Is $\text{Trace}(e^{XA+A^TX})$ a convex function of $X$? Is $\text{Trace}(e^{XA+A^TX})$ a convex function of $X$?
$X$ is diagonal and positive definite, $A$ is symmetric negative definite definite.
And by the way, what is the best way to solve a problem of the form:
$\text{min}.$ $\text{Trace}(e^{XA+A^TX})$
s.t. $X$ diagonal, $X_{ii}>0$ and $1^T X 1<x$
where $X$ is the variable and $A$ is symmetric negative definite, $x>0$, $x \in R$
 A: Let $D$ (resp. $D^+$) be the set of diagonal matrices (resp. diagonal positive matrices).
Let $f:X\in D^+\rightarrow tr(e^{XA+AX})$. We seek $\min_X f(X)$ under the condition $tr(X)=\sum_i x_{i,i}=x$. 
Fortunately the first derivative of $f$ is easy to calculate. For every $H\in D$, $f'_X(H)=tr(e^{XA+AX}(HA+AH))$ ; the Lagrange condition of our problem can be written: there is $\lambda\in\mathbb{R}$ s.t., for every $H\in D$,$f'_X(H)+\lambda tr(H)=0$ or $tr((Ae^{XA+AX}+e^{XA+AX}A+\lambda I)H)=0$. That is equivalent to the following $n$ equalities:
(*) for every $i$, $(Ae^{XA+AX}+e^{XA+AX}A+\lambda I)_{i,i}=0$.
Remark 1. If $tr(X)<x$, then put $\lambda=0$ in the previous formula. 
Remark 2. Unfortunately the second derivative of $f$ is difficult to calculate.
EDIT 1: @ questioner ,  Tim Notke (a highschool basketball coach) said 
« Hard work beats talent when talent fails to work hard ».
If we search the critical points, then necessarily $tr(X)=x$ and even $\lambda >0$. Indeed $XA+AX$ symmetric implies that $e^{XA+AX}$ is symmetric $>0$. Thus $spectrum(Ae^{XA+AX})\subset \mathbb{R}^{*-}$ and $tr(Ae^{XA+AX})<0$. Finally, the relations (*) imply that $\lambda>0$ and therefore $tr(X)=x$.
EDIT 2. Let $S_n$ be the set of real symmetric matrices. $X\in S_n\rightarrow e^X$ is not a matrix convex function. Yet $X\in S_n\rightarrow \log(tr(e^X))$ is a spectral matrix function and is convex. cf. convexity of matrix "soft-max" (log trace of matrix exponential)
Thus $g:X\in S_n\rightarrow tr(e^X)$ is also convex. Here we consider $f:X\in S_n\rightarrow g(XA+AX)$ and for every symmetric $H$, $f''_X(H,H)=g''_{XA+AX}(HA+AH,HA+AH)\geq 0$. Finally, for every symmetric $A$, $f$ is convex over $S_n$.
