If we consider the function $f(x)$ and the matrix A formed as follows
$$ A_{ij} = f(\frac{x_i + x_j}{2}) $$
and state that $f(x)$ is log-convex and non-negative
$$ f(x_1)^tf(x_2)^{(1-t)} \geq f(tx_1 + (1-t)x_2) $$
Then the Cholesky decomposition for when $A$ is a $2\times2$ matrix exists - if $A = LL^{T}$, then $L$ is given below $$ \left( \begin{matrix} f(x_1)^{\frac{1}{2}} & 0 \\ f(\frac{x1 + x2}{2}) / f(x_1)^{\frac{1}{2}} & \left(f(x_2) - f(\frac{x1 + x2}{2})^2 / f(x_1)\right)^{\frac{1}{2}} \end{matrix} \right) $$
Noting that by the definition of $f(x)$ as log-convex, the bottom right term will be non-negative within the square root, so the diagonals are real (as required).
By this answer on the inductive proof of the cholesky decomposition one can presumably generalise this, to show for $3\times3$ and so on for $(n+1)\times (n+1)$
My question is if there exist functions that are log-convex (or equivalently, where the log of the function is convex), that do not form a positive semi-definite matrix, as defined above? I can't see how this can be true, but in some papers and lectures and so on I've seen the statement that if a matrix is defined as above and is positive semi-definite, then f is log-convex, but the other way around isn't necessarily true.
