Like my previous question, I'm considering the same space and operator: Hilbertspace adjoint But this time I am trying to determine the spectrum of $T$. I feel like I'm messing up my definitions a little.
Again, let $(e_n)$ be a orthonormal basis of $l^2$, and fix arbitrary complex numbers $(\lambda_n)$ and define $T:l^2\to l^2 $ as $$T\left(\sum x_ne_n\right)=\sum \lambda_nx_ne_n, $$ and let $$D(T) = \left\{\sum x_ne_n: \sum |\lambda_nx_n|^2 <\infty\right\}.$$
My thought process went as follows:
Let $x \in l^2$ be such that $x_j = 1$ and $x_k =0$ for $k\neq j$, then clearly $x\in D(T)$. With this $x$, $\lambda_j$ is an eigenvalue for $T$, since $$(T-\lambda )(x) = \sum_k (\lambda_k-\lambda)x_ke_k = (\lambda_j - \lambda)e_j $$ which vanishes if and only if $\lambda = \lambda_j$. Hence $\lambda_j \in \sigma_p(T)$. Since $\sigma(T)$ is closed, it follows $\overline{\{\lambda_k\}_{k\geq 1}}\subset \sigma(T)$. Let $\lambda$ be such that $\lambda$ is no accumulation point of $\{\lambda_k\}_{k\geq 1}$, then there exists $\delta >0$ such that $|\lambda-\lambda_k | > \delta$ for all $k$. Then the map $R:l^2\to l^2$ given by $$ \sum x_ke_k \mapsto \sum (\lambda-\lambda_k)^{-1}x_ke_k$$ is a bounded on $l^2$, since $\left\|Rx\right\|\leq \frac{1}{\delta}\left\|x\right\|$.
But wait a minute..., I thought for a second there: Ok then $\lambda\in \rho(T)$, since $T-\lambda$ has a bounded inverse $R$. But the resolvent set is defined as: $$\rho(T) = \{\lambda: T-\lambda: D(T)\to l^2 \text{ bijective}\} $$ And I can't see how $T-\lambda$ would be surjective for any $\lambda$, Does it realy follow that $\sigma(T)=\mathbb{C}$?
The spectrum of $T$ is $\overline{\{\lambda_k,\ k\in\mathbb N\}}.$ Your arguments are correct.
$(T-\lambda):D(T)\to l^2$ is surjective if $|\lambda-\lambda_k | > \delta$ for all $k.$ Indeed, take $x=\sum_k x_ke_k\in l^2.$ Then $(T-\lambda)y=x$ where $y=\sum (\lambda_k-\lambda)^{-1}x_ke_k.$ It is easy to see that $y\in l^2.$ We check that $y\in D(T)$: $$ \sum_k\left|\frac{\lambda_k}{(\lambda_k-\lambda)}x_k\right|^2=\sum_k\left|1+\frac{\lambda}{(\lambda_k-\lambda)}\right|^2|x_k|^2\leq\left(1+\frac{|\lambda|}{\delta}\right)^2\sum_k|x_k|^2<\infty. $$