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.

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    $\begingroup$ Just a note about terminology. In the context you are considering, the word is "spectrum" not "spectral." $\endgroup$ – Cameron Williams Jul 23 '16 at 19:49
  • $\begingroup$ @CameronWilliams many thanks. I edited it. $\endgroup$ – John S Jul 23 '16 at 19:53
  • $\begingroup$ For each $i$, Can you find a nontrivial element $x$ so that $Tx = \alpha_i x$? (You are essentially just finding the eigenvalue of $T$). $\endgroup$ – user99914 Jul 23 '16 at 20:02
  • $\begingroup$ On the other hand, for $\lambda \notin \{ 0, \alpha_1, \alpha_2, \cdots \}$, one can very easily write down the inverse of $T - \lambda I$. $\endgroup$ – user99914 Jul 23 '16 at 20:05
  • $\begingroup$ Recall Fredholm theorem and use it to verify the claim (along with the fact that compact operators are never surjective). $\endgroup$ – user228113 Jul 23 '16 at 20:31

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

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


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

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 $m$?

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!

  • $\begingroup$ One might also overkill it by the bounded inverse theorem. $\endgroup$ – user99914 Jul 23 '16 at 20:56
  • $\begingroup$ @ArcticChar Yeah I thought about using some more sophisticated machinery, but I figured keeping it as simple as possible was best. $\endgroup$ – Cameron Williams Jul 23 '16 at 20:57
  • $\begingroup$ Thanks I really appreciate to your kind answer. I understood all the arguments above the sentence "The only accumulation point in ......" . In fact I could not understand that how the inequality $\|x\|_p^p \le \sup \frac{1}{|\alpha_m - \lambda|^p} \sum|y_m|^p$ implies the spectrum is $\{0, \alpha_1, ...\}$. Could you give me some more detail for this? $\endgroup$ – John S Jul 23 '16 at 22:17
  • $\begingroup$ The stated inequality shows that the inverse mapping $y\mapsto x$ is a bounded operator, if $\sup \frac{1}{|\alpha_m - \lambda|^p} <\infty$. @JohnS $\endgroup$ – user99914 Jul 24 '16 at 4:31
  • $\begingroup$ @ArcticChar Thanks, I think if $\lambda = 0$ then I can wrtie the inverse as $x_m = y_m/\alpha_m $. Why is this statement false? would you explain me for this? thanks : ) $\endgroup$ – John S Jul 25 '16 at 15:18

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