Are the off-diagonal elements of exp(At) log-concave in t, for nonnegative matrices? Let $A$ be a Metzler matrix, i.e. a real matrix (not necessarily symmetric) whose off-diagonal elements are all non-negative. Then, for $t\ge 0$, the matrix exponential $\exp(At)$ will have all non-negative elements.
Numerically, it seems that given a single off-diagonal element of $\exp(At)$, it is always a log-concave (i.e. log-convex downward) function of $t$, for $t\ge 0$, and that the diagonal elements are always log-convex upward functions.
That is, it looks like the function
$$
f(t) = \log\Big(\big(\exp(At)\big)_{ij}\Big)
$$
is a concave function of $t$ for $t\ge 0$ if $i\ne j$, and a convex function of $t$ if $i=j$. My question is whether this indeed is the case. Note that in this expression $\exp$ is a matrix exponential, but $\log$ is just an ordinary logarithm of a positive number.
Here is a typical result. The plot shows each element of $\exp(At)$ as a function of $t$, where $A$ is a matrix whose off-diagonal elements are independently uniformly sampled from $[0,1]$ and whose diagonal elements are independently uniformly sampled from $[-1,0]$. The diagonal elements of $\exp(At)$ are shown in red.

Showing this should be a simple case of finding the second derivative and showing that it can't be positive, but I haven't seen a way to do that. I haven't been able to find a counterexample either.
 A: For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find
$$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$
where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is 
$$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$
Notice that the assumption is implicitely used in that it implies $g>0$.
Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have 
$$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$
Let us form the Hankel matrix $M_h(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M_h(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$.
Remark that a smooth function $h$ satisfies a linear ODE of order $n-1$ with some constant coefficients if, and only if, $\det M_h\equiv0$.
A: Try e.g. $$ A = \pmatrix{0 & 0\cr 1 & 0\cr},\ \exp(At) = \pmatrix{1 & 0\cr t & 1\cr} $$
where $\log(t)$ is not convex.
Or did you mean log-concave?
