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This is a part of a theorem in Rudin's Functional Analysis, in the chapter on unbounded operators. Let $\mathcal M$ be a $\sigma$-algebra in a set $\Omega$, $H$, a Hilbert space and $E:\mathcal M\rightarrow \mathscr B(H)$, a resolution of identity. Let $f:\Omega\rightarrow \mathbb C$ be measurable and let $\mathscr D_{f}$={$x\in H:\int_D |f|^{2}dE_{x,x}< \infty$}, where $E_{x,y}(\omega)=<E(\omega)x,y>$ for each $\omega \in \mathcal M$.

The question is to show that $\mathscr D_{f\overline{f}}\subseteq \mathscr D_{f}$.

I would be grateful for hints on this.

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1 Answer 1

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You want to show that $\int_\Omega|f|^4\,dE_{x,x}<\infty$ implies $\int_\Omega|f|^2\,dE_{x,x}<\infty$. and $$ \int_\Omega|f|^2dE_{x,x}=\int_{\Omega\cap\{|f|<1\}}|f|^2dE_{x,x} +\int_{\Omega\cap\{|f|\geq1\}}|f|^2dE_{x,x}\leq E_{x,x}(\Omega)+\int_{\Omega\cap\{|f|\geq1\}}|f|^4dE_{x,x}\\<\|x\|^2+\int_{\Omega}|f|^2dE_{x,x}<\infty, $$ since the measure $E_{x,x}$ is finite (because $E(\Omega)=I$).

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