Convex conjugate of relative entropy

Question

The convex conjugate of a function, say, $f:X\to \mathbb{R}$ is a function $f^*:X^*\to \mathbb{R}$ 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) $D(\cdot||Q):\mathcal{P}(X)\to \mathbb{R}^+$, is defined for two probability measures $P$ and $Q$ $(P\ll Q)$ 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 $Q$.

Answer 1

If you want to get it directly from the source, corollary 3 from Rockafellar's Integrals which are convex functionals I will help you find the conjugate of $D(\cdot \| \nu)$, which is a convex integral functional.

You might also consult Proposition 4.2.4 of Optimal Bounds between f-Divergences and Integral Probability Metrics by Agrawal and Horel, which states that the convex conjugate of $D(\cdot \| \nu)$ (in their notation this is $D_{\phi, \nu}$ for $\phi \colon \mathbb R \to \mathbb R \cup \{ \infty \}$, $t \mapsto t \ln(t) - t + 1 + \iota_{[0, \infty)}(t)$ which is proper, convex and lower semicontinuous with $\phi(1) = 0$ and where $\iota$ is the convex indicator function and $0 \ln(0) := 0$) is $$ Y \ni h \mapsto \int_{\mathcal X} \phi^*(h(x)) \; \text{d}\nu(x), $$ where $\phi^* \colon \mathbb R \to \mathbb R$, $s \mapsto e^{s} - 1$ is the convex conjugate of $\phi$.

In the notation of the paper mentioned we have choosen $X := M(\mathcal X)$, where $\mathcal X$ is the space from the question, and $Y := X^* \cong C(\mathcal X; \mathbb R)$ (equipped with the appropriate topologies). Note futhermore that $$\phi'(\infty) = \lim_{t \to \infty} \frac{1}{t} \phi(t) = \lim_{t \to \infty} \ln(t) - 1 + \frac{1}{t} = \infty = - \phi'(- \infty) := - \lim_{t \to - \infty} \frac{1}{t} \phi(t).$$

In the end we arrive at $$D(\cdot \| \nu)^*(h) = \int_{\mathcal X} e^{h(x)} \, \text{d}\nu(x) - 1.$$