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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 $\sigma (S_r \circ M)=[-|\alpha|, |\alpha|]$.

Could you explain how to solve (10) in case $\alpha \neq 0$? Below is my failed attempt.

We write $S_r M := S_r \circ M$. Then $$ S_r M x = (0, \alpha_1 x_1, \alpha_2 x_2, \ldots, \alpha_n x_n, \ldots). $$

Let $\lambda \in \mathbb R$. Then $$ (S_r M - \lambda I) x = (-\lambda x_1, \alpha_1 x_1 - \lambda x_2, \alpha_2 x_2 - \lambda x_3, \ldots, \alpha_n x_n - \lambda x_{n+1}, \ldots). $$

  • $\alpha=0$.

Then $M$ and thus $S_r M$ are compact operators by exercise 6.1 (in the same book). Because $\dim E = \infty$, it follows from Theorem 6.8 (in the same book) that $0 \in \sigma (S_r M)$ and $\sigma (S_r M) \setminus \{0\} = EV (S_r M) \setminus \{0\}$. By (9), we have $EV (S_r M) \subset \{0\}$. Then $\sigma (S_r M) = \{0\}$.

  • $\alpha \neq 0$.

Clearly, $M - \alpha I$ is compact.

Akira
  • 17,367

1 Answers1

0

First, we consider the case $|\lambda| > |\alpha|$. Then $T :=\alpha S_r- \lambda I$ is bijective by (3). Then $T^{-1} \in \mathcal L(E)$ by open mapping theorem. We have $$ \begin{align*} S_rM - \lambda I &= S_r ( \underbrace{M-\alpha I}_{K}) + \alpha S_r - \lambda I \\ &= S_rK + T \\ &\overset{(*)}{=} T(I + \underbrace{T^{-1} S_r K}_{K_1}) \\ &= T(I+K_1). \end{align*} $$

Clearly, $K$ and thus $K_1$ are compact. Our goal is to prove that $S_rM - \lambda I$ is bijective. Because $T$ is invertible, it suffices to prove that $I+K_1$ is bijective. By Fredholm alternative, it suffices to prove that $I+ K_1$ is injective. Because $T$ is invertible, it suffices to prove that $S_rM - \lambda I$ is injective. This is indeed true by (9) where I showed that $EV (S_r M) \subset \{0\}$.

Second, we consider the case $0<|\lambda| \le |\alpha|$. Then $T$ is injective but not surjective by (2, 3). Then there is a (not necessarily continuous) injective linear map $\tilde T^{-1} : R(T) \to E$ such that $\tilde T^{-1} T = I$. As in $(*)$, we have $$ S_rM - \lambda I \overset{(**)}{=} T(I + \tilde T^{-1} S_r K). $$

Clearly, $R(S_rM - \lambda I) \subset R(T) \subsetneq E$. Then $S_rM - \lambda I$ is not surjective. This completes the proof.

Update: I have found a critical mistake in my proof for the case $0<|\lambda| \le |\alpha|$, i.e., we have $\tilde T^{-1} T = I$ but I incorrectly used $T\tilde T^{-1} = I$ in $(**)$. Below is Brezis' proof.

Second, we consider the case $0<|\lambda| \le |\alpha|$. Assume the contrary that $S_rM - \lambda I = S_r (M - \alpha I)+ \alpha S_r - \lambda I = S_rK +T$ is bijective. Then $$ \begin{align*} T &= (S_rM - \lambda I) - S_r K \\ &= (S_rM - \lambda I) ( I - \underbrace{(S_rM - \lambda I)^{-1} S_r K}_{K_2}). \end{align*} $$

Clearly, $K$ and thus $K_2$ are compact. Because $S_rM - \lambda I$ is bijective, we get

  • $T$ is injective IFF $I-K_2$ is injective.
  • $T$ is surjective IFF $I-K_2$ is surjective.

By Fredholm alternative, $I-K_2$ is injective IFF $I-K_2$ is surjective. It follows that $T$ is injective IFF $T$ is surjective, which is a contradiction.

Akira
  • 17,367