1
$\begingroup$

Let $\mathcal{H}$ be a separable Hilbert space equipped with a strongly continuous unitary representation of a locally compact group $G$. Denote by $\mathcal{L}^{\infty}(H)$ the space of the bounded Borel measurable real valued functions on $\mathcal{H}$ (with respect to the Borel $\sigma$-algebra of the norm topology) equipped with the sup norm.

Then the representation of $G$ on $\mathcal{H}$ induces, via composition, a representation of $G$ on $\mathcal{L}^{\infty}(H)$. Let $f \in \mathcal{L}^{\infty}(H)$ be fixed and set $$ \Psi_{f}(g):=g\cdot f, \quad g\in G, $$ where $\cdot$ stands for the latter representation. Is the map $\Psi_{f}: G \rightarrow \mathcal{L}^{\infty}(H)$ Borel measurable (with respect to the Borel $\sigma$-algebra of the sup norm topology) ?

If not, is the function $\ell\circ \Psi_{f}$, where $\ell$ is any (fixed) continuous linear functional on $\mathcal{L}^{\infty}(H)$, measurable?

Any reference would be greatly appreciated.

Improve this question
$\endgroup$
32
  • $\begingroup$ How is Ψ_f(g) an element of L^∞(H)? The definition you wrote down seems to make it an element of Aut(L^∞(H)). $\endgroup$
    – Dmitri Pavlov
    Commented Mar 12, 2021 at 19:45
  • $\begingroup$ Each fixed $f$ gives a map from $G$ to $\mathcal{L}^{\infty}(H)$. $\endgroup$
    – S.Z.
    Commented Mar 12, 2021 at 20:06
  • $\begingroup$ What σ-algebra of measurable sets is being used for L^∞(H)? $\endgroup$
    – Dmitri Pavlov
    Commented Mar 13, 2021 at 0:10
  • $\begingroup$ So to be more explicit, let $\varphi : G \to U(H)$ be the representation. Then $\Psi_f(g) = f(\varphi(g) \cdot)$, i.e. the function $h \mapsto f(\varphi(g) h)$. Is that right? $\endgroup$
    – Nate Eldredge
    Commented Mar 13, 2021 at 0:37
  • $\begingroup$ Dimitri Pavlov, I edited the question to clarify. Nade Eldredge, yes, that's correct. $\endgroup$
    – S.Z.
    Commented Mar 13, 2021 at 16:05

0

Reset to default

You must log in to answer this question.