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I find this queistion in my textbook:

QUESTION) Let $T \in \mathcal{L}(\mathcal{B})$, where $\mathcal{B}$ is a Banach Space. If exists $n$ with $||T^{n}|| < 1$, show that, given $\eta \in \mathcal{B}$, the equation $x - Tx = \eta$ has a unique solution.


So, I think I can solve this question using Banach fixed point theorem, setting $A(x) = Tx + \eta $. However, I can't show that $A$ is a contraction. Help me with it!

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Observe that $A^n(x) = T^n x + \beta_n$, where $\beta_n = \sum_{k=1}^{n-1}T^k \eta + \eta$. Thus,

$$ \|A^n(x) - A^n(y)\| = \|T^n(x - y)\| \leq \|T^n\|\|x - y\| $$

which implies that $A^n$ is a contraction. From here you can use the result that if $A^n$ is a contraction, then $A$ has a unique fixed point. For example, see the proof of this result here.

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  • $\begingroup$ Yeah, follow of proof of Banach fixed point theorem that, if $T^{k}$ has a fixed point $x_{0}$, then $T^{n}(x) \rightarrow x_{0}$ for any $x$. I just see it now. Thank you so much. $\endgroup$ – BBVM Nov 17 '15 at 23:07
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If $\|T^n\| < 1$, then $I-T^n$ is invertible with inverse $\sum_{k=0}^{\infty}(T^n)^k$. Then \begin{align} I-T^n & = (I-T)(I+T+T^2+\cdots+T^{n-1}) \\ & = (I+T+T^2+\cdots+T^{n-1})(I-T). \end{align} It follows that $I-T$ is invertible because it must be injective and surjective by the above. And that's what you need.

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