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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. Let $I:E \to E$ be the identity operator. 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.
  8. Prove that the spaces $R(S_r \pm I)$ and $R(S_{\ell} \pm I)$ are dense and that they are not closed.

Consider the multiplication operator $M$ defined by $$ M x = (\alpha_1 x_1, \alpha_2 x_2, \ldots, \alpha_n x_n, \ldots), $$ where $(\alpha_n)$ is a bounded sequence in $\mathbb{R}$.

  1. Determine $E V (S_r \circ M)$.
  2. Assume that $\alpha_n \to \alpha$ as $n \to \infty$. Prove that $\sigma (S_r \circ M)=[-|\alpha|, |\alpha|]$.
  3. Assume that for every integer $n$, we have $\alpha_{2 n}=a$ and $\alpha_{2 n+1}=b$ with $a \neq b$. Determine $\sigma (S_r \circ M)$.

Could you explain how to solve (11) in case $|\lambda| \le \sqrt{|ab|}$? Below is my failed attempt.

We write $S_r M := S_r \circ M$. Clearly, $S_r M$ is not surjective. So $0 \in \sigma (S_r M)$. We have $$ \begin{align*} S_r M x &= (0, \alpha_1 x_1, \alpha_2 x_2, \ldots, \alpha_n x_n, \ldots), \\ (S_r M)^2 x &= (0, 0, \alpha_1 \alpha_2 x_1, \alpha_2 \alpha_3 x_2, \ldots, \alpha_n \alpha_{n+1} x_n, \ldots), \\ &= ab (0, 0, x_1, x_2, \ldots, x_n, \ldots), \\ &= ab (S_r)^2 x \end{align*} $$

By (9), we have $$ EV (S_r M) = \begin{cases} \{0\} &\text{if} \quad ab \neq 0, \\ \emptyset &\text{if} \quad ab = 0. \end{cases} $$

Let $\lambda \in \mathbb R \setminus \{0\}$. Then $S_r M +\lambda I$ is injective and $$ \left \| \left ( \frac{S_r M}{\lambda} \right )^2 \right \| =\frac{\|(S_r M)^2\|} {\lambda^2} = \frac{|ab|}{\lambda^2}. $$

We need the following result, i.e.,

Brezis' exercise 6.16.4 Let $T \in \mathcal L(E)$. If there is an integer $n\ge 2$ such that $\|T^n\| < 1$, then $I-T$ is bijective.

If $\frac{|ab|}{\lambda^2}<1$ then $\frac{S_r M}{\lambda} + I$ is bijective. Consequently, if $|\lambda| > \sqrt{|ab|}$ then $S_r M +\lambda I$ is bijective. Let $|\lambda| \le \sqrt{|ab|}$. We will prove that $S_r M +\lambda I$ is not surjective.

Akira
  • 17,367

2 Answers2

3

You have $(S_rM)^*=MS_\ell$. It is easy to show that if $|\lambda|<\sqrt{|ab|}$ then $\lambda$ is an eigenvalue for $MS_\ell$ with eigenvector $$ \Big(1,\frac\lambda a,\frac{\lambda^2}{ab},\frac{\lambda^3}{a^2b},\frac{\lambda^4}{a^2b^2},\ldots\Big). $$ So, since the spectrum is closed, $$ \sigma(S_rM)=\sigma(MS_\ell)=\{\lambda:\ |\lambda|\leq\sqrt{|ab|}\}. $$

Martin Argerami
  • 205,756
0

Let $\beta := \sqrt{|ab|}$. If $\frac{\beta^2}{\lambda^2}<1$ then $\frac{S_r M}{\lambda} + I$ is bijective. Consequently, if $|\lambda| > \beta$ then $S_r M +\lambda I$ is bijective. Let $|\lambda| \le \beta$. We will prove that $S_r M -\lambda I$ is not surjective. Let $y := (-a, 0, 0, \ldots)$. Assume the contrary that there is $x \in E$ such that $(S_r M -\lambda I)x=y$. Then for $n \ge 0$, we have $$ \begin{align*} -\lambda x_1 &= -a, \\ bx_1-\lambda x_2 &= 0, \\ ax_2-\lambda x_3 &= 0, \\ \cdots \\ bx_{2n+1}-\lambda x_{2n+2} &= 0, \\ ax_{2n+2}-\lambda x_{2n+3} &= 0, \\ \cdots \end{align*} $$

It follows that for $n \ge 0$, we have $$ \begin{align*} x_1 &= \frac{a}{\lambda}, \\ x_2 &= \frac{ab}{\lambda^2} = \left ( \frac{ab}{\lambda^2} \right )^1, \\ x_3 &= \frac{a^2b}{\lambda^3}, \\ x_4 &= \frac{a^2b^2}{\lambda^4} = \left ( \frac{ab}{\lambda^2} \right )^2, \\ x_5 &= \frac{a^3b^2}{\lambda^5}, \\ x_6 &= \frac{a^3b^3}{\lambda^6} = \left ( \frac{ab}{\lambda^2} \right )^3, \\ \cdots \end{align*} $$

Then $x_{2n}=\left ( \frac{ab}{\lambda^2} \right )^n$ and thus $|x_{2n}| = \left ( \frac{|ab|}{\lambda^2} \right )^n = \left ( \frac{\beta}{\lambda} \right )^{2n} \ge 1$ for $n\ge 1$. Then $|x|_2=\infty$, which is a contradiction.

Akira
  • 17,367