Htykuut

Let $mathcal{A}$ be a unital Banach algebra and define $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n}$.

It is possible to show that $r^{text{Gelf}}(A) = lim_{n rightarrow +infty} | A^n |^{1/n} = inf_{n rightarrow +infty} | A^n |^{1/n}$.

I need to show that if $A$ is invertible and $B$ is such that $r^{text{Gelf}}(BA^{-1}) < 1$, then $(A - B)$ is invertible.

What I did is the following: consider the element $C = A^{-1} + A^{-1}sum_{n=1}^{+infty}(BA^{-1})^n$. This is known as the Neumann series.

If this series converges, it follows from a direct computation that $(A - B)^{-1} = C$, but I am struggling to show that this series in fact converges.

I tried to show if $r^{text{Gelf}}(BA^{-1}) < 1$, then $| BA^{-1} | < 1$, but I did not succeed (and maybe this isn't even true).

I would appreciate very much if someone could help me!

HINT:

Consider the Sequence $D_n=sum_{k=1}^n(BA^{-1})^n$. Since the series $sum_{k=1}^infty|(BA^{-1})^n|$ converges by the root test, the sequence ${D_n}$ converges to some $Dinmathcal A$. Then the $C$ you write in your question is $A^{-1}D$. Can you finish?

Side note: It is not true that $r^{text{Gelf}}(A) < 1$ implies $| A| < 1$. For example, consider $mathcal A=M_2(mathbb C)$ and

$$A=begin{pmatrix}0&frac{3}{2}\frac{3}{8} &0end{pmatrix}.$$

Then $r^{text{Gelf}}(A)=frac{3}{4}<1 $ while $|A|=frac{3}{2}>1$.