11
$\begingroup$

Is it possible to have a nonseparable $L^2$ Hilbert space for which the underlying measure space is sigma finite? I appreciate any example but prefer one built on the Borel sigma algebra of some topological space, if possible.

$\endgroup$

1 Answer 1

8
$\begingroup$

Let $G$ be a compact abelian group whose dual group $\Gamma$ is uncountable, and let $m$ be the normalized Haar measure on $G$ (which is a probability measure on the Baire sigma-algebra of $G$). By Plancherel's theorem, $L^2(G,m)$ is isometric with $\ell^2(\Gamma)$, and hence it is nonseparable.

You can say the same thing a little bit differently. Take any uncountable set $I$ and set $G:= \mathbb T^I$ (with the product topology). Let $m_0$ be the Lebesgue measure on $\mathbb T$, and let $m$ be the corresponding product measure on $G$. For each $i\in I$, denote by $\pi_i:G\to \mathbb T$ the $i$-th coordinate function on $G$. Then $(\pi_i)_{i\in I}$ is an uncountable orthonormal set in $L^2(G,m)$, and hence $L^2(G,m)$ cannot be separable.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .