So I've been working on the Spectral Theorem for self adjoint unbounded Operators with the Book by Rudin and got to a problem: Let $(X,\mathcal{A})$ be a measure space, $H$ a complex Hilbert space and $P:\mathcal{A}\rightarrow B(H)$ a resolution of the identity. Then, to every measurable function $f:X\rightarrow\mathbb{C}$ there exists a densely defined operator $\Psi(f)$ in $H$, with domain $D(\Psi(f)) = \{x\in H: \int_{X} \vert f(\lambda) \vert^{2} \ d\langle P(\lambda)x,x\rangle < \infty\}$, which is characterized by $$\langle \Psi(f)x,y\rangle = \int_{X} f(\lambda) \ d\langle P(\lambda)x,y\rangle$$ for all $x\in D(\Psi(f))$ and $y\in H$.
My problem is the following theorem: In the above situation, if $D(\Psi(f)) = H$ then $f$ is essentially bounded.
$\textbf{Proof:}$ Since $\Psi(f)$ is a closed operator, the closed graph theorem implies $\Psi(f)\in B(H)$. If $f_{n} = f\chi_{A_{n}}$ for $A_{n} = \{x\in X : \vert f(x)\vert \leq n\}$ and $n\in\mathbb{N}$, then it follows $$\Vert f_{n}\Vert_{\infty} = \Vert \Psi(f_{n})\Vert = \Vert \Psi(f)\Psi(\chi_{A_{n}}) \Vert \leq\Vert \Psi(f) \Vert,$$ since $\Vert\Psi(\chi_{A_{n}})\Vert = \Vert \chi_{A_{n}} \Vert_{\infty}\leq 1$. Thus $\Vert f \Vert_{\infty} \leq \Vert\Psi(f)\Vert$ and $f$ is essentially bounded.
I don't know how we can conclude that $\Vert f \Vert_{\infty} \leq \Vert \Psi(f)\Vert$, since $f_{n}\rightarrow f$ only pointwise. I also know that $\Vert \Psi(f_{n}) \Vert$ converges to $\Vert\Psi(f)\Vert$ but I can't see how that could help me. I would appreciate any help.


1 Answer 1


$|f_n(x)| \leq \|\Psi (f)\| $ for all $x$ outside some set $E_n$ of measure $0$. If $x \notin \cup_n E_n$ and $f_n(x) \to f(x)$ then $|f_n(x)| \leq \|\Psi (f)\| $ for all $n$ so $|f(x)| \leq \|\Psi (f)\| $. This proves that $\|f\|_{\infty} \leq \|\Psi (f)\| $.


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