A Spectrum of a compact operator in $\ell^p$

Question

Let $\alpha_n \in \mathbb C$ and $\lim_{n\to\infty}\alpha_n = 0$. Let $T$ be a linear continuous operator from $\ell^p \to \ell^p (1\le p\le \infty)$ defined by $$ T((x_1, x_2, \ldots)) = (\alpha_1 x_1, \alpha_2 x_2 , \ldots). $$ I could show that $T$ is compact operator in $\ell^p$. But how can I derive the spectrum of $T$? The answer was $\{0, \alpha_1, \alpha_2, \cdots\}$, but still I cannot know how to calculate this. The definition of the "spectrum of $T$" is that $\{\lambda \in \mathbb C : \lambda I - T \text{ is not invertible}\}$, but since I'm a newbie in spectral theory, I could not calculate directly from this definition.

Answer 1

Suppose that $T-\lambda I$ is invertible, then it has trivial kernel and is bounded. Particularly you can solve the equation

$$(T-\lambda I)x = y$$

for any $y\in\ell^p$. Writing $x = (x_m)$ and $y = (y_m)$, we see that

$$ (\alpha_m-\lambda)x_m = y_m,$$

i.e.

$$x_m = \frac{1}{\alpha_m-\lambda}y_m.$$

Note that $\lambda=\alpha_m$ is a serious problem here. We need $x$ to be in $\ell^p$. Computing its norm gives

$$\|x\|_p^p = \sum_{m=0}^{\infty} \left|\frac{1}{\alpha_m-\lambda}y_m\right|^p \le \sup_m\frac{1}{|\alpha_m-\lambda|^p}\cdot\sum_{m=0}^{\infty}|y_m|^p.$$

The only accumulation point in the spectrum of a compact operator is $0$. What does this tell you about $|\alpha_m-\lambda|$ when $\lambda\neq 0,\alpha_m$ for any $m$?

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It took me way longer than it should have to figure out the $\sup$ part. I kept trying to use reverse triangle inequality to get an upper bound even though I knew I needed to invoke accumulation points. Silly me!