TITLE: Strong continuity of the Borel functional calculus
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: $\def\norm#1{\left\|#1\right\|}\def\skp#1{\left<#1\right>}$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.