$\|(I+A)^{-1}\| \leq \frac{1}{1-\|A\|)}$

Question

I have the following problem, of which I have a slight problem to finish with the second part:

Let $X$ be a Banach space and let $A \in B(X)$, $\|A\| < 1$. Prove that $(I+A)^{-1}$ exists and is given by: \begin{equation} (I+A)^{-1} = \sum_{n=0}^{\infty}(-1)^nA^n \end{equation}

where the series is absolutely convergent in B(X). Show also that $$\|(I+A)^{-1}\| \leq \frac{1}{1-\|A\|)}$$ \begin{proof} We examine that the alternating series converges using the ratio-test. Now:Let's multiply

$$\sum_{n=0}^{\infty}(-1)^nA^n$$ by (I+A): \begin{align*} \sum_{n=0}^{\infty}(-1)^nA^n \cdot (I+A) &=\sum_{n=0}^{\infty}(-1)^nA^n + \sum_{n=0}^{\infty}(-1)^nA^{n+1}\\ &=(I+A)-(A+A^2)+(A^2+A^3)-\ldots\\ &=I \end{align*} since $\sum_{n=0}^{\infty}(-1)^nA^n < \infty$ we know $(I+A)^{-1}$ exists.

Now to show that $$\|(I+A)^{-1}\| \leq \frac{1}{1-\|A\|)}$$:

From the result that $\sum_{n=0}^{\infty}(-1)^nA^n < \infty$ we have $$\|(I+A)^{-1}\|=\|\sum_{n=0}^{\infty}(-1)^nA^n\|=\|I-A+A^2-A^3\ldots\|$$

and now I would expect something with geometric progression, but I failed to achieve the right inequality. I appreciate any help!

Answer 1

Note that,

$$ || I-A+A^2-\dots|| \leq 1+||A||+||A^2||+ \dots \leq 1+||A||+||A||^2+ \dots\,. $$

since $||A^n||\leq ||A||^n $. Now, if $||A||< 1$, then you can sum the series on the far right which gives you the result you want.