Let $x_{n+1} := \frac{x_n-y_{n+1}}{\lambda}$ for $n \ge 1$. Can we prove that $|x|_2 < \infty$?

Question

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.

1. Determine $\|S_r\|$ and $\|S_{\ell}\|$. Does $S_r$ or $S_{\ell}$ belong to $\mathcal{K}(E)$?
2. Prove that $EV(S_r) = \emptyset$.
3. Prove that $\sigma\left(S_r\right)=[-1,1]$.
4. Prove that $E V (S_{\ell}) = (-1, 1)$. Determine the corresponding eigenspaces.
5. Prove that $\sigma(S_{\ell})=[-1,1]$.
6. Determine the adjoints $S_r^*$ or $S_{\ell}^*$.
7. Prove that for every $\lambda \in(-1,1)$, the spaces $R (S_r-\lambda I)$ and $R (S_{\ell}-\lambda I)$ are closed. Give an explicit representation of these spaces.

I'm trying to prove $R (S_{\ell}-\lambda I)$ is closed in (7). In my below attempt, I failed to prove that $|x|_2 < \infty$? Could you please elaborate on how to fix my proof?

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I have proved in (6) that $S_r^* = S_\ell$ and $S_{\ell}^* = S_r$. Then $(S_r-\lambda I)^*= S_{\ell}-\lambda I$ and $(S_\ell - \lambda I)^*= S_r-\lambda I$. By Theorem 2.19 (in the same book), if $A$ is an unbounded linear operator which is densely defined and whose graph is closed, then $R(A)$ is closed iff $R(A^*)$ is closed. Let $\lambda \in(-1,1)$. It suffices to prove that $R(S_{\ell}-\lambda I)$ is closed.

The case $\lambda=0$ is clear. Let $\lambda \neq 0$. By Theorem 2.19 again, it suffices to prove $R(S_r-\lambda I) = N(S_{\ell}-\lambda I)^\perp$. By (4), $N(S_{\ell}-\lambda I) = \operatorname{span} \{(1, \lambda, \lambda^2, \ldots, \lambda^n, \ldots)\}$. It remains to verify $$ R(S_r-\lambda I) = F:= \left \{y \in E : \sum_{n=0}^\infty h^n y_{n+1} = 0 \right \}. $$

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 $$ R(S_r-\lambda I) = \{ (-\lambda x_1, x_1 -\lambda x_2, x_2 -\lambda x_3, \ldots, x_{n-1} -\lambda x_n, \ldots) : x\in E\}. $$

It's easy to see that $R(S_r-\lambda I) \subset F$. Let $y \in F$. Because $\lambda \neq 0$, we can pick $x_1 \in \mathbb R$ such that $y_1 = -\lambda x_1$. Then there is a unique $x_2 \in \mathbb R$ such that $y_2 = x_1 -\lambda x_2$. Recursively, for each $n \ge 1$ there is a unique $x_{n+1} \in \mathbb R$ such that $y_{n+1} = x_n -\lambda x_{n+1}$. Let $x := (x_1, x_2, \ldots)$. If we can verify that $|x|_2 < \infty$, then clearly, $(S_r-\lambda I) x=y$ and thus $y\in R(S_r-\lambda I)$. We have $x_{n+1} = \frac{x_n-y_{n+1}}{\lambda}$ for $n \ge 1$.

Answer 1

Writing $(\lambda I - S_r)x=y$ as $S_rx = \lambda x-y$, we get by taking norms $$ \|S_rx\| = \|x\| \le |\lambda| \cdot \|x\| + \|y\|, $$ so that $$ \|x\| \le \frac 1{1-|\lambda|}\|y\| = \frac 1{1-|\lambda|}\| (\lambda I - S_r)x\|.$$ This implies that $\lambda I - S_r$ has closed range.