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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.

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  • $\begingroup$ 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

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

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  • $\begingroup$ Thank you! Just a last question. What passage in my "proof" is then wrong? $\endgroup$
    – Leonardo
    Dec 8, 2022 at 17:49
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    $\begingroup$ 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
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    $\begingroup$ 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

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