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I was reading Kreyszig's book on functional analysis when I came across this theorem:

"Let $T\in{}B(X,X)$, where $X$ is a Banach space. If $||T||<1$, then $(I-T)^{-1}$ exists as a bounded linear operator on the whole space $X$ and $(I-T)^{-1}= \sum_{i=0}^{\infty}T^{i}=I+T+T^{2}+...$"

Here's my question:

The book proved this theorem by show that the series $\sum_{i=0}^{\infty}T^{i}$ converges absolutely, so that it is properly defined as an operator, and that $(I-T)(I+T+T^{2}+...T^{n}) \to I$ as $n\to\infty$.

But how can I prove that the operator $(I-T)$ is injective in the first place? And why can we say that $(I-T)^{-1}$ exists on the whole $X$ (which I believe would require the surjectivity of $(I-T)$ )?

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You either $\sum$ or write $I+T+\ldots$. You don't sum over the infinite sum. –  Asaf Karagila Apr 15 '12 at 13:58
    
Sorry @AsafKaragila that was a typo. –  Vokram Apr 15 '12 at 14:00
1  
A linear map is injective if and only if it has trivial kernel. So suppose there is some non-zero x with (I-T)x = 0. Then x = Tx so $||Tx||/||x|| \ge 1 \Rightarrow ||T|| \ge 1$, a contradiction. –  Conor Apr 15 '12 at 14:08
    
@Conor Oh I see... Thank you! But what about the surjectivity?.. –  Vokram Apr 15 '12 at 14:10

1 Answer 1

up vote 4 down vote accepted

Hint.

Let $\varphi :A\to B$ and $\psi:B\to A$ be maps between sets. Try to prove the following $$ \psi\varphi=Id_A\Longrightarrow \psi \text{ is surjective and }\varphi\text{ is injective} $$

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Oh I see what you mean now. Thank you so much! –  Vokram Apr 15 '12 at 14:23
    
@Vokram, not at all!${}{}{}$ –  userNaN Apr 15 '12 at 14:38

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