4

A silly question, but I don't see the answer. This question is from Rudin's Functional Analysis, Chapter 10, Exercise 1a).

It's obvious that $y$ has a left inverse as $((xy)^{-1}x)y=e$, where $e$ is the unit of the algebra. However, starting with $y((xy)^{-1}x)$, I cannot proceed further. Does this suggest that I need to 'venture beyond' the multiplicative group structure, maybe introduce ideas from the ring structure on the Banach Algebra? ( I'm thinking of something analogous to the proof where $e-xy$ is invertible if and only if $e-yx$ is invertible. There, if $z$ is the inverse of $e-xy$, then, $e+yzx$ is the inverse of $e-yx$.)

Malcolm King
  • 41
  • 1
    Hint: a product of two invertible elements is certainly invertible. Given the information you have, can you write y as the product of two invertible elements? – Manny Reyes Aug 13 '12 at 20:28

2 Answers2

6

I am guessing you meant Exercise 2(a)?

Let $p = y((xy)^{-1}x)$, then $xp = xy((xy)^{-1}x) = x$, since $x$ is invertible, it follows that $p=e$.

copper.hat
  • 172,524
2

The inverse of $x^{-1}xy$ is $(xy)^{-1}x$, since $x^{-1}xy\cdot (xy)^{-1}x=e$ and $(xy)^{-1}xx^{-1}xy=e$. Hence $y$ is invertible.

t.b.
  • 78,116
Davide Giraudo
  • 172,925