functional calculus and spectral measure

Question

Let $T$ be a normal operator and $f$ be a bounded borel function on ${\sigma}(T)$. If $E_{T}$ and $E_{f(T)}$ are the spectral decompositions of $T$ and $f(T)$ respectively, prove that for any borel set $w$ we have that $E_{f(T)}(w)=E_{T}(f^{-1}(w))$ for any borel subset $w$ of ${\sigma}(T)$

Answer 1

The point is that, at the function level, you have $$ 1_w\circ f=1_{f^{-1}(w)}. $$ So, using functional calculus, $$ E_{f(T)}(w)=1_w(f(T))=1_w\circ f(T)=1_{f^{-1}(W)}(T)=E_T(f^{-1}(w)). $$

Answer 2

The definition of $f(T)$ is $$ f(T) = \int_{\mathbb{R}} f(\lambda)dE(\lambda), $$ for any bounded Borel function $f$ on $\sigma(T)$. For convenience, extend $f$ to be $0$ on $\mathbb{C}\setminus\sigma(T)$, and suppose $f$ is strictly bounded by $M$. Then $\Re f$ and $\Im f$ are strictly bounded by $M$. Let $N$ be a positive integer, and divide the rectangle $\{ \lambda \in\mathbb{C} : -M \le \Re f < M,\;-M \le \Im f < M\}$ into $N^{2}$ equal squares $S_{j,k}$ (closed on the left and bottom, open on the right and top) with centers $c_{j,k}$ for $1 \le j,k \le N$. Then $$ f(\lambda) - \sum_{j,k}c_{j,k}\chi_{f^{-1}(S_{j,k})}(\lambda) $$ is uniformly bounded by $M/N$ on $\mathbb{C}$. Therefore, integrating with respect to $E$ gives $$ \left\|f(T)-\sum_{j,k}c_{j,k}E(f^{-1}(S_{j,k}))\right\| \le M/N. $$ Define $E_{f}(S)=E(f^{-1}(S))$. Then $E_{f}$ is a spectral measure, and the above implies $$ f(T)=\lim_{N}\sum_{j,k}c_{j,k}E_{f}(S_{j,k}). $$ Notice also that $$ \left\|\sum_{j,k}c_{j,k}E_{f}(S_{j,k})-\int \lambda dE_{f}(\lambda)\right\| \le M/N $$ Therefore, $$ f(T)= \int f(\lambda)dE(\lambda)=\int \lambda dE_{f}(\lambda). $$ By uniqueness of spectral measures $E_{f}$ is the spectral measure for $f(T)$. That is, $E_{T}(f^{-1}(S))=E_{f(T)}(S)$.