I have been looking around for an example of a general continuous (bounded) linear operator, who's spectrum is any compact set $K\subset\mathbb{C}$.

I have seen an example, where we take the set $\{\alpha_n\}$ to be a countable dense subset of $K$. Let $H$ be a hilbert space with orthonormal basis $\{e_n\}$, define the operator $T:H\to H$ such that:

$T(u)=\sum_n\alpha_n\langle u,e_n\rangle e_n.$ I can see that any eigenvalue of $T$ is some $\alpha_j\in\{\alpha_n\}$ since:

$Tu-\lambda u=0\Rightarrow \sum_n\alpha_n\langle u ,e_n \rangle e_n-\sum_n\lambda\langle u,e_n\rangle e_n=0$ Now each $i$-th term gives us:

$\alpha_iu_i-\lambda u_i=0\Rightarrow \lambda = \alpha_i$ so we know that every value in $\{\alpha_n\}$ is in the spectrum of $T$ so we see that the spectrum is at least a dense subset of $K$.

But this is where I am stuck and would appreciate help/ explaining what comes next. Thanks!

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    $\begingroup$ The spectrum is always closed, so $\sigma(T) \supset \overline{\{\alpha_n : n\in\mathbb{N}\}} = K$. On the other hand, it is not hard to see that if $\lambda\notin K$ then $\lambda I - T$ is invertible. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:00
  • $\begingroup$ Yes, but how do we go the other way i.e. $\sigma(T)\subset\overline{\{\alpha_n:n\in\mathbb{N}}\}$ $\endgroup$ – Ellya Apr 5 '14 at 15:02
  • $\begingroup$ As we then need to show that the resolvent exists, is bounded and is densely defined in $H$. I see existence and boundedness, but not necessarily densely defined? $\endgroup$ – Ellya Apr 5 '14 at 15:03
  • $\begingroup$ That's what the second sentence is about. If $\lambda \notin K$, then $\lambda \notin \sigma(T)$. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:03
  • $\begingroup$ Could you show me why? that is the bit I'm stuck on, thanks! $\endgroup$ – Ellya Apr 5 '14 at 15:04

By construction, every $\alpha_k$ is an eigenvalue of $T$, hence we have

$$K = \overline{\{\alpha_k : k\in\mathbb{N}\}} \subset \sigma(T).$$

On the other hand, if $\lambda\notin K$, then there is a $\delta > 0$ such that $\lvert\lambda - \alpha_k\rvert > \delta$ for all $k$, and then we find the inverse of $\lambda I - T$ explicitly:

$$\begin{align} (\lambda I - T)\left(\sum_{n\in\mathbb{N}} y_n\cdot e_n\right) &= \sum_{n\in\mathbb{N}} x_n e_n\\ \iff \sum_{n\in\mathbb{N}} (\lambda - \alpha_n)y_n e_n &= \sum_{n\in\mathbb{N}} x_n e_n\\ \iff \bigl(\forall n\in \mathbb{N}\bigr)\bigl( (\lambda-\alpha_n)y_n &= x_n\bigr)\\ \iff \bigl(\forall n\in\mathbb{N}\bigr)\biggl(y_n &= \frac{1}{\lambda-\alpha_n}x_n\biggr), \end{align}$$


$$(\lambda I - T)^{-1}\left(\sum_{n\in\mathbb{N}} x_n e_n\right) = \sum_{n\in\mathbb{N}} \frac{x_n}{\lambda-\alpha_n}e_n.$$

Since $\lvert\lambda - \alpha_n\rvert > \delta$ for all $n$, we have $\left\lvert\frac{1}{\lambda-\alpha_n}\right\rvert < \frac{1}{\delta}$, and so $(\lambda I - T)^{-1}$ is a globally defined continuous operator, hence $\lambda \notin \sigma(T)$, which yields the other inclusion $\sigma(T) \subset K$, altogether $\sigma(T) = K$.

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  • $\begingroup$ One thing, the sums and sets should all be over $\mathbb{Q}$, since $K\subset \mathbb{C}$ and $\{\alpha_n\}$ is dense subset of $K$ i.e. all values in $K$ with rational coeffiencents. Anyway, would the last series converge $\forall x\in H$? $\endgroup$ – Ellya Apr 5 '14 at 15:22
  • $\begingroup$ I assumed the $\alpha_n$ were $\mathbb{N}$-indexed. The bound $\left\lvert \frac{1}{\lambda-\alpha_n}\right\rvert < \frac{1}{\delta}$ ensures the convergence, since $$\sum \left\lvert \frac{x_n}{\lambda-\alpha_n}\right\rvert^2 \leqslant \sum \delta^{-2}\lvert x_n\rvert^2 < \infty$$ follows from $\sum \lvert x_n\rvert^2 < \infty$. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:28
  • $\begingroup$ So should I choose $H $ to be $l^2(\mathbb {Q})$? $\endgroup$ – Ellya Apr 5 '14 at 15:32
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    $\begingroup$ Careful. $K$ need not contain any point with rational coefficients, so taking $\{\alpha_n\}$ to be the points in $K$ with rational coefficients doesn't work then. What one does is take any countable dense subset of $K$, and countable means we can index by $\mathbb{N}$. We could also index by other sets, and we can even take a nonseparable Hilbert space, e.g. $\ell^2(K)$ (if $K$ is countable, that is separable), with all $\alpha\in K$ as eigenvalues. $\endgroup$ – Daniel Fischer Apr 5 '14 at 15:38
  • $\begingroup$ But often taking a countable set indexable by $\mathbb { N } $ may not be dense in $K $. Without specifying $H $, how did you justify the convergence of that last series? $\endgroup$ – Ellya Apr 5 '14 at 16:37

Suppose that a non-empty set $K$ in $\mathbb{C}$ is given and choose a dense sequence $(\lambda_n)$ in $K$. If $T$ is the corresponding diagonal operator, $T(e_n)=\lambda_n e_n$, then ${\rm sp}(T)=\overline{\{\lambda_n: n\in\mathbb{N}\}}=K$. Thus, the spectrum is an arbitrary non-empty compact set in general.

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