$\ell^2$ space, compute $\|L\|$.

Question

Consider $\ell^2$ space, define $L\in\mathcal{L}(\ell^2):(x_1,x_2,\ldots)\mapsto(x_2,x_3,\ldots)$.

1. Compute $\|L\|$.

2. Show that $\{\lambda\in C:|\lambda|<1\}\subseteq\sigma_p(L)$, $\sigma_p(L)$ is the point spectrum.

3. Show that $\{\lambda\in C:|\lambda|=1\}\subseteq\sigma_c(L)$ and use the results above to show that $\sigma_p(L)=\{\lambda\in C:|\lambda|<1\}$, $\sigma_c(L)=\{\lambda\in C:|\lambda|=1\}$. $\sigma_c(L)$ is the continuous spectrum.

Answer 1

The left shift and its eigenvalues and spectrum is a standard problem and can be found as exercises in a lot of books about functional analysis or Hilbert spaces, for example "Introduction to Hilbert Space" by Berberian (1976). Therefore I will only sketch the proofs.

1. Using the definition of the operator norm, it is easy to see that $\Vert Lx\Vert\leq\Vert x\Vert$ for any $x\in\ell_2(\mathbb N)$ so $\Vert L\Vert\leq 1$. On the other hand one quickly finds $x_0\in\ell_2(\mathbb N)$ such that $\Vert Lx_0\Vert=\Vert x_0\Vert$ so $\Vert L\Vert\geq 1$ and in total $\Vert L\Vert =1$.

2. We have to show that every $\lambda\in\mathbb C$ with $|\lambda|<1$ is eigenvector of $L$ so we consider

$$ Lx=(x_2,x_3,x_4\ldots)=(\lambda x_1,\lambda x_2,\lambda x_3,\ldots)=\lambda x. $$

Distinguishing the cases $\lambda=0$ and $\lambda\neq 0$, in each case we get an eigenvector $x\neq 0$ (reminder that we still have to show $x\in\ell_2(\mathbb N)$ so its entries have to be square-summable!). The latter is also the reason why any $\lambda\in\mathbb C$ with $|\lambda|\geq 1$ can not be eigenvalue of $L$ - we get an eigenvector $y$ but if its first entry is non-zero (and thus $y\neq 0$), $y\notin\ell_2(\mathbb N)$. Therefore we already showed

$$ \sigma_p(L)=\lbrace \lambda\in\mathbb C:|\lambda|<1\rbrace $$

3. Every eigenvalue is in the (continous) spectrum as can be easily seen from the definitions so we already have $\lbrace \lambda\in\mathbb C:|\lambda|<1\rbrace\subseteq\sigma_c(L)$. Another standard result says that the (continous) spectrum is a non-empty and closed set for any bounded operator so taking the closure gives

$$ \lbrace \lambda\in\mathbb C:|\lambda|\leq1\rbrace=\overline{\lbrace \lambda\in\mathbb C:|\lambda|<1\rbrace}\subseteq\overline{\sigma_c(L)}=\sigma_c(L). $$

Using the fact that the spectrum is bounded by the norm of the operator, we have $\sigma_c(L)\subseteq \lbrace \lambda\in\mathbb C:|\lambda|\leq\Vert L\Vert=1\rbrace$ so together with the upper inclusion

$$ \sigma_C(L)=\lbrace \lambda\in\mathbb C:|\lambda|\leq1\rbrace. $$