conjugate function of log trace of matrix exponential

Question

Given an n-by-n real symmetric matrix X, define $$e^X = \sum_{k=0}^{n}\frac{X^k}{k!}$$ Derive the Fenchel conjugate of $$f(X) = log(Tr(e^X))$$ as a function on n-by-n real symmetric matrices. Here Tr denotes the trace function.

Answer 1

By definition, the Fenchel conjugate of $f(X)$ is given by: \begin{equation} \label{eq:f}\tag{1} f^{*}(Y) = \sup_{X} X\circ Y - f(X). \end{equation}

It can be shown that $g(X) \overset{\text{def}}{=} X\circ Y - f(X)$ is a concave function in terms of $X$. (Actually, you can use the second derivative of $f(X)$ with respect to $X$ to show that $f(X)$ is a convex function.) To compute the supremum of $g(X)$, we take derivative of $g(X)$ with respect to $X$, we have \begin{equation*} \begin{aligned} Y= \frac{(\exp X) }{\text{Tr} (\exp X) }. \end{aligned} \end{equation*} Thus, $I\circ Y= 1$ and $y_{ii}>0$ for $1\le i\le n$. Using the fact $\log (AB) =\log (A) + \log (B) $ if $AB=BA$, it follows that \begin{equation} \label{eq:y}\tag{2} Y = \frac{(\exp X) }{\exp f(X)} \implies \log Y + f(X) I = X. \end{equation}

Substitute \eqref{eq:y} into \eqref{eq:f}, we have

\begin{align} \label{eq:f:final} f^{*}(Y) & = (\log Y + f(X) I ) \circ Y - f(X)\notag \\ & = \log Y \circ Y + f(X) I \circ Y - f(X) \notag\\ & = \log Y \circ Y. \tag*{using $I\circ Y= 1$.} \end{align}