It is a well known fact that if we have a fixed normal operator $T$ on $B(H)$, $f$ and $g$ are both continuous on $\sigma(T)$, then $||(f-g)(T)||=||f-g||_{\sigma(T)}$. It tells us that functional calculus for a fixed operator is continuous, which means if $f_n$ converges to $f$ uniformly in $\sigma(T)$, then $f_n(T)\rightarrow f(T)$ in norm.
My question is, if we have a fixed continuous function $f\in C(E)$, a sequence $T_n\rightarrow T$ in norm, where $T$ and $T_n$ are both normal, satisfying that $\cup_{n=1}^{\infty}\sigma(T_n)\cup \sigma(T)\subset E$, do we know that $f(T_n)\rightarrow f(T)$ in norm?
How about the case for Bounded Borel functional calculus?
For the continuous functional calculus the result is as you say. This is a typical $\varepsilon/2$-argument. You have that any continuous function is a uniform limit of polynomials, and it is easy to show that for a fixed polynomial $p$ you have $p(T_n)\to p(T)$.
For Borel functions, the answer is no even if you require sot-convergence instead of norm, even in finite-dimension. For this take $e_n=\frac1n\,1$, then $e_n\to0$ in all relevant topologies. If we take $f(t)=1_{(0,1]}$, then $f(e_n)=1$ for all $n$, but $f(0)=0$.