How to prove that withouting using Gelfand spectral radius theorem?

Question

Let $X$ be a Banach space and $T$ be a bounded linear operator on $X$. Assume that there is an integer $n\ge 1$ so that $\|T^n\|<1$. Show that $I-T$ is invertible and $$ (I-T)^{-1}=\sum_{j=0}^\infty T^j $$

I apply Neumann series property to $T^n$: let $S=I-T^n$. Since $\|I-S\|<1$, then $S$ is invertible.

Since $$S=(I-T)(I+T+T^2+\dots+T^{n-1}) $$ and $(I-T)$ commutes with $(I+T+T^2+\dots+T^{n-1}) $, then $I-T$ is invertible.

Moreover, $$ S^{-1}=\sum_{k=0}^\infty (I-S)^k $$ that is $$ (I-T^n)^{-1}=\sum_{k=0}^\infty T^{nk} $$

But I am stuck here...

geetha290krm · Answer 1 · 2023-12-03T09:11:14.420

1

Show that $\sum\limits_{j=0}^{\infty} \|T^{j} \|<\infty$ by observing that any $m$ can be written as $kn+r$ for some non-negative integer $k$ and some $r \in\{0,1,2...,n-1\}$. This implies that $\sum\limits_{j=0}^{\infty} T^{j}$ is a bounded operator. Now it is just a simple verification that $(I-T) (\sum\limits_{j=0}^{\infty} T^{j})=(\sum\limits_{j=0}^{\infty} T^{j}) (I-T)=I$.

[$\sum\limits_{j=0}^{\infty} \|T^{jn+r} \|\leq \|T^{r}\|\sum\limits_{j=0}^{\infty} \|T^{n} \|^{j}<\infty$. Sum this over $r$].

edited Dec 03 '23 at 09:11

answered Dec 03 '23 at 08:24

geetha290krm

36,632

So $\sum_{r=0}^{n-1}|T|^r$ is bounded because $|T|$ is bounded? – H.Y Duan Dec 06 '23 at 06:17
Yes, that is correct. @H.YDuan – geetha290krm Dec 06 '23 at 06:18

How to prove that withouting using Gelfand spectral radius theorem?

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