Spectrum and compactness of $\ell ^1$ operator Consider complex sequence $(a_n)$, $|a_n| \leq 1$. On $\ell ^1$ space consider the operator
$$ T(x_n) = (a_n x_n).$$
Find the norm. Decide for what sequences is $T$ compact. Find point spectrum and spectrum.

I think the norm is equal to $\lVert a_n \rVert_\infty$. Since $|a_n|\leq 1 $ implies $(a_n) \in \ell ^\infty$. Then
$$\lVert T(x_n) \rVert = \lVert (a_1 x_1, a_2 x_2, \ldots) \rVert = 
\sum_{n=1}^{\infty} |a_n x_n| \leq \sup_{n \in \mathbb{N}} |a_n| \sum_{n=1}^{\infty} |x_n| = \lVert a_n \rVert_\infty \lVert x_n \rVert_1.$$
Because of this we have $\lVert T \rVert \leq \lVert a_n \rVert_\infty$. Now we should find $(x_n) \in \ell ^1$ such that $\lVert x_n \rVert_1 = 1$ and $ \lVert T(x_n) \rVert_1 = \lVert a_n \rVert_\infty $. How can we do that?
For point spectrum we need to find nontrivial solutions of equation
$$ (\lambda I - T)(x_n) = 0.$$
We have
$ x_1(\lambda - a_1) = 0$, $x_2(\lambda - a_2) = 0$, $\ldots $
For $|\lambda| > 1 $ we do not have any nontrivial solution. (Since $|a_n| \leq 1$.) For $|\lambda| \leq 1$ we have solutions $\lambda = a_n$. From this the point spectrum $\sigma_p (T) = \{a_n, n \in \mathbb{N} \}$. Is this correct?
To find spectrum $\sigma(T)$ we want to check if the mapping $(\lambda I - T)$ is invertible.
$(\lambda I - T)$ is injective ($\ker T = \{0\}$). We need to check if it is surjective. That is we need to find if this equation always has a solution
$$ (\lambda I - T)(x_n) = (y_n).$$
For $\lambda = 0$ we see that the equation does not have any solution for sequences $(a_n)$ such it contains at least one zero. Let's say $(a_n)$ has a zero in $m$-th position.
For $(y_n) = (0,\ldots, 0,1,0,\ldots) $ where $1$ is in the $m$-th position. For such sequence $(y_n)$ this equation has no solution. Hence $0 \in \sigma(T)$. Which means that $T$ could be compact for such sequences.
How do we proceed from here? Thanks.
 A: $Te_n=a_ne_n$ and this gives $\|T\| \geq |a_n|$ for all $n$
Since $\sigma (T)$ is a closed set it contains the closure of $\{a_n: n \geq 1\}$. To prove the reverse inclusion suppose $\lambda \notin \overline {\{a_n: n \geq 1\}}$. Then there exists $r>0$ such that $|\lambda -a_n| \geq r$ for all $n$. The unique solution of $(T-\lambda I)(x_n)=(y_n)$ is given by $x_n=\frac {y_n} {a_n-\lambda}$.  Note that this sequence $(x_n)$ does belong to $\ell^{1}$.
Also, $\|(x_n)\|\leq \frac1  r \|(y_n)\|$ so $T-\lambda I$ is invertible. Thus $\sigma (T)=\overline {\{a_n: n \geq 1\}}$.
Compactnes: If $T$ is compact then non-zero eigen values cannot have any limit point other than $0$. Hence $a_n \to 0$. Conversely, if $a_n \to 0$ define $T_N(x_n)=(a_1x_1,a_2x_2,...,a_Nx_N,0,0,0...)$. Check that these finite rank operators converge in operator norm to $T$. Hence $T$ is compact.
A: The spectrum of $T$ contains clearly all the $a_n$'s, since $Te_n=a_ne_n$, and since the spectrum is closed, then it contains all the limit points of $\{a_n\}$. Hence
$$
\overline{\{a_n\}}\subset\sigma(T)
$$
On the other hand, if $\lambda\not\in\overline{\{a_n\}}$, then
$$
r=\mathrm{dist}\big(\lambda,\overline{\{a_n\}}\big)=\inf_{n\in\mathbb N}|\lambda-a_n|>0,
$$
and hence
$$
\frac{1}{|\lambda-a_n|}\ge\frac{1}{r},\quad\text{for all $n$.}
$$
Thus, for every $(x_n)\in \ell^1$, if we set
$$
y_n=\frac{x_n}{\lambda-a_n}
$$
then $(y_n)\in\ell^1$, and in particular, $\|(y_n)\|_{\ell^1}\le\frac{1}{r}\|(x_n)\|_{\ell^1}$, and
$$
(\lambda I-T)(y_n)=(x_n).
$$
Thus, if  $\lambda\not\in\overline{\{a_n\}}$, then $\lambda\not\in\sigma(T)$, and therefore
$$
\sigma(T)\subset\overline{\{a_n\}}.
$$
Altogether
$$
\sigma(T)=\overline{\{a_n\}}.
$$
