Let $S\subset\mathbb{R}$ be a convex set and $\mathbb{S}^{n}$ be the set of real symmetric matrices of order $n\times n$. 


A matrix valued function $\Gamma: S \rightarrow \mathbb{S}^{n}$ is said to be convex if for all $x\_1,x\_2 \in S$ and for all $\lambda \in (0,1)$ one has
$$\Gamma\left(\lambda x\_{1}+\left(1-\lambda\right)x\_{2}\right)\preceq\lambda\Gamma\left(x\_{1}\right)+\left(1-\lambda\right)\Gamma\left(x\_{2}\right),$$
where $\prec$ denotes the Loewner partial order, i.e. $A \prec B$ if $A - B$ is negative definite.


Is the matrix valued function $f: [0,c] \rightarrow \mathbb{S}^{n}$ given by
$$ f(t) = e^{At} e^{A^{T} t} $$
convex for any $A \in \mathbb{R}^{n\times n}$ and $c>0$?


Numerical experiments suggest that this statement is true, but I could not prove.



Convexity of the product of two exponential matrices
I think the $3\times 3$ Jordan block
$$
A=\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{pmatrix} 
$$
is a counterexample.


For this $A$, we have 
$$
\exp{At}=\begin{pmatrix} 1 & t & \frac{t^2}{2} \\ 0 & 1 & t \\ 0 & 0 & 1\end{pmatrix} 
$$
so
$$
f(t) = \begin{pmatrix} 1+t^2+\frac{t^4}{4} & t+\frac{t^3}{2} & \frac{t^2}{2} \\ t+\frac{t^3}{2} & 1+t^2 & t \\ \frac{t^2}{2} & t & 1\end{pmatrix}
$$
Take any $c>0$. The last column of $\frac{1}{2}(f(c)+f(0)) - f(\frac{c}{2})$ is
$$
\begin{pmatrix} \frac{c^2}{8} \\ 0 \\ 0\end{pmatrix}
$$
so the difference $\frac{1}{2}(f(c)+f(0)) - f(\frac{c}{2})$ cannot be positive for any $c$.


**EDIT:** In fact it is possible to obtain a necessary and sufficient condition for the convexity of $f$ in terms of $A$. Namely, I claim that if the matrix
$$
A^2 +A^{\intercal 2}+ 2A A^\intercal
$$
is positive semidefinite, then $f$ is convex on $\mathbb R$, and if not, then $f$ is not convex on any interval $[a,b]$. To see this, one can first check that $f(t)$ is convex on $[a,b]$ if and only if the scalar functions 
$$
f\_u(t):=\langle f(t)u, u\rangle
$$
are convex on this interval for all $u\in\mathbb R^n$. But since the functions $f\_u$ are smooth, they will be convex if and only if the second derivatives are nonnegative. Computing we find
$$
f\_u^{\prime\prime}(t) = \langle (A^2 +A^{\intercal 2} + 2AA^\intercal)\exp(tA^\intercal)u, \exp(tA^\intercal)u\rangle
$$
and the claim follows. 


