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