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In the middle of a proof of the Inverse Function Theorem (namely, the proof of Baby Rudin), we use the fact that if $A$ is invertible and:

$$ ||B-A||~||A^{-1}|| <1$$

then $B$ is invertible. The proof for this, however, relies on the fact that the vector spaces are finite dimensional (because it concludes that $B$ is bijective by using that it is injective). How do we circumvent this, if we want to prove that IFT in banach spaces?

Aloizio Macedo
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    This is true in general for continuous operator between Banach spaces. You can find a proof in Murphy's C$\hspace{0pt}^$-algebras and Operator Theory*, Theorem 1.2.3. – Luiz Cordeiro May 26 '15 at 19:57
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    For $\lVert X\rVert < 1$, the Neumann series $\sum_{n=0}^\infty X^n$ is convergent, and gives you the inverse of $I - X$. Apply to $X = I - BA^{-1}$. – Daniel Fischer May 26 '15 at 20:18

1 Answers1

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If $X$ is any Banach space, with $A, B \in L(X)$ bounded linear operators and $A$ invertible, we have, in the operator norm, using the hypothesis presented in the text of the question,

$\Vert I - A^{-1}B \Vert = \Vert A^{-1}(A - B)\Vert \le \Vert A^{-1} \Vert \Vert A - B \Vert < 1. \tag{1}$

Since (1) shows that

$\Vert I - A^{-1}B \Vert < 1, \tag{2}$

we have

$A^{-1}B = I - (I - A^{-1}B) \tag{3}$

is invertible; this follows from the well-known result that $\Vert C \Vert <1$, $C \in L(X)$, implies $I - C$ invertible; a detailed proof may be found in my answer to this question. Writing

$A^{-1}B = K, \tag{4}$

with $K$ invertible, also allows us to write

$B = AK; \tag{5}$

$B$, being the product of invertible operators, is itself invertible; indeed, we have

$B^{-1} = K^{-1}A^{-1}. \tag{6}$

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Robert Lewis
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