Spectrum of $S+S^{*}$ where $S$ is the shift operator on $l^{2}(\mathbb{N})$

Question

The problem is to show that the spectrum of $S+S^{*}$, denoted by $\sigma(S+S^{*})$ is equal to $[-2,2]$. All I have been able to show so far is the trivial direction, that is $\sigma(S+S^{*})\subset [-2,2]$. I really don't know how to proceed in the other direction, any hints would be appreciated.

Answer 1

Here is a hint: showing that $[-2,2]\subset \sigma$ is the same as showing that $|\lambda|>2$ then the resolvent is not a bounded operator $\ell^2(\mathbb{N})\rightarrow \ell^2(\mathbb{N})$. If the resolvent existed, we can see the form it must have. Define $R_\lambda=(S+S^*-\lambda)^{-1}$ (the resolvent of $S+S^*$ at $\lambda$ and note that $R_\lambda (y_n)$ is some sequence $(x_n)$ such that $(S+S^*-\lambda)(x_n)=(y_n)$. If we can always find a sequence satisfying this criteria, and $||(x_n)||\leq C||(y_n)||$, then we have shown that the resolvent is bounded. The expression $(S+S^*-\lambda)(x_n)=(y_n)$ gives us the recurrence relation $$x_{n+1}+x_{n-1}-\lambda x_n=y_n$$ For $n=1$, we see that $$x_2-\lambda x_1=y_1\implies x_2=y_1+\lambda x_1$$ for $n=2$ we get $$x_3+x_1-\lambda x_2=y_2\implies x_3=y_2+\lambda y_1+\lambda^2x_1-x_1$$ for n=3 we get $$x_4+x_2-\lambda x_3=y_3\implies x_4=y_3+\lambda y_2+\lambda^2y_1+\lambda(\lambda^2-1)x_1-(y_1+\lambda x_1)$$ and so forth so that you can always write $x_n$ in terms of $\{y_i\}_{i=1}^{n-1}$, $\lambda$, and $x_1$. See if you can identify a pattern that allows you to write down $x_n$ (and think about what $x_1$ should be) and show that the result is in $\ell^2(\mathbb{N})$ and that the operator it defines is bounded if and only if $\lambda\not\in[-2,2]$.