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 $EV(T) \subset \sigma(T) = \mathbb R \setminus \rho(T)$.
- for $\lambda \in EV(T)$, the set $N(T-\lambda I)$ is called the eigenspace corresponding to $\lambda$.
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 $\sigma\left(S_r\right)=[-1,1]$.
- Prove that $E V (S_{\ell}) = (-1, 1)$. Determine the corresponding eigenspaces.
- Prove that $\sigma(S_{\ell})=[-1,1]$.
There are possibly subtle mistakes that I could not recognize in below attempt of (4, 5). Could you please have a check on it? I'm also happy to see other approaches.
Clearly, $S_\ell$ is not injective and thus $0 \in EV(E_\ell)$. Let $\lambda$ be a non-zero real number. For $x \in E$, $$ (S_\ell - \lambda I)x = (x_{n+1} - \lambda x_n)_{n\ge 1}. $$
Then $x \in N (S_\ell - \lambda I)$ iff $x_{n+1} = \lambda x_n$ for all $n\ge 1$ iff $x_{n+1} = \lambda^n x_1$ for $n \ge 1$. We have $|x|_2 < \infty$ iff $\sum_{n=1}^\infty |x_n|^2 = |x_1|^2 \sum_{n=0}^\infty |\lambda|^n < \infty$ iff $x_1 =0$ or $|\lambda| < 1$. Then the existence of $x \in N (S_\ell - \lambda I) \setminus \{0\}$ is equivalent to $|\lambda| < 1$. Then $E V (S_{\ell}) = (-1, 1)$. Also, $$ N (S_\ell - \lambda I) = \{ (\lambda^n h)_{n \ge 0} : h \in \mathbb R\}. $$
5.
I already showed in (1) that $\|S_\ell\|=1$. By Proposition 6.7 (in the same book), $\sigma(S_\ell)$ is compact and $\sigma(S_\ell) \subset [-\|S_\ell\|, \|S_\ell\|]=[-1, 1]$. Let $\lambda \in \{\pm 1\}$. By (4), it remains to prove that $S_\ell - \lambda I$ is not surjective. For $x \in E$, $$ (S_\ell - \lambda I)x = (x_{n+1} - \lambda x_n)_{n\ge 1}. $$
Let $x, y \in E$ such that $y=(1, 0, 0,\ldots, 0, \ldots)$ and $(S_\ell - \lambda I)x=y$. Then $$ x_2= \lambda x_1 + 1, x_3= \lambda (\lambda x_1+1), x_4 = \lambda^2 (\lambda x_1+1), \ldots, x_{n+1} = \lambda^{n-1} (\lambda x_1+1), \ldots $$
Then the sequence $(x_n)_{n\ge 1}$ is not convergent. Hence $x \notin E$ and thus $y \notin R(S_\ell - \lambda I)$. This completes the proof.
