Let $T$ be a positive, compact operator on a Hilbert space $\mathscr H$. Now $T$ is diagonalizable i.e. there is an orthonormal basis $\{e_i:i\in \mathscr I\}$ and a bounded set of complex numbers (actually non-negative numbers) $\{\alpha_i:i\in \mathscr I\}$ such that $$T(x)=\sum_{i\in \mathscr I}\alpha_i\langle x,\alpha_i\rangle e_i,\forall x\in \mathscr H.$$
Now let $p$ be a positive real number. Now $\sigma(T)\subseteq [0,\infty)$. So we can define a continouos function $f:\sigma(T)\rightarrow \Bbb R$ like $f(x)=x^p$. Now Continuous Functional Calculus gives us an operator $f(T)$. Now $f(0)=0$. So the operator $f(T)$ is compact and $f(T)$ is self-adjoit as $f(T)^*=\bar f(T)=f(T)$. Therefore $f(T)$ is compact and normal, hence diagonalizable. My question is the operator $f(T)$ has diagonal representation as of $$f(T)(x)=\sum_{i\in \mathscr I}\alpha_i^p\langle x,\alpha_i\rangle e_i,\forall x\in \mathscr H.$$
Best Answer
It might help to think about this with $f$ being a polynomial, i.e. $f(x)=\sum_{k=0}^n a_kx^k$ (after all, every continuous real-valued function on a compact interval can be approximated arbitrarily well by polynomials, cf. the Weierstrass approximation theorem). Now given the diagonal structure of $T$, it should be quite evident that $T^k=\sum_{i=0}^\infty \alpha_i^k\langle e_i,\cdot\rangle e_i$. Applying $f$ to $T$ then gives the following computation: $$ f(T)=\sum_{k=0}^n a_k T^k=\sum_{i=0}^\infty \Big(\sum_{k=0}^n a_k \alpha_i^k\Big) \langle e_i,\cdot\rangle e_i=\sum_{i=0}^\infty f(\alpha_i)\langle e_i,\cdot\rangle e_i\tag{1} $$ hence why functional calculus on normal (i.e. in some sense "diagonal") operators is about applying a continuous function solely to its spectrum--the diagonal structure is not altered by $f$ in any way.