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Spring 2013 - Problem 2

Hilbert spaces, measure theory

Consider the Hilbert space $\ell^2\p{\Z}$ . Show that the Borel $\sigma$ -algebra $\mathcal{N}$ on $\ell^2\p{\Z}$ associated to the norm topology agrees with the Borel $\sigma$ -algebra $\mathcal{W}$ on $\ell^2\p{\Z}$ associated to the weak topology.

Solution.

Since linear functionals are continuous with respect to the strong topology, the strong topology is finer than the weak topology, so $\mathcal{W} \subseteq \mathcal{N}$ . It remains to show the other inclusion.

Notice that $\ell^2\p{\Z}$ is a separable metric space, so it is second countable. Thus, its topology is generated by open balls, so it suffices to show that norm-open balls are weakly open.

Let $F = \set{y \in \ell^2\p{\Z} \mid \norm{x - y}_{\ell^2} < r}$ . Then we can realize $F$ as a preimage of $f\p{y} = \norm{x - y}^2$ : $F = f^{-1}\p{\pco{0, r^2}}$ . Notice that if $\set{e_n}_n$ is an orthonormal basis for $\ell^2\p{\Z}$ , then by Parseval's identity,

f\p{y} = \norm{x}^2 + \norm{y}^2 - 2\Re{\inner{x, y}} = \norm{x}^2 + \sum_{n=1}^\infty \abs{\inner{y, e_n}}^2 - 2\Re{\inner{y, x}}.

We see that $f$ is a limit of $\mathcal{W}$ -measurable functions: $\norm{x}^2$ is a constant, $\abs{\inner{y, e_n}}$ is the absolute value of the (weakly-continuous) linear functional $\inner{\:\cdot\:, e_n}$ , and $\Re{\inner{y, x}}$ is the real part of the (weakly-continuous) linear functional $\inner{\:\cdot\:, x}$ . Thus, $f$ is $\mathcal{W}$ -measurable, which implies that $F \in \mathcal{W}$ and so $\mathcal{N} \subseteq \mathcal{W}$ , by definition of the Borel $\sigma$ -algebra.