# 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., $$$$R(x_1,x_2,...)= (0,x_1,x_2,...).$$$$ 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 $$$$\sigma(R)= \sigma(A)=\{\lambda\in \mathbb{C}:|\lambda|\leq 1\},$$$$ and $$$$\sigma_r(A) \subset \sigma_p(R)=\varnothing.$$$$ Therefore, $$\sigma_c(A)= \{\lambda\in \mathbb{C}: |\lambda|=1\}$$. Also, using (3), we deduce $$$$\sigma_c(R) \subset \sigma_c(A) =\{\lambda\in \mathbb{C}: |\lambda|=1\}.$$$$ But I have no idea how to determine $$\sigma_c(R)$$ explicitly. Any ideas would be greatly appreciated.

• I have found a solution below. Aug 4, 2021 at 12:51

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 $$$$\begin{cases} a_1=\lambda b_1,\\ a_2= \lambda b_2- b_1,\\ \ \ \ \ \cdots\\ a_n= \lambda b_n -b_{n-1},\\ \ \ \ \ \cdots \end{cases} \quad \Longleftrightarrow \quad \begin{cases} b_1=\overline{\lambda} a_1,\\ b_2= \overline{\lambda}(a_2 +b_1) ,\\ \ \ \ \ \cdots\\ b_n= \overline{\lambda} (a_n+ b_{n-1}) ,\\ \ \ \ \ \cdots \end{cases}$$$$ Hence this equation has the solution $$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 $$$$(\lambda I− R)b = a = a_{[\overline{\lambda}]} + \beta$$$$ is $$b = (b_1, b_2, b_3, ... )$$, with: $$$$b_n = n\overline{\lambda}^n +\sum_{j=1}^{n} \overline{\lambda}^{n+1-j} \beta_j,$$$$ which implies that: $$$$\left| b_n - n \overline{\lambda}^n\right| = \left| \sum_{j=1}^{n}\overline{\lambda}^{n+1-j} \beta_j \right|\leq \sum_{j=1}^{n} |\beta_j|<\frac{n}2.$$$$ Using the triangle inequality $$n = \left|n\overline{\lambda}^n\right| \leq \left|b_n − n\overline{\lambda}^n\right| + |b_n|$$, we deduce that $$$$|b_n| \geq n − \left|b_n − n\overline{\lambda}^n \right| > n −\frac{n}2 = \frac{n}2.$$$$ Thus $$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 $$\sigma_c(R)=\{z\in\mathbb{C}:|z|= 1\}$$.

Remark:

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".