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Let $T$ be a bounded linear operator on $E$ with norm strictly less than 1. Do we need $E$ to be Banach to define $(I - T)^{-1}$ as $$ \sum_{k=0}^\infty T^k ?$$ The proof I am aware of for $(I - T)^{-1} = \sum_{k=0}^\infty T^k$ pointwise requires Cauchy sequences. Can't we get away with using the fact that $E^*$ is Banach (regardless of the completeness of $E$) and using the fact that the sum considered above converges absolutely and by completeness (of $E^*$) converges in the operator norm to some operator and hence converges pointwise?

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    What is meant by pointwise convergence in this case? Are we really guaranteed pointwise convergence? AFAIK the proof guarantees exactly that it converges in norm. If it converges only in $E^*$ that would mean that the limit is "outside" $E$, I'm not sure if this means that $I-T$ is non-invertible since we can find an inverse "outside" $E$. – 2017-02-27
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    If $T_n \to T$ in operator norm, then we get $T_n(x) \to T(x)$ for free, right? – 2017-02-27
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    The sum converges in operator norm to some $S \in E^*$ since it converges absolutely and $E^*$ is Banach. Therefore the sum converges pointwise, right? – 2017-02-27
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    But $T_n(x)\to T(x)$ indeed, but it is convergence in norm (since $|T_n(x)-T(x)|\le|T_n-T|$ by definition). That's why I wondered what is meant by convergence in point. – 2017-02-27
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    To be honest, I'm not sure what you're saying in your first comment. Can you rephrase? – 2017-02-27
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    $E^* = L(E,\Bbb R)\ne L(E,E)$. – 2017-02-27
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    The first question is what is meant by pointwise convergence in this case. The rest of the comment is somewhat dependent on what the meaining of pointwise convergence is. – 2017-02-27
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    @Martín-BlasPérezPinilla Well, normally that's what one mean perhaps, but from the given context I interpret $E^*$ here to mean the completion of $E$. – 2017-02-27
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    @Martín-BlasPérezPinilla wow I'm not sure why I was conflating those two... thanks! – 2017-02-28

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$E$ = polynomials in $[0,1]$ with the $\sup$ norm, $Tf(x) = \frac12\int_0^x f$ is bounded, namely: $$\|Tf\|\le\frac12\|f\|.$$ But $\sum_{k=0}^\infty T^k$ is not convergent in $L(E,E)$ because transforms polynomials in infinite series.

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    For this counter example to be complete one should of course note that there in fact is an inverse to $I-T$ and that $(I-T)^{-1}\in L(E,E)$. Also note that if $T$ is extended to $E^*$ we would have that the series converges to an inverse in $L(E^*, E^*)$ which is not the same as the inverse in $L(E,E)$. – 2017-02-27
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    @skyking, the solution of $f(x) - (1/2)\int f(x)dx = x$ is an exponential... – 2017-02-27
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    That would explain a lot - I don't know where I got $I-T$ to be invertible in $L(E,E)$. It ought to be invertible in $L(E^*, E^*)$ and the inverse there ought to be $\sum T^k$. It then looks like my comment to the question holds. I guess that the only complication is where the series fails to converge inside the space is exactly the cases where the inverse also happens to fall outside the space. – 2017-02-27