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I am reading the book of Walter Rudin Functional Analysis and the page 235 was given theorem 10:12 which is as follows:

Theorem: If $A$ is a Banach algebra, then $G(A)$ is a open subset of $A$, and the mapping $x\rightarrow x^{-1}$ is a homeomorphism of $G(A)$ onto $G(A)$.

After Theorem Proving also given her, but I can not understand very well. Please, if someone has the opportunity to help by giving a proof of this theorem other.

Previously thank you.

Madrit Zhaku
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1 Answers1

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You can follow this outline :

  1. If $\|x\| < 1$, then $(1-x)$ is invertible, because the series $$ \sum_{n=0}^{\infty} x^n $$ converges (because it converges absolutely, and you are living in a Banach space)

2.If $x\in G(A)$, and $\|h\| < \frac{1}{2}\|x^{-1}\|^{-1}$, then $(x+h) \in G(A)$ because $$ x+h = x(1 + x^{-1}h); \quad \text{ and } \|x^{-1}h\| < 1 $$ 3. Define a linear operator $u(a) = -x^{-1}ax^{-1}$ then $$ \|(x+h)^{-1} - x^{-1} - u(h)\| = \|(1+x^{-1}h)^{-1}x^{-1} - x^{-1} + x^{-1}hx^{-1}\| $$ $$ \leq \|(1+x^{-1}h)^{-1} - 1 +x^{-1}h\|\|x^{-1}\| $$ $$ \leq 2\|x^{-1}\|^3\|h\|^2 $$ (That last inequality again uses the series from above). Therefore the map $a \mapsto a^{-1}$ is differentiable at $a=x$ with derivative $u$.

In particular, it is continuous. Furthermore, it is its own inverse, and hence it is a homeomorphism.

Prahlad Vaidyanathan
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