4
$\begingroup$

I am struggling with the second part of this problem:

Let $A$ be a unital Banach Algebra, and let $x$ and $y$ be elements in A such that that $xy=1_A$ and $yx\neq1_A$.

(i) Let $z$ be an element in $A$, such that $\|x-z\|<\frac{1}{\|y\|}$. Show that $z$ is not invertible.

(ii) Let $H$ be an infinite dimensional Hilbert space. Let $S\in B(H)$ be a non-unitary isometry (i.e. $S^*S=I$ and $SS^* \neq I$). Show that $$1=\|S\|=\textrm{dist}(S,G(B(H))),$$ where $G(B(H))$ is the subset of invertible elements in $B(H)$.

This is the progress, I have made: I have finished the proof of part (i) using the fact that if $\|1_A-x\|<1$, for some $x\in A$, then $x$ is invertible . Further, it is easy to see that $\|S\|=\|S^*\|=1$. It therefore follows from part (i) that $\textrm{dist}(S^*, G(B(H))\geq 1$. Finally, it is evident that neither $S$ nor $S^*$ are invertible.

Any hints would be appreciated.

  • 1
    I'm still trying to understand this myself, but it may help to know that you can show $\textrm{dist}(S, G(B(H))\geq 1$ using a statement analogous to(and just as easy to prove as) part (i). – 2017-02-17
  • 0
    @Aweygan. You are right about that. Thanks. Now we only need to show the other inequality sign. – 2017-02-17
  • 0
    You're welcome. May I ask where this problem comes from? Is it an exercise from a book? – 2017-02-17
  • 0
    They did this exercise in class in a course on operator algebra, I am following at the moment. But I was out travelling, so I did not see the solution. If it is from a book, I don't know which. – 2017-02-17
  • 1
    You are done, because $\|S-A\|=\|S^*-A^*\|$, and $A$ is invertible if and only if $A^*$ is, so $S$ and $S^*$ have the same distance from the invertible elements. – 2017-02-18
  • 0
    @ChristianRemling. That is a useful observation. But how does this imply, that the common distance is 1? – 2017-02-18
  • 2
    @Kimarokko: I missed that part, but it's clear too: you already showed that the distance is $\ge 1$, and it's $\le 1$ because $\|S\|=1$ and $\epsilon I$ is invertible for all $\epsilon\not= 0$. – 2017-02-18

1 Answers 1

2

We have $\|\varepsilon I-S\|\geq1$, since $\|I-S\|<1$ would imply that $S$ is invertible. So $$1=\|S\|\leq\|\varepsilon I-S\|\leq \|S\|+\varepsilon$$ for all $\varepsilon>0$. Thus $\text{dist}\,(S^*,G(B(H)))\leq1+\varepsilon$ for all $\varepsilon >0$, and thus $$\tag{1} \text{dist}\,(S,G(B(H)))\leq1. $$ On the other hand, using $(i)$ we have that if $T$ is invertible, then $$\tag{2} \|S-T\|\geq\frac1{\|S\|}=1. $$ Combining $(1)$ and $(2)$, we get $$\text{dist}\,(S,G(B(H)))=1.$$