Spectrum of direct sum and direct integral included in the union of the spectra of its constituents

Question

Suppose that $A = \oplus_{n=0}^\infty A_n $. Where $A_n$ are some operators on some Hilbert spaces $\mathcal{H}_n$ and $A$ an operator on $\oplus H_n$. Let $\sigma(A)$ denote the spectrum of $A$. Does it hold that $\sigma(A) = \cup_{n \in \mathbb{N}} \sigma(A_n) $ or something like $\sigma(A) \subset \cup_{n \in \mathbb{N}} \sigma(A_n) $? Does the same thing generalise to direct integrals?

Answer 1

In the case where the operators are normal everything becomes easier, so let us only think about the non-normal case (and the normal case will follow from it).

In the non-normal case one good tool is the pseudospectrum. The $\varepsilon$-pseudospectrum of an operator $A$ is the set of $\lambda$ where $\Vert (A -  \lambda)^{-1} \Vert\ge\varepsilon^{-1}$ we call the set $\sigma_{\varepsilon}(A).$

Then from our paper https://arxiv.org/abs/2206.09879 we prove following result from here:https://ejde.math.txstate.edu/Volumes/2011/89/abstr.html :

Suppose that $\mathcal{H} = \oplus_{n \in \mathbb{N}} \mathcal{H}_n$ where each $ \mathcal{H}_n$ is a separable Hilbert space. Let $A_n \in \mathcal{B}( \mathcal{H}_n) $ for each $n \in \mathbb{N}$ and $A = \oplus_{n \in \mathbb{N}} A_n$ be a bounded operator on $\mathcal{H}$. Then for all $\varepsilon > 0 $ it holds that \begin{align*} \sigma_{\varepsilon}(A) = \bigcup_{n \in \mathbb{N}} \sigma_\varepsilon( A_n) \text{ and  } \sigma(A) = \bigcap_{\varepsilon > 0}  \bigcup_{n \in \mathbb{N}} \sigma_\varepsilon( A_n). \end{align*}

For the proof: Let first $\lambda \in \cup_{n \in \mathbb{N}} \sigma_{\varepsilon}(A_n)$, then there exists an $n_0 \in \mathbb{N}$ such that $\vert \vert{(A_{n_0} - \lambda)^{-1}} \geq \frac{1}{\varepsilon} $. Thus, $ \sup_{n \in \mathbb{N}} \vert \vert (A_{n} - \lambda)^{-1} \vert \vert \geq \frac{1}{\varepsilon} $ and hence $\lambda \in \sigma_{\varepsilon}(A)$.

For the converse inclusion, we use contraposition. So suppose that $\lambda \notin \cup_{n \in \mathbb{N}} \sigma_\varepsilon( A_n)$. Then for all $n \in \mathbb{N}$ it holds that $ \vert \vert (A_n - \lambda)^{-1} \vert \vert \leq \frac{1}{\varepsilon}$. Thus, $ \sup_{n \in \mathbb{N}} \vert \vert ( A_n - \lambda)^{-1} \vert \vert \leq \frac{1}{\varepsilon}$. It follows that $ \oplus_{n \in \mathbb{N}} (A_n - \lambda)^{-1}$ is a well-defined bounded operator. Since \begin{align*} \left( \oplus_{n \in \mathbb{N}} (A_n - \lambda)^{-1} \right) (A- \lambda) = \oplus_{n \in \mathbb{N}} (A_n - \lambda)^{-1} (A_n - \lambda) = 1, \end{align*} we see that $ (A- \lambda)$ is invertible. Thus $ \lambda \notin \sigma(A)$. The second relation follows immediately.

This is also the statement that generalizes well to the direct integral case. This we also prove in the paper https://arxiv.org/abs/2206.09879.