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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.

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    $\begingroup$ 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
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    $\begingroup$ @DanieleTampieri thank you for the suggestion I think I am close to proving it $\endgroup$ – kroner Oct 14 at 4:30
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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.)

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  • $\begingroup$ 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

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