Let $X$ be an infinite dimensional Banach space, and let $T \in L(X)$ such that the set of accumulation points of $T$ is non empty, i-e $\mbox{acc}\,\sigma(T)\neq 0.$\ Is there a Banach space $Y$ and an operator $S_{T} \in L(Y)$ (which is a function of $T$) with $\sigma(S_{T})=\mbox{acc}\,\sigma(T)$ and $\mbox{iso}\,\sigma(S_{T})=\emptyset.$
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1$\begingroup$ A subset of the corresponding field, real or complex, is the spectrum of a bounded linear operator if and only if it is compact. $\endgroup$– hordubalCommented Oct 31, 2021 at 9:04
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$\begingroup$ Thanks, but I am looking for how can I remove isolated points from an operator $T$ to find an operator $S_ {T}$ which is a function of $T$ with $\sigma(S_{T})=\mbox{acc}\,\sigma(T)$ and $\mbox{iso}\,\sigma(S_{T})=\emptyset.$ $\endgroup$– LuffyCommented Oct 31, 2021 at 17:58
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$\begingroup$ Maybe I don't understand this question correctly, but any diagonal compact operator K on Hilbert space H has only 0 as an accumulation point, but the only function of K which removes the isolated points is 0, and 0 is an isolated point of the spectrum of the 0 operator, with the eigenspace = H. $\endgroup$– DerekCommented Apr 1, 2022 at 9:27
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