Regarding essential spectrum of the unilateral shift operator

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?

• What is left (right) spectrum ? Spectrum as the left (right) multiplication operator on the space of bounded operators ? Commented Jun 27, 2019 at 13:44
• In Conway's book he defines the left (right) spectrum of an operator $a$ to be the set of $z \in \mathbb{C}$ such that $a - z$ is not left (right) invertible. The spectrum is therefore the intersection of the left and right spectrum. I had never heard of it before, so I don't know how widely used this terminology is. Commented Jun 27, 2019 at 17:59
• Does looking back at Proposition 4.3 of the same section help in showing that $\sigma_{le}(S) = \sigma_{re}(S)$? Commented Jun 27, 2019 at 18:18

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.

• This is a fine way to show that $\sigma_e(S) = \partial\mathbb{D}$ but I can't see how this shows that $\sigma_{le}(S) = \sigma_{re}(S)$. Commented Jun 27, 2019 at 18:19
• If $\lambda$ were in $\sigma_{le}(S)\setminus\sigma_{re}(S)$, then $S-\lambda I$ would have infinite dimensional kernel, and closed, finite codimensional range. And this is no true for any $\lambda$. Commented Jun 28, 2019 at 7:29
• Similarly, $\lambda \in\sigma_{re}(S)\setminus\sigma_{le}(S)$ is not possible. Commented Jun 28, 2019 at 7:30
• Antother argument: by the continuity of the index for semi-Fredholm operators, if both $\sigma_{le}(S)$ and $\sigma_{re}(S)$ have empty interior, they coincide. Commented Jun 28, 2019 at 7:33
• Being injective and bounded below DOES NOT imply that the map is not Fredholm; e.g., the identity operator. Commented Mar 7, 2022 at 18:58