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Let $(V, \| \cdot \|)$ be a Banach Space over $\mathbb{F}$ and let $T \in B(V)$ (i.e linear and continuous $V \to \mathbb{F}$). If $\|I - T \| < 1$ then prove

  • T bijective
  • $T^{-1}$ continuous
  • $S_n = \sum_{k=0}^{n} (I-T)^k$ converges to $T^{-1}$

Here is what I have done.

Assume wlog $n>m$ $$ \|S_n - S_m \| = \| \sum_{k=0}^n (I-T)^k - \sum_{k=0}^m (I-T)^k \| = \| \sum_{k=m+1}^n (I-T)^k \| \leq \sum_{k=m+1}^n \|I-T\|^k \leq \\ \leq \sum_{k=0}^n \|I-T \|^k = \frac{1}{1 - \|I -T \| } < \infty $$

Which proves that $S_n$ is Cauchy. Since $(V, \| \cdot \|)$ complete $\Rightarrow$ $B(V)$ complete, thus $S_n \to S$.

Now

$$ S_nT = \left( \sum_{k=0}^n (I-T)^k \right) (I - (I-T)) = (I-T)^{n+1} + I \to I $$ and similiar for $TS_n$. Thus $S$ is the inverse of $T$.

Since

$$ \|T^{-1} \| = \|S \| = \lim_{n \to \infty} \|S_n\| = \lim_{n \to \infty} \|\sum_{k=0}^n (I-T)^k \| \leq \lim_{n \to \infty} \sum_{k=0}^n \|I-T\|^k < \infty $$ $T^{-1}$ is bounded and hence continuous.

Since $TS = ST = I$ we get that $T$ is bijective.

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    See: http://math.stackexchange.com/questions/2080877/please-help-about-the-linear-operators-in-banach-spaces/2080897#2080897 or the corresponding duplicate. – 2017-01-12
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    Have you said anything about the spectrum of an operator yet? – 2017-01-12
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    No, that is not included in the course. @Omnomnomnom – 2017-01-12
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    I think you mean $T$ is a map $V \to V$ – 2017-01-12

1 Answers 1

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Hint: Use a "geometric series" to build its inverse. Any operator with a two-sided bounded inverse must be bijective.

Alternative approach: To show that $T$ is injective, show that $Tx = 0 \implies x = 0$. Note that for any $x$, $\|Tx - x\| < \|x\|$, and that (by the triangle inequality) $$ \|Tx\| \geq \|x\| - \|Tx - x\| $$ Not sure how to show surjectivity.

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    Actually later on im asked to prove that $S_n = \sum_{k=0}^n (I-T)^k$ converges to $T^{-1}$. Im not sure if that is what you were thinking. If this is what you are thinking of. I believe I am supposed to use something else to prove the bijectivity. – 2017-01-12
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    That's exactly what I was thinking. Okay, I'll think of something else. – 2017-01-12
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    @Olba12 A typical argument is to show that $(S_n)$ converges, call the limit $S$, and then you show that $ST = I = TS$, i.e. $S = T^{-1}$. You show the bijectivity of $T$ by producing an inverse. – 2017-01-12
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    Maybe it's useful to note that $\|Tx\| \geq 1 - \|T - I\|$ for all $x$, which is to say that $T$ is "bounded below". – 2017-01-12
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    I'm gonna give it a try using the series. With two sided bounded inverse, do you mean bounded above and below? @Omnomnomnom – 2017-01-12
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    @Obla no. I mean $TT^{-1}$ and $T^{-1}T$ are both the identity – 2017-01-12
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    I have updated my post. Would you care to look at it? @Omnomnomnom – 2017-01-12
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    Looks good to me – 2017-01-12