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Let $E = \ell^2$, and consider the multiplication operator $M$: $$Mx = (\alpha_1x_1,\dots,\alpha_nx_n,\dots), \forall x\in E$$ where $(\alpha_n)$ is a bounded sequence in $\mathbb{R}$.

And consider the right shift operator $S_r:(x_1,x_2,\dots)\to(0,x_1,x_2)$

Assume $\alpha_n\to \alpha$, as $n\to\infty$. Prove that the spectrum of $S_r\circ M$ $$\sigma(S_r\circ M) = [-|\alpha|,|\alpha|]$$ Here is the solution of the of this problem. I don't know why $M-\alpha I = K$, is compact. And also what is the explict expression of $K_2$. Can someone help me with this? Thanks.

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Martin Argerami
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1 Answers1

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You have $$ (M-\alpha I)x=(\beta_1x_1,\beta_2x_2,\ldots), $$ where $\beta_n=\alpha_n-\alpha\to0$. By truncating, you may write $M-\alpha I$ as a limit of finite-rank operators, and that makes it compact.

Then you take $K_2=(\alpha S_r-\lambda I)^{-1}K_1$.

Martin Argerami
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