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.

  • $\begingroup$ 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. $\endgroup$
    – DanielC
    Dec 8, 2022 at 14:19
  • $\begingroup$ 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 $\sigma(f(T))=f(\sigma(T))$. $\endgroup$
    – Leonardo
    Dec 8, 2022 at 14:29

1 Answer 1


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.


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

  • $\begingroup$ Thank you! Just a last question. What passage in my "proof" is then wrong? $\endgroup$
    – Leonardo
    Dec 8, 2022 at 17:49
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 8, 2022 at 17:50
  • $\begingroup$ Oh my...you are absolutely right. $\endgroup$
    – Leonardo
    Dec 8, 2022 at 17:54
  • $\begingroup$ Don't worry! :) $\endgroup$ Dec 8, 2022 at 17:54
  • 4
    $\begingroup$ 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. $\endgroup$ Dec 8, 2022 at 19:04

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.