A Question on Convex Conjugate Duality for KL Divergence The convex conjugate of a function, say, $f:X\mapsto \mathbb{R}$ is a function $f^*:X^*\mapsto \mathbb{R}$ defined as
$$f^*(x^*):=\sup_{x\in X} ~\langle  x, x^*\rangle-f(x),$$ where $X^*$ is the topological dual of $X$ and $\langle\cdot, \cdot\rangle$ is dual pairing between $X$ and $X^*$.

The relative entropy (aka Kullback-Leibler divergence) $D(\cdot||Q):\mathcal{P}(X)\mapsto \mathbb{R}^+$, is defined for two probability measures $P$ and $Q$ (P&lt;&lt; Q)  as 
$$D(P||Q)=\int_XdP \log\frac{dP}{dQ}.$$

I have been trying to calculate the convex conjugate of map $P\mapsto D(P||Q)$ but I have failed. I know that the answer is $\log \mathbb{E}_{Q}[e^{f}]$ where $\mathbb{E}_Q[\cdot]$ is the expectation operator with respect to probability measure $Q$.
To make it easier let's assume $X$ is finite, of size $n$ and associate the density of $Q$  with an $n$-dimensional vector $q$. Assume also that $q$ is everywhere positive - otherwise replace $X$ with the support of $q$. Then the conjugate is
$$
f^*_q(x) = \sup_p\  \langle x, p \rangle - \sum_{i = 1}^n{p_i\log(p_i/q_i)}.
$$
where the supremum is over the probability simplex $\{p\geq 0: \sum_i p_i = 1\}$. Since the simplex is compact and the function inside the supremum is continuous, the supremum is achieved at some $p$. Using Lagrange multipliers you get that for some real value $\lambda$ an optimal $p$ must satisfy $x_i - 1 - \log(p_i/q_i) = - \lambda$ for all $i$, which gives $p_i = q_ie^{x_i + \lambda-1  }$. Since $1 = \sum_i p_i$, we have $\lambda - 1 = -\log\left(\sum_i{q_i e^{x_i}}\right)$. Substituting, we get
$$
\begin{align}
f^*_q(x) &amp;= \sum_i{q_i e^{x_i + \lambda- 1 }x_i} - \sum_i{q_i e^{x_i+ \lambda-1 }(x_i+ \lambda-1)}  \\
&amp;= -(\lambda-1)\sum_i{q_ie^{x_i+ \lambda-1 }}  \\
&amp;= -(\lambda-1)\sum_i{p_i}\\
&amp;= -(\lambda-1) = \log\left(\sum_i{q_i e^{x_i}}\right),
\end{align}
$$
which is exactly the logarithm of the expectation of $e^{x_i}$ under $q$.