# Find the spectrum of the operator $T: \ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined by $(Tx)_n = \frac{x_n}{n}$

Consider the linear operator $T:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ defined as $$(Tx)_n = \frac{x_n}{n}, \quad x \in \ell^2(\mathbb{C}).$$ I can show that it is bounded with norm $\|T\|=1$, which tells me that $$\sigma(T) \subseteq \{ \lambda \in \mathbb{C} : \,\, |\lambda| \le 1 \}.$$ I am also able to show that the point spectrum is $$\sigma_p(T) = \left\{ \frac{1}{n} \right\}_{n \in \mathbb{N} }.$$ Finally I know that the residual spectrum is empty because $T$ is self-adjoint.

1. How can I go on and find the continuous spectrum of $T$?
2. Is there a "more direct" way to show that the residual spectrum of $T$ is empty, without using the self-adjointness property?
• ad 2.: Maybe in some particular cases you can rule out the residual spectrum in another way, but as you see that even the continuous spectrum is identified by 'not point spectrum, not resolvent set'. – Roland Mar 31 '15 at 23:23
• – glS Jun 15 '15 at 7:53

## 2 Answers

One can show that $T$ is a compact operator: define $S_n:\ell^2(\mathbb{C}) \to \ell^2(\mathbb{C})$ by $$(S_nx)_m = \left\{ \begin{matrix} \frac{x_m}{m} & m \leq n \\ 0 & m > n \end{matrix} \right.$$ Note that the $S_n$ are finite-rank and that $$(T-S_nx)_m = \left\{ \begin{matrix} 0 & m \leq n \\ \frac{x_m}{m} & m > n \end{matrix} \right.$$ so $\lVert T-S_n\rVert_2^2 = \frac{1}{n+1} \to 0$ as $n \to \infty$. Hence, $T$ is compact. By the Fredholm Alternative, the non-zero spectrum of $T$ consists purely of eigenvalues (i.e. the point spectrum). Since the spectrum must be closed, it must also contain $0$; as $T$ is injective and bounded, $0$ cannot belong to the point or residual spectra so it belongs to the continuous spectrum.

• Is there a 'except for the possible point 0' missing in your last sentence? Because $0 \notin \sigma_p(T)$, but the spectrum is closed, so 0 must be a point of the continuous spectrum, right? – Roland Mar 31 '15 at 23:21
• @Roland: Ah yes good point. Fredholm Alternative works only for non-zero lambdas, and since the spectrum should be closed, it must contain zero as well. T itself is injective so zero doesn't belong to the point spectrum; the residual spectrum is empty so zero must belong to the continuous spectrum – The Ginger Prince Apr 1 '15 at 3:52

For $\lambda \notin \{0\} \cup \sigma_p(T)$, you can explicitly write down $(\lambda I - T)^{-1}$.

• all right, but the problem is finding the domain of definition of $(\lambda - T)^{-1}$. Is it correct to say that showing that $\lambda \notin \sigma(T)$ amounts to show that $$\forall x \in \ell^2, \,\, (\lambda - T)^{-1} x \in \ell^2 \,?$$ – glS Mar 29 '15 at 19:17