In my research group in functional analysis and operator theory (where we do physics and computer science as well), we saw in an old Russian combination paper/PhD thesis in our library a nice claim about the spectral mapping theorem's possible proof. Let me attempt to bring the context here. I should mention there are some nice results in this paper that I wanted to use and generalize for my own research, I hope to accurately bring the context below.

They bring up the continuos functional calculus $\phi: C(\sigma(A)) \rightarrow L(H)$ for a bounded, self-adjoint operator on a Hilbert space A. This is an algebraic *-homomorphism from the continuous functions on the spectrum of $A$ to the bounded operators on $H$. The paper's spectral mapping theorem basically says in this context $$ \sigma(\phi(f)) =f(\sigma(A)) $$ and the paper says something nice about this. It does not actually give a proof but it says there is a nice way to prove it using both inclusions with the inclusion $ f(\sigma(A)) \subseteq \sigma(\phi(f)) $ sketched in the following way: the author supposes $ \lambda \in f(\sigma(A)) $ and says "it is very obvious" that there exists a vector $h \in H$ with $\|h\|=1$ such that $\|\phi(f)-\lambda)h\|$ is arbitrarily small which shows $\lambda \in \sigma(\phi(f))$ which shows the desired inclusion.

The author says that it is "very obvious" to show this but I am a bit stumped. The way I would construct the continuous functional calculus is to start with polynomials and then generalize to $ C(\sigma(A)) $ based on the Weierstrass approximation theorem on the real compact set $\sigma(A)$ and the BLT theorem. The inclusion $\sigma(\phi(f)) \subseteq f(\sigma(A))$ is, I think, quite obvious but the other one in the above context has me stumped. Since I am already working on generalizing some results, I would really love to know how the author proves the inclusion with the method of showing the mentioned vector exists. Maybe use approximation in some way, but even though I suspect it is simple, I still do not see the author's proposed proof. Can someone here please help me recover it? I thank all interested persons.

  • 2
    $\begingroup$ Does he/she continued to work on his thesis problem? Perhaps looking at the Author's closely following works would help you: this can include short notes, abstract of congresses presentations and the like. The All Russian Mathematical Portal can be very effective in this kind of researches. $\endgroup$ – Daniele Tampieri Oct 14 at 4:29
  • 2
    $\begingroup$ @DanieleTampieri thank you for the suggestion I think I am close to proving it $\endgroup$ – kroner Oct 14 at 4:30

It is quite hard to answer this question, as I do not know exactly how $\phi$ is defined, nor what we "know" about the spectrum of a self-adjoint operator. I think standard presentations of this circle of ideas tend to be quite "tight", in the sense that you have to be careful not to get into the situation of giving a circular argument.

So... With that said, you could argue as follows. Let's assume:

  • $\phi$ is continuous, and does what we expect on polynomials. This is enough to define $\phi$ completely.
  • With $A$ a bounded self-adjoint operator on $H$, every $\mu\in\sigma(A)$ is an eigenvalue, or in the continuous spectrum. So, for any $\epsilon>0$ we can find $h\in H$ with $\|h\|=1$ and $\|(A-\mu)h\|<\epsilon$.

Given $g\in C(\sigma(A))$ we can approximate $g$ by a polynomial $f$, and so by a 3 $\epsilon$ argument, we can assume we just have a polynomial (with real coefficients). To be more precise, given $\lambda\in g(\sigma(A))$ we can find a polynomial $f$ with $\|\phi(f) -\phi(g)\|<\epsilon$ and with $\|f-g\|_\infty<\epsilon$, so if $\lambda=g(\mu)$ for some $\mu\in\sigma(A)$, then $|f(\mu)-\lambda| < \epsilon$. If the result holds for $f$, say we have $h\in H$ with $\|h\|=1$ and $\| (\phi(f)-f(\mu))h \|<\epsilon$, then \begin{align*} \| (\phi(g)-\lambda)h \| &\leq \| (\phi(g)-\phi(f))h \| + \|(\phi(f)-f(\mu))h\| + \|(f(\mu)-\lambda)h\| \\ &\leq \| \phi(g)-\phi(f) \| + \epsilon + |f(\mu)-\lambda| \\ &< \epsilon + \epsilon + \epsilon. \end{align*}

We are now done, because for a polynomial (with real coefficients) $f$, we have that $\phi(f) = f(A)$ is a bounded self-adjoint operator, and we know that $f(\sigma(A)) = \sigma(f(A))$ from just algebraic arguments. The result follows from my second assumption.

But again I warn that without seeing the rest of your source, I cannot be sure if this is not a circular argument.

(Alternative argument, which doesn't use spectral mapping for polynomials: Suppose $$ f(t) = \sum_{i=0}^n a_i t^i. $$ Then given $\lambda\in\sigma(f(A))$ we have that $\lambda=f(\mu)$ for some $\mu\in\sigma(A)$. Then $$ \phi(f) = f(A) = \sum_{i=0}^n a_i A^i. $$ Choose $h$ with $\|h\|=1$ and $\|(A-\mu)h\|<\epsilon$. Then $$ \|(\phi(f)-\lambda)h\| = \|(f(A)-f(\mu))h\| \leq \sum_{i=1}^n |a_i| \|(A^i - \mu^i)(h)\|. $$ Now use that $$ A^i-\mu^i = \big( A^{i-1} + \mu A^{i-2} + \cdots + \mu^{i-1}\big)(A-\mu). $$ So $$ \|(\phi(f)-\lambda)h\| \leq \sum_{i=1}^n |a_i| \big( \|A\|^{i-1} + \|A\|^{i-2} |\mu| + \cdots + |\mu|^{i-1} \big). $$ As $f$ is fixed, by choosing $\epsilon>0$ small we can make $\|(\phi(f)-\lambda)h\|$ small, as desired.)

| cite | improve this answer | |
  • $\begingroup$ Thank you very much $\endgroup$ – kroner Oct 14 at 9:34
  • $\begingroup$ Please do check that my argument works! $\endgroup$ – Matthew Daws Oct 14 at 9:38
  • $\begingroup$ I do think it works. Thank you very much. $\endgroup$ – kroner Oct 14 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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