If $a\in\text{Inv}\mathcal{A}~$ And $~b\in\mathcal{A}~;~\|a-b\|\leq\frac{\|a\|^{2}}{\|a^{-1}\|}$ then $~b\in\text{Inv}\mathcal{A}$

Question

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Exercices :

Let $\mathcal{A}$ be a banach algebra , $a\in\text{Inv}\mathcal{A}~$ And $~b\in\mathcal{A}$

Suppose that :

$$\|a-b\|<\frac{\|a\|^{2}}{\|a^{-1}\|}$$

Then prove that :

1. $~~~~b\in\text{Inv}\mathcal{A}$

2. $~~||b^{-1}-a^{-1}||\leq\frac{||a^{-1}||^{2}~||b-a||}{1-||a^{-1}||~||b-a||}$$

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My attempts

I can't show the fist question but I have I try for Second

If $~~b\in\text{Inv}\mathcal{A}$ then $b=a-(a-b)=a(e-a^{-1}(a-b))$ invertible and :

$$b^{-1}=[e-a^{-1}(a-b)]^{-1}a^{-1}$$ $$~~~~~~=\displaystyle\sum\limits_{n=0}^{\infty}(a^{-1}(a-b))^{n}a^{-1}$$ $\implies b^{-1}-a^{-1}=\displaystyle\sum\limits_{n=1}^{\infty}(a^{-1}(a-b))^{n}a^{-1}$ $$\implies \|b^{-1}-a^{-1}\|=||a^{-1}(a-b)||~||a^{-1}||\frac{1}{1-||a^{-1}(a-b)||}$$ $$~~~~~~~~~~~~~~\leq\frac{||a^{-1}||^{2}~||b-a||}{1-||a^{-1}||~||b-a||}$$ ....so I need a help for fist questions!!

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Thanks for your time

Answer 1

The first part is wrong. If $b=0$ and $a=2e$ then $\|a-b\|=2$ and $\frac {\|a\|^{2}} {\|a^{-1}\|}=\frac 4 {1/2}=8>\|a-b\|$ but $b$ is not invertible.