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?

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.

  • $\begingroup$ 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? $\endgroup$ – Roland Mar 31 '15 at 23:21
  • $\begingroup$ @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 $\endgroup$ – The Ginger Prince Apr 1 '15 at 3:52

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

  • $\begingroup$ 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 $$ \forall x \in \ell^2, \,\, (\lambda - T)^{-1} x \in \ell^2 \,?$$ $\endgroup$ – glS Mar 29 '15 at 19:17

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