Consider on a Hilbert space $\mathcal{H}$ a self-adjoint operator $T$ with spectrum given by $\sigma(T)=\{\lambda_n\}_{n \in \mathbb{N}} \subseteq \mathbb{R}$ (let's suppose for simplicity that the spectrum is discrete).
Let it be $\{|\lambda_n,d_n\rangle\}_{n,d_n}$ a Hilbert basis of $\mathcal{H}$ formed by the eigenvectors of $T$, where $|\lambda_n,d_n\rangle$ is the eigenvector of $T$ such that:
$$
T|\lambda_n,d_n\rangle=\lambda_n|\lambda_n,d_n\rangle\,.
$$
Here $d_n$ is an index that represents the degeneration of $\lambda_n$.
Consider the continuous function $f\colon \mathbb{R} \to \mathbb{R}$. We can define the self-adjoint operator
$$
f(T) \equiv \sum_n\sum_{d_n} f(\lambda_n)|\lambda_n,d_n\rangle\langle\lambda_n,d_n|\,.
$$
Obviously:
$$
f(T)|\lambda_n,d_n\rangle=f(\lambda_n)|\lambda_n,d_n\rangle\,,
$$
i.e., every eigenvector of $T$ associated to $\lambda_n$ is also an eigenvector of $f(T)$ associated to $f(\lambda_n)$. So we have that:
$$
f(\sigma(T)) \subseteq \sigma(f(T))\,,~~~\mbox{where}~~~f(\sigma(T))=\{f(\lambda_n) \in \mathbb{R} \mid \lambda_n \in \sigma(T)\}\,.
$$
My question is: is it true in general that $f(\sigma(T))=\sigma(f(T))$?
My attempt:
Let's suppose that there exists an eigenvalue $\mu \in \sigma(f(T)) \setminus f(\sigma(T))$.
Let it be $|\mu\rangle$ an eigenvector of $f(T)$ associated to $\mu$, namely $f(T)|\mu\rangle=\mu|\mu\rangle$.
So we have that $\langle \lambda_n,d_n \mid \mu \rangle=0$ for every $n,d_n$, being $f(T)$ self-adjoint (then eigenvectors corresponding to different eigenvalues are orthogonal).
So we have that $|\mu\rangle=0$ ($\{|\lambda_n,d_n\rangle\}$ is a Hilbert basis), and this is a contradiction.
Then we have that $f(\sigma(T))=\sigma(f(T))$.
Is my approach correct? Thank you in advance.
Best Answer
The answer is negative already for continuous functions $f: \mathbb{R} \to \mathbb{R}$.
Elementary conterexample. Consider the Hamiltonian $H$ of the harmonic oscillator and then focus on $H^{-1}$. The spectrum of the latter includes the further point $0$ which does not belong to $1/\sigma(H)$. As a matter of fact, the spectrum of $H^{-1}$ admits also $0$ as (unique) element of the continuous part of spectrum $\sigma_c(H^{-1})$. $\qquad\blacksquare$
As a general fact, if 𝑇 is selfadjoint and $f: \mathbb{R} \to \mathbb{C}$ is Borel-measurable, then 𝑓(𝑇) is closed and normal. In particular 𝑓(𝑇) is selfadjoint as well, if 𝑓 as above is real-valued. Obviously continuous functions are Borel-measurable so everything applies to that case.
For continuous functions, the general relation is $$\sigma(f(T))=\overline{f(\sigma(T))}\:.$$
𝜎(𝑓(𝑇)) may be different from 𝑓(𝜎(𝑇)). The former is always closed for the very definition of spectrum, the latter may not, even if 𝑓 is continuous. The conterexample above is an elementary case.
However,
PROPOSITION. $$\sigma(f(T))=f(\sigma(T))$$ if $T=T^*$ is bounded and $f: \mathbb{R} \to \mathbb{R}$ is continuous.
Proof. 𝑇 bounded is equivalent to 𝜎(𝑇) is bounded since $||T||= \sup |\sigma(T)|$. In that case 𝜎(𝑇) is compact (closed and bounded set in $\mathbb{R}$), and thus 𝑓(𝜎(𝑇)) is compact as well because $f$ is continuous, hence 𝑓(𝜎(𝑇)) is closed since it is a compact subset of $\mathbb{R}$. In that case, $$\sigma(f(T))=\overline{f(\sigma(T))}=f(\sigma(T))\:.$$ QED
It is therefore clear that, when $f: \mathbb{R} \to \mathbb{R}$ is continuous, problems may pop up only for unbounded (seldafjoint) operators.
Restricting to the point part of the spectrum, as a general fact we have that $f(\sigma_p(T)) \subset \sigma_p(f(T))$, the proof essentially is the one you wrote. It is valid for every (Borel-measurable) function $f: \mathbb{R} \to \mathbb{R}$. The converse inclusion is false in general. To this end consider the position operator $X$ whose point spectrum is empty. Next consider the map $f: \mathbb{R}\ni x \mapsto 1 \in \mathbb{R}$. Evidently $f(X)=I$. Therefore $\sigma_p(f(X)) = \sigma(I) = \{1\}$, but $f(\sigma_p(X))= \emptyset$ since $\sigma_p(X)= \emptyset$. Therefore $f(\sigma_p(X)) \subsetneq \sigma_p(f(X))$.