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******** PROGRESS : so thanks to Ian's great comment I can get by the proof and that completeness is necessary but I need to know does this hold for general Banach spaces that are not Hilbert spaces?

I have this assignment question from Functional Analysis class stating:

Let $ \mathcal{H} $ be a Hilbert Space with a sequence $ \{x_n\}_{n=1}^{\infty} $ of elements in $ \mathcal{H} $. There is a constant $ A \in R $ such that for every sequence $ \{a_n\}_{n=1}^{\infty} $ satisfying $ 0 \leq |a_n| \leq 1 $ all zero except a finite number of elements, we have ||$ \sum_{n} a_nx_n $|| $ \leq $ A. We are asked to prove $\lim_{N\to\infty} {\sum_{n=1}^{N} x_n}$ exists in the norm. Also, does this hold if the space H is not complete? What about a norm space where the norm is not induced from an inner product?

Last part before this I proved:

For all $ \{ x_n \}_{n=1}^N \in \mathcal{H} $ there are scalars $ \{ a_n \}_{n=1}^N $ on the complex unit circle such that : $||\sum_{n=1}^{N} a_nx_n ||^2 \geq \sum_{n=1}^{N} ||x_n||^2 $

I don't exactly know how to incorporate that previous part or if I even need to for that matter. I would really appreciate help.

PROGRESS : so thanks to Ian's great comment I can get by the proof and that completeness is necessary but I need to know does this hold for general Banach spaces that are not Hilbert spaces?

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    $\begingroup$ In the first part, you want to show that $y_N=\sum_{n=1}^N x_n$ is Cauchy in the norm. This means that for every $\varepsilon > 0$ there exists a $B \in \mathbb{N}$ such that if $N \geq M \geq B$ then $\| \sum_{n=M}^N x_n \| < \varepsilon$. Of course it is the same to have $\| \sum_{n=M}^N x_n \|^2 < \varepsilon$. Can you control this quantity using your previous result and the boundedness assumption involving $A$? $\endgroup$ – Ian Sep 20 '15 at 17:45
  • $\begingroup$ @Ian : no Ian that's just it I cannot get over the last result having squares (powers of two) but the task at hand does not $\endgroup$ – kroner Sep 20 '15 at 17:47
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    $\begingroup$ As I edited into the previous comment, you can force the square into the problem at no loss. $\endgroup$ – Ian Sep 20 '15 at 17:48
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    $\begingroup$ The argument I suggested fails if you don't have completeness, but to be sure that the conclusion also fails if you don't have completeness, you should use some example. Consider $X \subset \ell^2$ where $x \in X$ if and only if all but finitely many entries of $x$ are zero. I'm not so sure about the last part. $\endgroup$ – Ian Sep 20 '15 at 17:54
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    $\begingroup$ The address the last question, consider the space $\ell^{\infty+}$ of bounded sequences over $\mathbb{N}$ with the standard $\max$ norm and $x_n=e_n$ the canonical unit vectors with all zeros except unity at index $n$. $\endgroup$ – A.Γ. Sep 20 '15 at 19:44

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