If H is hilbert space and $T\in B(H)$ normal then $(g\circ f)(T)=g(f(T)$

Question

How to prove the following claim (theorem): If $H$ is a Hilbert space and $T\in B(H)$ is normal and if $f\in C(\sigma(T))$ and $g\in f(C(\sigma(T))$ then $(g\circ f) (T)=g(f(T))$.

Answer 1

If $f(t)=t^n$ and $g(t)=t^m$, then $$ (g\circ f)(T)=T^{nm}=g(f(T)). $$ If now $f(t)=\sum_{k=1}^m\alpha_k t^k$, $g(t)=\sum_{j=1}^n\beta_jt^j$, writing $g\circ f(t)=\sum_\ell \gamma_\ell\,t^\ell$, $$ (g\circ f)(T)=\bigg(\sum_{k=1}^m\alpha_k\big(\sum_{j=1}^n\beta_jt^j\big)^k\bigg)(T) =\sum_\ell \gamma_\ell T^\ell=f(g(T)). $$ Now if $g\in C(\sigma(T))$ and $f$ is a polynomial, there exist polynomials $\{g_k\}$ with $g_k\to g$. Then $$ (g\circ f)(T)=\lim_n(g_n\circ f)(T)=\lim_n g_n(f(T)=g(f(T)). $$ If now $f,g\in C(\sigma(T))$, choose polynomials $f_j$ with $f_j\to f$. Then $$ (g\circ f)(T)=\lim_j (g\circ f_j)(T)=\lim_j g(f_j(T))=g(f(T)). $$