Let $E$ be a real Banach space. Let $\mathcal L(E)$ be the space of bounded linear operators on $E$ and $\mathcal K(E)$ its subspace consisting of compact operators. For $T \in \mathcal L(E)$,
- we denote by $N(T)$ its kernel and by $R(T)$ its range.
- we denote by $\rho(T)$ its resolvent set, by $\sigma(T)$ its spectrum, and by $EV(T)$ its set of eigenvalues. Then $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,
Let $E := \ell^2$ with its norm $|\cdot|_2$. An element $x\in E$ is denoted by $x=(x_1, x_2, \ldots, x_n,\ldots)$. Consider the operators $$ \begin{align*} S_r x &= (0, x_1, x_2, \ldots, x_{n-1}, \ldots), \\ S_{\ell} x &= (x_2, x_3, x_4, \ldots, x_{n+1}, \ldots), \end{align*} $$ respectively called the right shift and left shift.
- Determine $\|S_r\|$ and $\|S_{\ell}\|$. Does $S_r$ or $S_{\ell}$ belong to $\mathcal{K}(E)$?
- Prove that $EV(S_r) = \emptyset$.
- Prove that $\sigma\left(S_r\right)=[-1,1]$.
There are possibly subtle mistakes that I could not recognize in below attempt of (2, 3). Could you please have a check on it? I'm also happy to see other approaches.
Clearly, $S_r$ is injective, so $0 \notin EV(S_r)$. Let $\lambda$ be a non-zero real number. For $x \in E$, $$ (S_r - \lambda I)x = (-\lambda x_1, x_1 -\lambda x_2, x_2 -\lambda x_3, \ldots, x_{n-1} -\lambda x_n, \ldots). $$
Then $(S_r - \lambda I)x=0$ iff $$ 0= -\lambda x_1 = x_1 -\lambda x_2 = x_2 -\lambda x_3 = \ldots = x_{n-1} -\lambda x_n = \ldots $$
It follows from $\lambda\neq 0$ that $0=x_1=x_2=\cdots=x_n=\cdots$. Then $(S_r - \lambda I)x=0$ iff $x=0$. Then $EV(S_r) = \emptyset$.
3.
By (1), $\|S_r\|=1$. By Proposition 6.7 (in the same book), $\sigma(S_r)$ is compact and $\sigma(S_r) \subset [-\|S_r\|, \|S_r\|]=[-1, 1]$. Let $\lambda \in [-1, 1]$. It remains to prove that $S_r - \lambda I$ is not bijective. By (2), it suffices to prove that $S_r - \lambda I$ is not surjective. Clearly, $S_r$ is not surjective. WLOG, we assume $\lambda \neq 0$. Assume $x, y\in E$ such that $(S_r - \lambda I)x =y$. Then $$ \begin{align*} -\lambda x_1&= y_1, \\ x_n-\lambda x_{n+1} &= y_{n+1} \quad \forall n \ge 1. \end{align*} $$
Then $x_{n+1} = \frac{x_n-y_{n+1}}{\lambda}$ for $n\ge1$. We pick $y_1 :=0, y_2 :=1$ and $y_n :=0$ for $n \ge 3$. Then $$ x_1=0, x_2=\frac{-1}{\lambda}, x_3 = \frac{-1}{\lambda^2}, \ldots, x_{n+1} = \frac{-1}{\lambda^n}, \ldots $$
Clearly, $|x|_2 = \infty$, which is a contradiction. Hence $y \notin R(S_r - \lambda I)$. This completes the proof.
