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I have sometimes heard that the Borel functional calculus maps bounded pointwise convergent sequences of Borel functions to strongly convergent sequences of operators. I gather "sequence" is important here, due to the measure theory aspect, we can't use nets. But why must it converge strongly? I am only familiar with the result of weak convergence.

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up vote 5 down vote accepted

$\def\norm#1{\left\|#1\right\|}\def\skp#1{\left<#1\right>}$Suppose $T \in L(X)$ is a normal operator on the Hilbert space $X$ and $f_n$, $f \colon \mathbb C \to \mathbb C$ are bounded and measurable such that $f_n \to f$ pointwise. As you remark, we have $f_n(T)x \rightharpoonup f(T)x$ weakly for each $x \in H$. We moreover have \begin{align*} \norm{f_n(T)x}^2 &= \skp{f_n(T)x, f_n(T)x}\\ &= \skp{f_n(T)^*f_n(T)x, x}\\ &= \skp{(\bar f_n f_n)(T)x, x}\\ &\to \skp{(\bar f f)(T)x,x}\\ &= \norm{f(T)x}^2 \end{align*} Hence, as for Hilbert spaces $x_n \rightharpoonup x$ weakly plus $\|x_n\| \to \norm x$ implies $x_n \to x$, we have $f_n(T) \to f(T)$ strongly, as wished.

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