Continuous spectrum $\sigma_c(R)$ of the right-shift operator $R$ on $\ell^\infty$. Let $R$ be the right-shift operator on $\ell^\infty$, i.e.,
\begin{equation}
R(x_1,x_2,...)= (0,x_1,x_2,...).
\end{equation}
I know the following results from Functional
Analysis Proposition 14.11 on p. 316:
Proposition. Let $X$ be a Banach space, $T\in \mathcal{L}(X)$, and $T^\star$ is the adjoint of $T$. Then
(1) $\sigma(T^\star) =\sigma(T)$;
(2) $\sigma_r(T)\subset \sigma_p(T^\star) \subset \sigma_r(T)\cup \sigma_p(T)$;
(3) $\sigma_c(T^\star) \subset \sigma_c(T)$.
Clearly, $\sigma_p(R)=\varnothing$, and $R= A^\star$, where $A$ is the left-shift operator on $\ell^1$.
Since the spectral radius of $A$ is $r(A)=1$, $\sigma(A)\subset \{\lambda\in \mathbb{C}:|\lambda|\leq 1\}$. Note that if $|\lambda|<1$, then $\lambda\in \sigma_p(A)$ (an eigenvector is $(1,\lambda,\lambda^2,\lambda^3,...)\in \ell^1$). Moreover, if $|\lambda|=1$, then $\lambda\notin\sigma_p(A)$ (otherwise any eigenvector must form a geometric sequence with a common ratio of $\lambda$, which is not in $\ell^1$). Hence $\sigma_p(A)= \{\lambda\in \mathbb{C}:|\lambda|<1\}$. Since $\sigma(A)$ is closed, we have $\sigma(A)= \{\lambda\in \mathbb{C}:|\lambda|\leq 1\}$. Using the above
results, we obtain
\begin{equation}
\sigma(R)= \sigma(A)=\{\lambda\in \mathbb{C}:|\lambda|\leq 1\},
\end{equation}
and
\begin{equation}
\sigma_r(A) \subset \sigma_p(R)=\varnothing.
\end{equation}
Therefore, $\sigma_c(A)= \{\lambda\in \mathbb{C}: |\lambda|=1\}$. Also, using (3), we deduce
\begin{equation}
\sigma_c(R) \subset \sigma_c(A) =\{\lambda\in \mathbb{C}: |\lambda|=1\}.
\end{equation}
But I have no idea how to determine $\sigma_c(R)$ explicitly. Any ideas would be greatly appreciated.
 A: Today, I have found a solution from Spectral Theory step f) on p. 15--p.16:
In fact $\sigma_r(R) = \overline{B(0, 1)}$: because of the previous observations it suffices to show that
any $\lambda\in \mathbb{C}$ s.t. $|\lambda| = 1$ is also in $\sigma_r(R)$.
Let $\lambda$ be such a value. We start by computing
a formal inverse of $\lambda I− R$ : if $a \in \ell^\infty$ and if $b$ is another sequence with values in $\mathbb{C}$, the
equation $(\lambda I− R)(b) = a$ reads
\begin{equation}
\begin{cases}
  a_1=\lambda b_1,\\
  a_2= \lambda b_2- b_1,\\
  \ \ \ \ \cdots\\
  a_n= \lambda b_n -b_{n-1},\\
  \ \ \ \ \cdots
\end{cases} \quad \Longleftrightarrow  \quad
\begin{cases}
  b_1=\overline{\lambda} a_1,\\
  b_2= \overline{\lambda}(a_2 +b_1)  ,\\
  \ \ \ \ \cdots\\
  b_n= \overline{\lambda} (a_n+ b_{n-1})  ,\\
  \ \ \ \ \cdots
\end{cases}
\end{equation}
Hence this equation has the solution $b_n = \overline{\lambda} a_n + \overline{\lambda}^2 a_{n−1} + \cdots + \overline{\lambda}^n a_1$.
We can already
see that $\lambda I− R$ is not onto because, for
$a = a_{[\overline{\lambda}]} := \left(1, \overline{\lambda} , \overline{\lambda}^2,...\right) \in \ell^\infty$, the solution is
$b_n = n \overline{\lambda}^n$
and this sequence cannot be in $\ell^\infty$: thus $a_{[\overline{\lambda}]}\notin \text{Im} \left(\lambda I− R\right)$.
But we   need a stronger result, i.e. that $\text{Im} \left(\lambda I− R\right)$ is not dense in $\ell^\infty$. For
that purpose we show that $B\left(a_{[\overline{\lambda}]}, 1/2\right) \cap  \text{Im} \left(\lambda I− R\right) =\varnothing$
in $\ell^\infty$. Let $a \in B\left(a_{[\overline{\lambda}]}, 1/2\right)$, we
can write $a = a_{[\overline{\lambda}]} + \beta$, where $\|\beta\|_\infty < 1/2$. The solution in the space of complex valued
series of the equation
\begin{equation}
(\lambda I− R)b = a = a_{[\overline{\lambda}]} + \beta 
\end{equation}
is $b = (b_1, b_2, b_3, ... )$, with:
\begin{equation}
 b_n = n\overline{\lambda}^n  +\sum_{j=1}^{n} \overline{\lambda}^{n+1-j} \beta_j, 
\end{equation}
which implies that:
\begin{equation} 
\left| b_n  - n \overline{\lambda}^n\right| = \left| \sum_{j=1}^{n}\overline{\lambda}^{n+1-j} \beta_j \right|\leq \sum_{j=1}^{n} |\beta_j|<\frac{n}2.
\end{equation}
Using the triangle inequality $n = \left|n\overline{\lambda}^n\right| \leq \left|b_n − n\overline{\lambda}^n\right| + |b_n|$, we deduce that
\begin{equation}
|b_n| \geq n − \left|b_n − n\overline{\lambda}^n \right| > n −\frac{n}2 = \frac{n}2. \end{equation}
Thus $b$ is not in $\ell^\infty$.
A: You can show that $\sigma_c(R)=\sigma(R)\backslash(\sigma_p(R)\cup \Gamma(R))$ where $\Gamma(R)=\{\lambda\in \mathbb{C}:(R-\lambda)$ doesn't have dense range$\}$. Furthermore, you have one more relation such that $\sigma_p(T^*)=\Gamma(T)^*:=\{\overline{z}:z\in \Gamma(T)\}$ for any $T:X\rightarrow Y$, hence, $\Gamma(R)^*=\sigma_p(A)=\{z\in\mathbb{C}:|z|< 1\}$. This clearly gives you that $\Gamma(R)=\{z\in\mathbb{C}:|z|< 1\}$.
Again, you already know that $\sigma_p(R)=\emptyset$. Hence $\sigma_c(R)=\{z\in\mathbb{C}:|z|= 1\}$.
Remark:

*

*The first identity I have used can be derived from the proposition "$T-\lambda$ is invertible $\Leftrightarrow$ $T-\lambda$ is bounded below and has dense range".

