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Let $H$ be a separable Hilbert space with orthonormal basis $\{e_i\}$. Let $(c_n)$ be a bounded sequence of complex numbers and consider the bounded linear operator $T$ on $H$ defined by $$Tx = \sum_{n\geq 1} c_n(x,e_n)e_n$$

a) What is the spectrum of $T$?

b) Give an example of a selfadjoint operator on $H$ whose spectrum is $[-1,1]$.

Looking at $(T - \lambda I)x$ where $x = \sum_{n\geq 1} (x,e_n)e_n$ by Parsevall we get the eigenvalues $\lambda_i = c_i$ one guess is that the spectrum $\sigma(T) = \overline{\{\lambda_i\}}$.

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The spectrum is the closure of the set $\{c_n,n\geqslant 1\}$. To see that, take an element $\lambda$ which is not in the closure of this set, and show that $T-\lambda I$ is invertible. To see that, fix $\delta$ such that for all $n$, $|\lambda-c_n|>\delta$. So we can define the operator $T\colon e_n\mapsto \frac 1{\lambda-c_n}e_n$.

Then part b) is easily solvable from that.

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Davide, please help me to see why the boundary elements of $\{c_n\}$ are in the spectrum. – Tomás Dec 19 '12 at 12:32
The spectrum is closed and contains the set of eigenvalues. – Davide Giraudo Dec 19 '12 at 12:33
Thanks! So say we have $\mu \not\in cl(\lambda_n)$ is it enought to only look whats happens to the basis elements? I get that they are bounded bellow. How can I proceed? – Johan Dec 20 '12 at 14:43

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