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.

  • $\begingroup$ I have found a solution below. $\endgroup$
    – Z. Zhu
    Aug 4, 2021 at 12:51

2 Answers 2


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 $\ell^\infty$.


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 $\sigma_c(R)=\{z\in\mathbb{C}:|z|= 1\}$.


  1. 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".

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