# Problem of the Week #18 - October 1st, 2012

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#### Chris L T521

##### Well-known member
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Here's this week's problem.

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Problem: Suppose that $T\in L(X)$ is a bounded linear operator in a Banach space $X$ such that $\|T\|<1$. Show that $I-T$ is invertible, i.e. has a bounded inverse linear operator and
$(I-T)^{-1}=\sum_{k=0}^{\infty}T^k.$

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#### Chris L T521

##### Well-known member
Staff member
This week's question was correctly answered by girdav. You can find his solution below.

Let $S_n:=\sum_{j=0}^nT^j$. We have $$\lVert S_{m+n}(x)-S_n(x)\rVert=\lVert\sum_{j=n+1}^{n+m}T^jx\rVert\le \sum_{j=n+1}^{n+m}\lVert T^j\rVert \lVert x\rVert\le \lVert x\rVert \sum_{j=n+1}^{n+m}\lVert T\rVert^j,$$
which proves that $\{S_nx\}$ is Cauchy for each $x$ ($\lVert T\rVert<1$) hene it converges to some $Sx$. We have $S(I-T)x=\lim_{n\to \infty}(I-T^{n+1})x=x$ as $\lVert T\rVert<1$ and $(I-T)S=I$. This proves that $I-T$ is invertible with inverse $S$. This one is bounded as $\lVert Sx\rVert\leq \frac 1{1-\lVert T\rVert}\lVert x\rVert$.

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