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Given a bounded sequence $\pi = (\lambda_n)$ in $\mathbb{C}$ consider the continuous linear map $M_\pi:\ell^2\rightarrow \ell^2$ defined by $$M_\pi(x_n) = (\pi_nx_n)$$ a) determine the spectrum. b) Characterize the sequences $\pi$ for which $M_\pi$ is compact.

my try: a) The spectrum is defined where $(M_\pi - I\lambda)$ is invertible. $M_\pi$ is bounded iff $\pi$ is bounded. hence we get the solution $\sigma(M_\pi) = {\lambda_i} \;\; \forall i$ b) If $\pi_i \neq 0$ we have sequences $(x_n) = ne_n$ which has no cauchy subsequences, hence $M_\pi$ is only compact if $\pi_i = 0\;\; \forall i$

it feels like im missing something, is there any other $\pi$ that makes $M_\pi$ compact?

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Yes, there are. For (a) note, that the spectrum is closed. And the set you gave, namely $\Pi := \{\pi_n \mid n \in \mathbb N\}$ need be closed. As the spectrum is closed, we obtain $\overline\Pi \subseteq \sigma(M_\pi)$, we will now show the other inclusion: Let $\lambda \not\in \overline\Pi$, then $\mathrm{dist}(\lambda, \Pi)> 0$, define $\psi_n := (\pi_n - \lambda)^{-1}$, then $|\psi_n| \le \frac 1{\mathrm{dist}(\lambda,\Pi)}$, hence $\psi := (\psi_n)$ is bounded and $M_{\psi}$ is inverse to $M_\pi - \lambda$. So $\sigma(M_\pi) = \overline\Pi$.

For (b) note, that if $M_\pi$ is compact, by $M_\pi e_n = \pi_n e_n$ we need $(\pi_n e_n)$ to have a convergent subsequence, with 0 as only possible limit (as $\ell^2$-convergence implies coordinatewise convergene). Applying this argument to each subsequence, we obtain $\pi_n \to 0$ if $M_\pi$ is compact. On the other hand, suppose $\pi_n \to 0$. Consider for $n \in \mathbb N$ the sequence $$ \pi^{(n)} := (\pi_1, \ldots, \pi_n, 0, \ldots) $$and the corresponding multiplication operator $M_{\pi^{(n)}}$. $M_{\pi^{(n)}}$, having finite dimensional range, is compact, and for $x \in \ell^2$ we have \begin{align*}\def\norm#1{\left\|#1\right\|} \norm{M_\pi x - M_{\pi^{(n)}}x} &= \left(\sum_{k=n+1}^\infty |\pi_kx_k|^2\right)^{1/2}\\ &\le \sup_{k \ge n+1}|\pi_k| \cdot \left(\sum_{k=n+1}^\infty |x_k|^2\right)^{1/2}\\ &\le \sup_{k\ge n+1} |\pi_k| \cdot \|x\| \end{align*} Hence $M_{\pi^{(n)}} \to M_\pi$ in $L(\ell^2)$, and $M_\pi$, being the norm-limit of compact operators is compact. So $M_\pi$ is compact iff $\pi$ converges to zero.

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What is $\overline\Pi$ is that not equal to $\Pi$? –  Johan Dec 10 '12 at 10:03
    
$\overline\Pi$ denotes the closure of $\Pi = \{\pi_n \mid n \in \mathbb N\}$, $\Pi$ is seldom closed ... exactly if $(\pi_n)$ contains its accumulation points. If, for example $\pi = (\frac 1n)$, then $\overline \Pi = \Pi \cup\{0\}$. –  martini Dec 10 '12 at 10:06
    
I understand! Can you clarify a little bit why having finite dimensional range implies compactness? –  Johan Dec 10 '12 at 10:23
    
Let $T\colon X \to Y$ a linear operator having finite dimensional range. Let $(x_n)$ a bounded sequence in $X$. Then $(Tx_n)$ is a bounded sequence in the finite dimensional space $TX \cong \mathbb K^d$. As we now from calculus, in $\mathbb K^n$, each bounded sequence has a convergent subsequence, so has $(Tx_n)$. –  martini Dec 10 '12 at 10:26
    
Thanks! Looking more closely, how do we know that the $\psi$ is bounded in $\ell^2$, And what does it mean that we have an inverse to $M_\pi - \lambda$, it means that lambda is not in the spectrum, hence a contradictions? –  Johan Dec 10 '12 at 12:14

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