an example, any sequence of the form $$A = (1,x,1)$$ is log concave if $$x\geq 1$$, and $$A\ast A$$ is log concave if $$x\geq \sqrt{2/3}$$. In general, I have checked numerically that $$A^{\ast k}$$ is log concave if $$x\geq \sqrt{2/(k+1)}$$. I have also checked this statement numerically for many other sequences of longer length, but I have no idea how to go about proving anything.

• To make sure, how do you define the convolution of finite sequences? By extending them (by zeroes) to functions on $\mathbb Z$? – Iosif Pinelis Mar 6 at 18:03
• If so, your conjecture seems somewhat plausible, in view of the central limit theorem of probability theory. Perhaps, the Fourier transform can be used here. – Iosif Pinelis Mar 6 at 18:06
• That's right, one can imagine extending the sequences by zeros to functions on $\mathbb{Z}$. I agree that the central limit theorem supports this conjecture, although I don't know of a way to recover log-concavity from the Fourier transform. – felipeh Mar 6 at 18:19