# Spectrum of $f(T)$, where $T$ is a self-adjoint operator

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.

• There are technicalities that set limits on "$T$ is selfadjoint -> $f(T)$ is selfadjoint". But in general, the statement is true. One proves this with the so-called functional calculus. @Valter Moretti. Dec 8, 2022 at 14:19
• Thank you for your comment. I read that if $T$ is self-adjoint and $f\colon \mathbb{R} \to \mathbb{R}$ is a real continuous function, then it can be shown that $f(T)$ is self-adjoint. See for example Theorem 5.9 on Konrad Schmudgen, unbounded self-adjoint operators on Hilbert space. I see also that in general, under the same hypothesis, we have that $\sigma(f(T))=\overline{f(\sigma(T))}$. But this contraddicts the fact that $\sigma(f(T))=f(\sigma(T))$. Dec 8, 2022 at 14:29

is it true in general that 𝑓(𝜎(𝑇))=𝜎(𝑓(𝑇))?

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 $$f(\sigma_p(X)) \subsetneq \sigma_p(f(X))$$.

• Thank you! Just a last question. What passage in my "proof" is then wrong? Dec 8, 2022 at 17:49
• Actually I do not understand it: why there should be an eigenvector $|\mu\rangle$ of $T$? $\mu$ is an eigenvalue of $f(T)$ not of $T$, as far as I understand. Dec 8, 2022 at 17:50
• Oh my...you are absolutely right. Dec 8, 2022 at 17:54
• Don't worry! :) Dec 8, 2022 at 17:54
• It is correct now, but it is valid for selfadjoint operators which admit a Hilbert basis of eigenvectors and concerns the identity of the point spectrum parts only. As I pointed out $f(T)$ may include also a continuous part even if $\sigma(T)$ does not. In finite dimension it is valid for the whole spectrum obviously. Dec 8, 2022 at 19:04