TITLE: Is $C_n$ infinitely log-convex?
QUESTION [5 upvotes]: A sequence $a_n$ is called log-convex if $\mathcal{L}(a_n):=a_{n+1}a_{n-1}-a_n^2\geq0$ for all $n$; it is infinitely log-convex provided that all the iterates $\mathcal{L}^k(a_n)$ are still log-convex, $k\geq1$. Here $\mathcal{L}^2(a_n)=\mathcal{L}(\mathcal{L}(a_n))$, etc.
Consider in particular the well-known sequence of Catalan numbers $C_n=\frac1{n+1}\binom{2n}n$. It has been established that both $C_n$ and some of its "quantum" versions are log-convex.
I would like to ask:

QUESTION. Is it true that $C_n$ is infintely log-convex? It appears to be so.

REPLY [11 votes]: I think so. If $a_n=\int_X f^n(x)d\mu(x)$ for a certain positive function $f$ on a measure space $(X,\mu)$, then $$a_{n-1}a_{n+1}-a_n^2=\int_{X\times X} f^{n-1}(x)f^{n+1}(y)d\mu(x)d\mu(y)-\int_{X\times X} f^{n}(x)f^{n}(y)d\mu(x)d\mu(y)\\=\frac12\int_{X\times X} f^{n}(x)f^{n}(y)\left(\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-2\right)d\mu(x)d\mu(y)$$
has also a similar representation.
But
$$
C_n=4^n\frac{(2n-1)!!}{(2n)!!}\cdot \frac1{n+1}=\frac2{\pi}\int_0^{\pi/2} (2\sin x)^{2n}dx\int_0^{1} t^n dt.
$$