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If $X$ is a Banach space, and $T:X \to X$ is a bounded linear operator with norm < $1$, then $I-T$ has a bounded inverse defined by $(I-T)^{-1} = \sum_{n=0}^\infty T^n$.

Thinking in terms of a converse, if $T$ is any bounded linear operator defined on $X$, then does the existence of a bounded inverse $S=(I-T)^{-1}$ imply that $S$ can be represented as $S=\sum_{n=0}^\infty T^n$?

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    Just added some tags.2012-01-22

2 Answers 2

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No, not even in the finite-dimensional case. If $T$ is a linear map from $\mathbb{R}^n$ to itself with all eigenvalues $>1$ in absolute value, then $I-T$ is invertible, but $\sum T^n$ certainly does not converge.

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    not even in dimension 1: 1-2 is invertible but $\sum_n 2^n$ does not converge2010-11-11
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Well there are slightly weaker condition that improves the result you cite.

Suppose $T$ is a bounded operator with spectral radius $r(T)< 1$ Then by the spectral radius formula we have $$\lim_{n\to\infty}\|T^n\|^{1/n} =\inf_n\|T^n\|^{1/n}=r(T)< 1$$ which ensure the convergence of $$\sum_{n=0}^\infty\|T^n\|$$ which in turn, by the triangle inequality, bounds $$\left\|\sum_{n=0}^\infty T^n\right\|$$ Now, it is a standard exercise to show that $$\sum_{n=0}^\infty T^n=(I-T)^{-1}$$ If there is any doubt at all do not hesitate to ask.. **Edit:** By the Banach algebra inequality we have $\|T^n\|\le\|T\|^n$, which means that $$r(T)=\inf_k\|T^k\|^{1/k}\le\|T^n\|^{1/n}\le\|T\|$$ Hence $\|T\|<1$ implies not only $r(T)<1$, but also $r(T)\le\|T\|<1$. Also, this is not the case in the example of Robin above, because we also have $r(T)=\sup{|\lambda|:\lambda\in\sigma(T)}$ where $\sigma(T)$ is the spectrum of $T$ (the set of all $\lambda\in\mathbb{C}$ such that $\lambda I-T$ is not invertible) and the eigenvaules of $T$ is certainly in the spectrum. (Note that $r(T)$ is the radius of the smallest closed disc that contain $\sigma(T)$ - hence the name spectral radius).

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    Where is the text???2010-11-11
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    An extreme example is a nilpotent operator (i.e. $T^n=0$ for some $n$).2010-11-11
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    @AD: Fixed. I don't know why but it seems you have to type `<` for `<`.2010-11-11
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    @KennyTM♦ Great - Thanks!2010-11-11
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    Wow, that's pretty cool. Can we get directly from the spectral radius formula that $r(T) < 1$ implies $||T|| < 1$?2010-11-11
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    @user1736: No, but conversely because $\|T^n\|\le\|T\|^n<1$ if $\|T\|<1$ -- we also have $r(T)=\lim_n\|T^n\|^{1/n}=\inf_n\|T^n\|^{1/n}$, waybe I should add that?.2010-11-11
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    @user1736: I just added this (and a bit more - I couldn't stop myself ;) ) into the answer.2010-11-11
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    Hm, isn't it always true that $\|T\| \geq r(T)$? I thought that if a number is in your spectrum, then the norm of your operator must always be greater than or equal to that number. I don't see why we need the Banach algebra inequality for that, but perhaps I'm missing something again..2010-11-11
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    The Banach algebra inequality is indeed useful here as seen above, but you are free to use a different argument for this.2010-11-12
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    BTW - I do not claim that this is *my* spectrum. This set called the spectrum is used in linear algebra, Fourier analysis, operator theory and functional analysis etc. I suggest that you look it up in Wikipedia.2010-11-12