Minimum over Probability Measures

Question

Let $f$ and $g$ be polynomials in $\mathbf x \in \mathbb R^n$. Let $X$ be a compact subset of $\mathbb R^n$. Finally, Let $\mathcal M(X)$ be the set of probability measures over $X$.

Can the optimization problem

\begin{equation} \inf_{P\in \mathcal M(X)} \left ( \int f \ dP\right )^2 - \left ( \int g \ dP\right )^2 \end{equation}

be expressed exclusively in terms of the polynomials $f$ and $g$, without reference to the measure $P$?

For example, my hunch is that the above optimization problem is equivalent to

\begin{equation} \min_{\mathbf x \in X} f(\mathbf x)^2 - g(\mathbf x)^2, \end{equation}

but I don't know how to prove this. Anyone have any ideas?

Answer 1

Put $n=1$, $X=\{-1,1\}$, $f(x)=x$, and $g(x)=0$. Then the infimum equals $0$, whereas the minimum equals $1$.

Answer 2

You could think of Problem 2 instead as $\min_{P \in \mathcal M(X)} \int [f^2 - g^2] dP$. This has the same value as the original problem; the objective function is linear in the control ($P$), so there will be a solution on the boundary, i.e. a degenerate distribution.

Then write $\phi = f+g$ and $\psi = f - g$ and factor,

• Problem 1: $\min_P \int \phi~ dP \cdot \int \psi dP$.
• Problem 2: $\min_P \int\phi \psi ~dP$

So for example if either $\phi$ or $\psi$ crosses $0$, then Problem 1 has value $\leq 0$, whereas if $\phi \psi$ crosses $0$ then Problem 2 has value $\leq 0$. You could say more by looking at cases.