Trouble finishing a (direct) proof that $\ell^2(A)$ is a complete metric space

Let $A$ be any non-empty set. We can define summations of non-negative numbers over this index set by using a supremum of summations over finite subsets of $A$. That is,

$$\sum\limits_{\alpha \in A} x_\alpha = \sup \{\sum\limits_{\alpha \in F} x_\alpha \vert \ F \subseteq A \ \text{is finite}\}.$$

Define

$$\ell^2(A) = \{f: A \to \mathbb{C} \ \vert \sum\limits_{\alpha \in A} \lvert f(\alpha) \rvert^2< \infty \}.$$

Clearly $\ell^2(A)$ is a $\mathbb{C}$-vector space with pointwise addition and scaling. We give this vector space a norm according to $\|f\|^2= \sum\limits_{\alpha \in A} \lvert f(\alpha) \rvert^2$. This is the same as the standard definition of $L^2(\mu)$ where $\mu$ is counting measure on $A$. It is a well-known result that this norm gives $\ell^2(A)$ the structure of a Banach space. The main issue here is to show that the metric induced by this norm is complete. Following a comment in Gerald Folland's excellent text on real analysis, it should be "rather easy" to prove completeness directly.

To that end, we take a Cauchy sequence $(f_n) \in \ell^2(A)^\mathbb{N}$ and hope to show convergence with respect to the metric. It isn't hard to show that for each $\alpha \in A$, $f_n(\alpha)$ is a Cauchy sequence in $\mathbb{C}$ with the usual distance metric. By completeness of this metric space, we know that $f_n(\alpha)$ converges to some number $z_\alpha \in \mathbb{C}$.

We define a function $f: A \to \mathbb{C}$ by $f(\alpha)=z_{\alpha}$. Of course, $f$ is our candidate for the limit of the original sequence. To show that $f \in \ell^2(A)$, let $F \subseteq A$ be a finite subset of A and let $\epsilon >0$. Take $N \in \mathbb{N}$ so that $\|f-f_n\|<\epsilon \ \ \ \forall n \geq N$. Thus, if $n \geq N$, then $\|f_n\| < \|f_N\| + \epsilon$. Let $B = \max\{\|f_1\|, \|f_2\|, ..., \|f_{N-1}\|, \|f_N\|+\epsilon\}$. Then evidently $\|f_n\| \leq B \ \ \ \forall n \in \mathbb{N}$.

Using this upper bound on the sequence, we conclude that $$\sum\limits_{\alpha \in F} \lvert f(\alpha) \rvert^2 = \lim_{n \to \infty} \sum\limits_{\alpha \in F} \lvert f_n(\alpha) \rvert^2 \leq \limsup_{n \to \infty} \sum\limits_{\alpha \in A} \lvert f_n(\alpha) \rvert^2 = \limsup_{n \to \infty} \|f_n\|^2 \leq B^2$$ Since this is true for every finite subset $F \subseteq A$, it follows that $$\|f\|^2= \sum\limits_{\alpha \in A} \lvert f(\alpha) \rvert^2=\sup \{\sum\limits_{\alpha \in F} \lvert f(\alpha) \rvert^2 \vert \ F \subseteq A \ \text{is finite}\} \leq B^2 < \infty$$

Now comes the part that is troubling me. We would like to say that $$\lim_{n \to \infty} \sum\limits_{\alpha \in A} \|f(\alpha)-f_n(\alpha) \|^2 = 0$$

Indeed, each term in the summation converges to $0$, but the trouble is in trying to exchange the two limiting operations $\lim$ and $\sup$.

Here is what I thought I should do, but I haven't been successful. For each $n \in \mathbb{N}$, we can choose a sequence $(F_{n,m})$ of finite subsets of A such that $$\lim_{m \to \infty} \sum\limits_{\alpha \in F_{n,m}} \|f(\alpha)-f_n(\alpha) \|^2 = \sum\limits_{\alpha \in A} \|f(\alpha)-f_n(\alpha) \|^2$$

Then the statement above that we wanted to prove becomes $$\lim_{n \to \infty} \lim_{m \to \infty} \sum\limits_{\alpha \in F_{n,m}} \|f(\alpha)-f_n(\alpha) \|^2 =0$$

I wasn't successful in justifying that we can exchange these limits. I would very much appreciate some help in completing this proof with a minimum of hand-waving. I understand that we can characterize completeness in a normed vector space according to convergence of absolutely convergent series, and that this can be used to prove completeness of $L^p(\mu)$ in a more general sense. However, it seems like a good lesson in basic analysis to carry this direct proof through to the end.

As a side note, I kicked around another idea which might be relevant: If $\sum\limits_{\alpha \in A} x_\alpha < \infty$, then I believe we can control the "tail" of this series in the sense that $\forall \epsilon >0, \exists$ finite $F \subseteq A$ such that $\sum\limits_{\alpha \in E^C} x_\alpha < \epsilon \ \ \forall E \subseteq A$ with $E \supseteq F$.

Thanks for your time and suggestions.

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the last idea is the correct one, try to develop it further – Norbert Apr 19 '14 at 0:01

1 Answer

Fix $\varepsilon\gt 0$: there exists an integer $N$ such that if $n,m\geqslant N$ and $F\subset A$ is finite, then $$\sum_{\alpha\in F}|f_n(\alpha)-f_m(\alpha)|^2\leqslant\varepsilon.$$ What is important here is that $N$ depends only on $\varepsilon$ but no on the finite set $F$ we are considering. We thus obtain, taking the limit $m \to \infty$, that for each $n\geqslant B$, and each finite $F\subset A$, $$\sum_{\alpha\in F}|f_n(\alpha)-f(\alpha)|^2\leqslant\varepsilon.$$

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It is rather embarrassing that the solution was so simple. With a little more patience, I would have been likely to understand this. But time has been at a premium lately, so I appreciate your response. Also, just a note that there is a small typo in your answer which I am not able to edit; you seem to have used B rather than N near the conclusion. Otherwise, this issue is completely settled for me. Thanks again. – James Staff Apr 21 '14 at 19:15