Progressed : Convergence problem in Hilbert Space and necessity of inner product

******** 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?

• 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 $A$? – Ian Sep 20 '15 at 17:45
• @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 – kroner Sep 20 '15 at 17:47
• As I edited into the previous comment, you can force the square into the problem at no loss. – Ian Sep 20 '15 at 17:48
• 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. – Ian Sep 20 '15 at 17:54
• 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 $n$. – A.Γ. Sep 20 '15 at 19:44