Interchange of the integral and bounded linear map on separable Hilbert spaces

18 Views Asked by At

Let $\mathcal{H}_1, \mathcal{H}_2, \mathcal{H}_3$ be separable Hilbert spaces and $d\mu$ be a Borel probability measure on $\mathcal{H}_1$ such that \begin{equation} \int_{\mathcal{H}_1} \lVert u \rVert_{\mathcal{H}_1}^2 d\mu(u) < \infty \end{equation}

Now, suppose that $f : \mathcal{H}_1 \to \mathcal{H}_2$ be a (nonlinear) measurable mapping with \begin{equation} \lVert f(u) \rVert_{\mathcal{H}_2 } \leq \lVert u \rVert_{\mathcal{H}_1} \end{equation} and $T : \mathcal{H}_2 \to \mathcal{H}_3$ be a bounded linear map.

Then, is it true that \begin{equation} T\Biggl(\int_{\mathcal{H}_1} f(u) d\mu(u) \Biggr) = \int_{\mathcal{H}_1} T\bigl(f(u)\bigr) d\mu(u) \in \mathcal{H}_3 \end{equation} ?

I tried to use the dominated convergence theorem together with continuity of $T$ to justify the above equality, but I am quite lost and confused due to possible infinite-dimensionality of the Hilbert spaces..

Could anyone please help me?