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Spring 2018 - Problem 6

measure theory

Let $\T$ denote the unit circle in the complex plane and let $\mathcal{P}\p{\T}$ denote the space of Borel probability measures on $\T$ and $\mathcal{P}\p{\T \times \T}$ denote the space of Borel probability measures on $\T \times \T$ . Fix $\mu, \nu \in \mathcal{P}\p{\T}$ and define

\mathcal{M} = \set{\gamma \in \mathcal{P}\p{\T \times \T} \st \iint_{\T \times \T} f\p{x}g\p{y} \,\diff\gamma\p{x, y} = \p{\int_\T f\p{x} \,\diff\mu\p{x}}\p{\int_\T g\p{y} \,\diff\mu\p{y}} \text{ for all } f, g \in C\p{\T}}.

Show that $\func{F}{\mathcal{M}}{\R}$ defined by

F\p{\gamma} = \iint_{\T \times \T} \sin^2\p{\frac{\theta - \phi}{2}} \,\diff\gamma\p{e^{i\theta}, e^{i\phi}}

achieves its minimum on $\mathcal{M}$ .

Solution.

Observe that $\mathcal{M} = \set{\mu \otimes \nu}$ . Indeed, given $\gamma \in \mathcal{M}$ , we see that $\gamma = \mu \otimes \nu$ (viewed as elements of $\p{C\p{\T \times \T}}^*$ ) agree on the set

A = \set{\sum_{k=1}^n c_k f\p{x} g\p{y} \st n \in \N,\ c_k \in \R,\ f, g \in C\p{\T}}.

By Stone-Weierstrass, this set is dense in $C\p{\T \times \T}$ , and so $\gamma = \mu \otimes \nu$ everywhere in $\p{C\p{\T \times \T}}^*$ , so by Riesz representation, it follows that $\gamma = \mu \otimes \nu$ , so $\mathcal{M}$ is a singleton. Thus, because

\abs{\sin^2\p{\frac{\theta - \phi}{2}}} \leq 1 \in L^1\p{\mu \otimes \nu},

it follows that $F$ trivially attains its minimum on $\mathcal{M}$ .