Finding a adjoint operator, spectrum, point spectrum and continuous spectrum Let $T:\ell_{1}(\mathbb{N})\rightarrow \ell_{1}(\mathbb{N})$ a operator such that:
\begin{equation}
 T(x_{1},x_{2},x_{3},\dots)=\left(x_{1},\frac{1}{2}x_{2},\frac{1}{3}x_{3},\dots\right)
\end{equation}
Compute the adjoint operator, spectrum $\sigma(T)$, point spectrum $\sigma_{p}(T)$, continuous spectrum $\sigma_{c}(T)$ of $T$.

My attempt

*

*$T$ is selft-adjoint, I considerer $x=(x_{1},x_{2},x_{3},\dots)\in \ell_{1}(\mathbb{N}), y=(y_{1},y_{2},y_{3},\dots)\in \ell_{\infty}(\mathbb{N})$ So for duality product
\begin{align}
 \langle Tx, y\rangle &= \langle \left(x_{1},\frac{1}{2}x_{2},\frac{1}{3}x_{3},\dots\right),(y_{1},y_{2},y_{3},\dots)\rangle \\
&=x_{1}y_{1}+\frac{1}{2}x_{2}y_{2}+\frac{1}{3}x_{3}y_{3}+\dots \\
&= \langle (x_{1},x_{2},x_{3},\dots),\left(y_{1},\frac{1}{2}y_{2},\frac{1}{3}y_{3},\dots\right)\rangle\\
&= \langle x,Ty\rangle
\end{align}


*I tried compute then norm operator $\left\|T \right\|$ this way:
\begin{align}
 \left\|T \right\| = \sup_{x\in \ell_{1}(\mathbb{N}),\left\|x \right\|=1} \left\|Tx \right\|
\end{align}
So by definition of $T$
\begin{align}
 \left\|Tx \right\|&=\sum_{n=1}^{\infty}\frac{|x_{n}|}{n}\\
&\leq\left(\left(\sum_{n=1}^{\infty}\frac{1}{n^2}\right)\left(\sum_{n=1}^{\infty} |x_{n}|^2\right)\right)^\frac{1}{2}\\
&< \frac{\pi}{\sqrt{6}}
\end{align}
Can I conclude that $\left\|T \right\|=\frac{\pi}{\sqrt{6}}$? So the spectrum is $\sigma(T)\subset D\left(0,\frac{\pi}{\sqrt{6}}\right)$. For the point spectrum, I read that I have to analyse when $\ker(T-\lambda I)\neq \{0\}$, If $x\in \ell_{1}(\mathbb{N})$ then
\begin{align}
 T-\lambda I(x)=0\\
\sum_{n=1}^{\infty}|x_{n}|\left|\frac{1}{n}-\lambda\right|=0\\
\end{align}
But I'm not sure when $\left|\frac{1}{n}-\lambda\right|=0$. This is correct? Can you give me some advice or hints to find the continuous spectrum ? Thank you.
 A: The norm of $T$ is one. Indeed,
$$\|Tx\| = \sum_n \frac{|x_n|}{n}\le \sum_n |x_n| = \|x\|$$
so that $\|T\| \le 1$ and for $x=(1,0 , 0 ,\dots)$ one has $\|Tx\|=1$.
The eigenvalues $\lambda \in \mathbb C$ must satisfy $Tx -\lambda x =0$ for some nonzero $x \in \ell^1$. That is,
$$ \tfrac 1n x_n =\lambda x_n , \ \ \ \forall n\ge1. $$
This implies that $\lambda = \tfrac 1n$ for some $n$ and its eigenvector is the sequence $x$ with $x_i=0$ for $i \neq n$ and $x_n=1$. In other words, the point spectrum (set of eigenvalues) is $\{\tfrac 1n \colon n \ge 1\}.$
Since the spectrum is closed,  $\overline{\{\tfrac 1n \colon n \ge 1\}} \subseteq \sigma(T)$. Try and and show the other inclusion. Hint: if $z \notin \overline{\{\tfrac 1n \colon n \ge 1\}}$ then there exists $δ>0$ such that $|z-\tfrac 1n|\ge \delta >0$ for all $n$. Show that $zI-T$ is invertible.
For the continuous spectrum, try and show that it is equal to $\{0\}$, that is, $\overline{\{\tfrac 1n \colon n \ge 1\}}  \setminus \{\tfrac 1n \colon n \ge 1\}$.
Edit
The same logic applies to any multiplication operator on $\ell^1$. Let $a \in \ell^{\infty}$ and define $M_a\colon \ell^1 \to \ell^1$ by
$$ M_a(x) = (a_n x_n)_{n \ge 1}.$$
Then $\|M_a\|=\|a\|_{\infty}$. Indeed, that $\|M_a\| \le \|a\|_{\infty}$ is clear. For the opposite inclusion, for any $ε>0$ pick $N\ge 1$ such that $|a_N|> \|a\|_{\infty}-ε$ and let $x$ be the sequence with $x_i =0$ for $i \neq N$ and $x_N=1$. Then
$$\|M_a\| \ge \|M_a(x)\|= \|a\|_{\infty} -ε.$$
The point spectrum of $M_a$ is exactly the elements of the sequence $a$, that is,
$$\sigma_p(M_a) = \{ a_n \colon n \ge 1\}$$ and its spectrum is $$\sigma(M_a) = \overline { \{ a_n \colon n \ge 1\} }.$$
The proof is essentially the same as above. The continuous spectrum is $$\sigma_c(M_a) = \overline { \{ a_n \colon n \ge 1\} } \setminus \{ a_n \colon n \ge 1\} .$$
Indeed, if $\lambda \in \sigma_c(M_a)$ then $M_a -\lambda I$ is injective and so $\lambda \notin \{a_n \colon n \ge1\}.$ For the other inclusion, if $ \lambda \in  \overline { \{ a_n \colon n \ge 1\} } \setminus \{ a_n \colon n \ge 1\} $ we have to show that $M_a - \lambda I$ is injective, has dense range, and is not surjective. Injectivity is clear. Since $c_{00}$, the space of compactly supported sequences, is dense in $\ell^1$, it suffices to show that $c_{00} \subset Range(M_a-\lambda I)$. But that is easy: If $x= (x_1,\dots, x_N,0 ,0 ,\dots) \in c_{00}$ then for
$$u=\big(\tfrac{1}{a_1-\lambda} x_1, \tfrac{1}{a_2-\lambda} x_2,\dots, \tfrac{1}{a_N-\lambda}x_N,0,0, \dots\big) \in \ell^1,$$
we have that $x=(M_a-\lambda I)(u)$. Finally, as $\lambda \in \overline {\{ a_n\colon n \ge 1\}}$ we see that the sequence $\big( \frac{1}{a_n-\lambda} \big)_{n\ge1}$ is not in $\ell^{\infty}$ and thus there exists $u \in \ell^1$ such that
$$ \sum_n \big|u_n \tfrac{1}{a_n-\lambda}\big |= \infty.$$
Then $u \notin Range(M_a-\lambda I)$.
