Let $T: \ell^2 \rightarrow \ell^2$ be an operator $T(x)= (0, \frac{x_1}{1}, \frac{x_2}{2}, \frac{x_3}{3},…)$, $\forall x = (x_n)_{n \in \mathbb{N}} \in \ell^2$. $T$ is a compact, bounded linear operator and $\sigma_p (T)= \emptyset$, so $(T- \lambda I)^{-1}$ exists for all $\lambda$. Now I have to prove that residual spectrum consists only of $0$. Residual spectrum are those $\lambda \in \mathbb{C}$ for which $(T- \lambda I)^{-1}$ exists but is not defined on the dense subset of $\ell^2$.
It can be proved that $0 \in \sigma_r (T)$ and now we consider $\lambda \neq 0$. My attempt is to find the vector $x =(x_n) \in \ell^2$ such that
$(T- \lambda I )(x)= (\alpha_1, \alpha_2, …,\alpha_n,0,0,0….)$,
since the set of such sequences is dense in $\ell^2$.
$ – \lambda x_1 = \alpha_1$
$ x_1- \lambda x_2= \alpha_2$
$\frac{x_2}{2} – \lambda x_3 = \alpha _3$
…
$\frac{x_{n-1}}{n-1} – \lambda x_{n} = \alpha _n$
$\frac{x_{n}}{n} – \lambda x_{n+1} = 0$
…
How can one prove from this system that $x \in \ell ^2$?
Best Answer
For $k>n$ we have $$x_k={\lambda^n\over \lambda^k}{(n-1)!\over (k-1)!}x_n$$ The sequence $\lambda^{-k}((k-1)!)^{-1}$ is absolutely summable (by for example the ratio test), hence square summable.
Another argument: for $k>n$ and $k>2|\lambda|^{-1}$ we have $$|x_{k+1}|={1\over |\lambda| k}|x_k|\le {1\over 2}|x_k|$$ Thus the sequence is square summable.