Regarding essential spectrum of the unilateral shift operator

Question

This is with context to Example 4.10 in Section 11 of the book : A course in functional Analysis by J.B Conway. Let $\sigma_{le}(S)$ and $\sigma_{re}(S)$ denote the left and right essential spectrum of the unilateral shift operator $S$ respectively. Let $\partial{\mathbb{D}}$ be the boundary of the open unit ball in the Complex plane. I can understand in the example why $\partial{\mathbb{D}}\subseteq \sigma_{le}(S)\cap\sigma_{re}(S)$. And can prove that $\sigma_{le}(S)\cap\sigma_{re}(S)\subseteq\partial{\mathbb{D}} $. But I do not understand how $\partial{\mathbb{D}}=\sigma_{le}(S)=\sigma_{re}(S)$? Can anyone explain how?

Answer 1

If $S(x_1,x_2,x_3,\ldots)=(0,x_1,x_2,\ldots)$ is the unilateral shift, it is easy to see that $S-\lambda I$ is bounded below for $|\lambda|<1$: $\|(S-\lambda I)x\| \geq \|Sx\|-\|\lambda x\|= (1-|\lambda|)\|x\|$.

And considering the adjoint operator $S^*(x_1,x_2,x_3,\ldots)=(x_2,x_3,x_4,\ldots)$, it is easy to check that $\dim \ker(S^*-\lambda I)=1$ for $|\lambda|<1$.

Moreover $\dim\ell^2/Im(S-\lambda I)= \dim \ker(S^*-\lambda I)=1$ for $|\lambda|<1$, hence $S-\lambda I$ is a Fredhom operator with index equal to $-1$ for $|\lambda|<1$.

For $|\lambda|>1$, $S-\lambda I$ is invertible, hence a Fredhom operator with index equal to $0$.

The continuity of the index implies that $S-\lambda I$ is not a Fredhom operator for $|\lambda|=1$. This fact admits a direct proof by showing that for $|\lambda|=1$, $S-\lambda I$ is injective but not bounded below.