Let $X$ be a Banach space and $A :X \rightarrow X$ be a linear mapping with norm $\|A\|<1$. Let $I$ be the unit mapping $I(x)=x$.

a) Using the Contraction Mapping Theorem to show $I-A$ is a bounded mapping and is one-to-one and onto.

b) Show that $I-A$ is an isomorphism and $(I-A)^{-1}=I+\sum_{n=1}^{\infty}A^n$

Thoughts: Well first off I am a little confused of what part b is asking, would part a not already show that it is an isomorphism? Or must I show that $(I-A)^{-1}$ exists and is bounded? I was also thinking for part b would it work to take the limit of summation as n approaches infinity? I know what the contraction mapping theorem states but how can I apply this here to do part a? Help is much appreciated!


As tattwamasi amrutam remarked, the formula for $(I-A)^{-1}$ solves both (a), (b). Just for "completeness", I'll show how Banach fixed-point theorem can answer (a).

$I-A$ is obviosly bounded. In order to show that $I-A$ is one-to-one and onto, fix $y \in X$. We have $(I-A)x = y$ exactly when $y + Ax = x$, i.e. if $x$ is a fixed point of $z \mapsto y+Az$. By Banach fixed-point theorem, this mapping has exactly one fixed point $x \in X$. This shows that $I-A$ is a bijection.

It can be shown using the open-mapping theorem that $I-A$ is an isomorphism (i.e. $(I-A)^{-1}$ is bounded), but let us do it by hand:

if $(I-A)^{-1}y = x$, then $x = y + Ax$ and \begin{align*} \| x \| = \| y + Ax \| & \le \| y \| + \| A \| \cdot \| x \|, \\ \| x \| & \le (1-\|A\|)^{-1} \|y\|, \end{align*} therefore $(I-A)^{-1}$ is bounded, $\|(I-A)^{-1} \| \le (1-\|A\|)^{-1}$.


I don't know what contraction mapping is. But everything in your claim follows once you show that $(I-A)$ is invertible. And the inverse has been given to you already in part(b):

Hint : Consider the series $\sum_{n=0}^\infty A^n$. Show that the series is absolutely convergent. Let $T=\sum_{n=0}^\infty A^n$. Show that $T(I-A)=(I-A)T=I$.

The boundedness of $(I-A)$ follows from the given conditions.

  • $\begingroup$ So don't use the Banach fixed-point theorem (contraction mapping theorem) for part a) ? Also I must show is that the inverse of $(I-A)$ exists and it implies that $(I-A)$ is a bounded mapping and is one-to-one and onto? $\endgroup$ – Rick Owens Apr 2 '17 at 21:14
  • $\begingroup$ thanks for the hint as well $\endgroup$ – Rick Owens Apr 2 '17 at 21:14
  • $\begingroup$ Showing that the series converges ensures that $T$ is continuous. One-one and Onto follows from showing that $(I-A)T=T(I-A)=I$ $\endgroup$ – tattwamasi amrutam Apr 2 '17 at 21:18

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