Is the mapping from a Schatten-class compact operator to its spectrum continuous?

Question

On an infinite dimensional Hilbert space, let $S_p$ be the Banach space of compact operators that are $p$ Schatten class. Since for $A\in S_p$ its spectrum $\sigma(A)$ is at most countably infinite, we can naturally consider it as an element of the Banach space $\ell_p$

Is the mapping from an operator to its spectrum $\sigma: S_p \to \ell_p$ continuous for any $p\in [1,\infty]$? This is true for finite-dimensional Hilbert spaces but I'm wondering how it generalizes.

In particular, what if we restrict things to the subspace of $S_p$ consisting of self-adjoint operators?

Answer 1

This is less of an answer and more of a related result that came to mind: Lemma 5 in Chapter XI.9 of the Dunford & Schwartz classic "Linear operators. Part II. Spectral theory" reads

Let $T_n,T$ be compact operators, and let $T_n\to T$ in the uniform operator topology. Let $\lambda_m(T)$ be an enumeration of the non-zero eigenvalues of $T$, each repeated according to its multiplicity. Then there exist enumerations $\lambda_m(T_n)$ of the non-zero eigenvalues of $T_n$, with repetitions according to multiplicity, such that $$ \lim_{n\to\infty}\lambda_m(T_n)=\lambda_m(T),\qquad m\geq 1, $$ the limit being uniform in $m$.

Once the spectral mapping $\sigma:\mathcal K(\mathcal H)\to\ell^\infty(\mathbb N)$ (resp. the restriction to some Schatten class) is well-defined${}^1$ this result may allow to generalize the result from finite dimensions to the general setting.

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_{${}^1$: In a comment you specified that $\sigma$ shall order the eigenvalues of the input decreasingly according to its absolute value, but this leaves room for problems. For example if the input has eigenvalues $i$, $-i$ then $\sigma$ fails to give a unique ordering to them. Of course this problem vanishes for the simpler special case of self-adjoint operators you mentioned in your post; so you may want to try to rigorously formulate and prove that result first and then try to "make sense" of the case of arbitrary input operators.}