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 $\mathcal{K}(E)$ ?
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it? I'm also happy to see other approaches.
We have $$ \begin{align*} |S_r x|_2 &= \sqrt{\sum_{n=1}^\infty |x_n|^2} = |x|_2, \\ |S_{\ell} x|_2 &= \sqrt{\sum_{n=2}^\infty |x_n|^2} \le |x|_2. \end{align*} $$
Then $\|S_{\ell} \| \le \|S_r\| =1$. If we pick $x=(0, 1, 0, 0, \ldots, 0, \ldots)$. Then $|x|_2= |S_{\ell} x|_2 =1$. Then $\|S_{\ell} \|=1$. We have $S_r$ is an isometry, so it is an isometric isomorphism. Then $S_r$ is certainly not compact.
We define $x^n = (x^n_m)_{m\ge 1} \in E$ by $x^n_m= 1$ if $n=m$ and $0$ otherwise. Then $|x^n|_2=1$ for all $n$. Also, $|x^n-x^m|_2 = \sqrt{2}$ for all $1 \le m <n$. Then $(x^n)$ does not have any convergent subsequence in $E$. On the other hand, $S_\ell x^{n+1} = x^n$ for all $n\ge 1$. Hence $( S_\ell x^n)_n$ does not have any convergent subsequence in $E$. So $S_\ell$ is not compact.
