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?

1$\begingroup$ What is left (right) spectrum ? Spectrum as the left (right) multiplication operator on the space of bounded operators ? $\endgroup$– InfiniteLooperCommented Jun 27, 2019 at 13:44

$\begingroup$ 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. $\endgroup$– Robert FurberCommented Jun 27, 2019 at 17:59

$\begingroup$ Does looking back at Proposition 4.3 of the same section help in showing that $\sigma_{le}(S) = \sigma_{re}(S)$? $\endgroup$– Robert FurberCommented Jun 27, 2019 at 18:18
1 Answer
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.

$\begingroup$ This is a fine way to show that $\sigma_e(S) = \partial\mathbb{D}$ but I can't see how this shows that $\sigma_{le}(S) = \sigma_{re}(S)$. $\endgroup$ Commented Jun 27, 2019 at 18:19

$\begingroup$ 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 $\lambda$. $\endgroup$ Commented Jun 28, 2019 at 7:29

$\begingroup$ Similarly, $\lambda \in\sigma_{re}(S)\setminus\sigma_{le}(S)$ is not possible. $\endgroup$ Commented Jun 28, 2019 at 7:30

$\begingroup$ Antother argument: by the continuity of the index for semiFredholm operators, if both $\sigma_{le}(S)$ and $\sigma_{re}(S)$ have empty interior, they coincide. $\endgroup$ Commented Jun 28, 2019 at 7:33

1$\begingroup$ Being injective and bounded below DOES NOT imply that the map is not Fredholm; e.g., the identity operator. $\endgroup$ Commented Mar 7, 2022 at 18:58