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I am supposed to prove that for two bounded commuting operators $A, B$ on a banach space, the hausdorff distance of the spectra $d_H(\sigma(A),\sigma(B))$ is less than or equal to the spectral radius of the difference $r(A-B)$. I figured that it would suffice to show that for every $a\in \sigma(A)$, there is a $b \in \sigma(B)$ such that $a-b \in \sigma(A-B)$. This is the case in the finite dimensional case and can be shown using matrix decompositions unavailable in the infinite case (see here). However, I don't know if this holds true in the infinite case, let alone how to prove it.

Julius Himmel
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    This is certainly true if $A$ and $B$ are normal operators on a Hilbert space, by a standard argument via Gelfand transform. Not sure whether this works in general. – David Gao May 21 '24 at 06:00
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    A further remark: using Gelfand transform, one can show that $d_H(\sigma_K(A), \sigma_K(B)) \leq r(A - B)$, where $K$ is the unital, commutative Banach algebra generated by $A$, $B$, and $I$, the identity operator. However, there is no guarantee in general that $\sigma_K(A) = \sigma(A)$ or $\sigma_K(B) = \sigma(B)$. The case for normal operators on Hilbert spaces works because one may instead work with the commutative $C^\ast$-algebra generated by $A$, $B$, and $I$, and inclusions of $C^\ast$-algebras preserve spectra… – David Gao May 21 '24 at 06:33
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    … The finite-dimensional case works because in that case, both $\sigma(A)$ and $\sigma(B)$ are finite sets and therefore neither contains holes. And when the spectra have no holes, we do have $\sigma_K(A) = \sigma(A)$ and $\sigma_K(B) = \sigma(B)$. – David Gao May 21 '24 at 06:36
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    @DavidGao This may help: if $\mathcal{K}$ is a Banach subalgebra of a Banach algebra $\mathcal{L}$ and $A\in \mathcal{K}$ then $\partial\sigma_{\mathcal{K}}(A)=\partial\sigma_{\mathcal{L}}(A).$ – Ryszard Szwarc May 21 '24 at 13:50
  • @RyszardSzwarc I’m aware of the result. I tried to use it but unfortunately the Hausdorff distance between two sets does not reduce to the Hausdorff distance between their boundaries - you don’t even get an inequality in either direction, and counterexamples in both directions can be given by just filling in/hollowing out a region in the plane bounded by a closed curve, which is exactly what can happen when passing to a Banach subalgebra (the spectrum can become larger by filling in holes in the spectrum within the larger algebra). – David Gao May 21 '24 at 15:04
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    See here for the same question (with answer). – jd27 May 21 '24 at 19:09

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