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One of our problem sets this quarter in analysis asked us to show that $\ell^2$, the set of all square-summable sequences, is complete. At the time, I struggled to prove this, but I have an idea after reading on in Rudin's Principles. My idea is to use uniform convergence of a sequence of continuous functions and change the order of limits. Please let me know what I need to make the following go through; I have seen many questions here on StackExchange, similar to this, where the answerer skirts the question of this exchange of limits, and I would like if someone could just answer with as many epsilons as needed to get the thing settled once and for all.

We have shown that $\ell^2$ is an inner-product space, and it has a norm $\lVert \mathbf x\rVert^2 = \sum_{n\ge 1} x_n^2$. One step of the process is to come up with a candidate limit for a Cauchy sequence, and we can do this fairly easily. Let $(\mathbf x_n)$ be a Cauchy sequence in $\ell^2$ with $\mathbf x_n = x_{n,1},x_{n,2},x_{n,3},\dotsc$. We can arrange the sequences $\mathbf x_n$ into an infinite square matrix $(x_{i,j})$, and show without much trouble that for a fixed $j\in \mathbf N$ the sequence $(x_{i,j})_{i\in\mathbf N}$ is Cauchy (the columns of the infinite array are themselves Cauchy).

For each fixed $j\in \mathbf N$, and with $m,n$ sufficiently large, $|x_{n,j}-x_{m,j}|\le\lVert \mathbf x_n-\mathbf x_m\rVert < \varepsilon$, since $(\mathbf x_n)$ is Cauchy. Since $\mathbf R$ is complete, put $y_j = \lim_{n\to\infty}x_{n,j}$. We claim that $\mathbf y = y_1,y_2,y_3,\dotsc$ is in $\ell^2$ and $\mathbf x_n\to \mathbf y$. By the reverse-triangle inequality, for sufficiently large $m,n\in\mathbf N$, $|\lVert \mathbf x_n\rVert-\lVert\mathbf x_m\rVert| \le \lVert \mathbf x_n-\mathbf x_m\rVert < 1$ by the Cauchy criterion, so that since $\lVert \mathbf x_m\rVert = C < \infty$, we have $\lim_{n\to\infty}\lVert \mathbf x_n\rVert \le C + 1 < \infty$.

We want to show that $\sum_{j=1}^\infty y_j^2 < \infty$, and here is how I want to do that. For each $n=1,2,\dotsc$, put $f_n:\mathbf N\to \mathbf R$ to be the function $f_n(k) = \sum_{j=1}^kx_{n,j}^2$. Every function from $\mathbf N$ to $\mathbf R$ is continuous, so $(f_n)$ is a sequence of continuous functions. Now, $\sum_{j=1}^\infty y_j^2 = \lim_{k\to\infty}\lim_{n\to\infty}\sum_{j=1}^kx_{n,j}^2 = \lim_{k\to\infty}\lim_{n\to\infty}f_n(k)$. If only I could interchange these limits, then I would have $\sum_{j=1}^\infty y_j^2 = \lim_{n\to\infty}\sum_{j=1}^\infty x_{n,j}^2 = \lim_{n\to\infty}\lVert \mathbf x_n\rVert^2 < \infty$, as we already saw. My thoughts are that, as $k\to\infty$, $f_n(k)\to \lVert \mathbf x_n\rVert^2$, and as $n\to\infty, f_n(k)\to\sum_{j=1}^ky_j^2$, and that this latter convergence is uniform (we are dealing with a finite sum, so the uniform convergence shouldn't be hard to demonstrate is my gut reaction). Thus, if we put $f(k) = \sum_{j=1}^k y_j^2$, then $f_n\to f$ uniformly on $\mathbf N$, so we can exchange the order of the limits, just like in Rudin's theorem 7.11.

I am not sure if this is correct—particularly the part where I am asserting $f_n\to f$ uniformly on $\mathbf N$. If anyone could help straighten this out, I'd be much obliged.

Edit I am aware that this isn't all that has to be said to justify completeness. My main question is the rigorous justification of switching limits.

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  • $\begingroup$ Assume you have proved that the limits can be interchanges and $\|\mathbf x_n\|_2\to \|y\|_2$. It is not the whole thing, you still need to prove that ${\mathbf x_n}\to y$ in $\ell^2$. $\endgroup$ – A.Γ. Dec 24 '16 at 20:14
  • $\begingroup$ @A.G. Thanks for commenting on that. I know this, but I really wanted some help with the interchange of limits. Assuming that goes through, I can handle the rest. I'll update my question. $\endgroup$ – Alex Ortiz Dec 24 '16 at 21:49
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Right, if the uniform convergence of $(f_n)$ is shown, you can interchange limits per theorem 7.11 (using $\infty$ as an accumulation point of $\mathbf{N}$).

To show the uniform convergence, for $k \in \mathbf{N}$, let $P_k \colon \ell^2 \to \ell^2$ be the projection setting all components with index $> k$ to $0$, i.e.

$$(P_k(x))_j = \begin{cases} x_j, & j \leqslant k \\ 0, & j > k.\end{cases}$$

Then we note that $f_n(k) = \lVert P_k(x_n)\rVert^2$, and consequently

$$\lvert f_n(k) - f_m(k)\rvert = \bigl\lvert \lVert P_k(x_n)\rVert^2 - \lVert P_k(x_m)\rVert^2\bigr\rvert = \bigl\lvert\lVert P_k(x_n)\rVert - \lVert P_k(x_m)\rVert\bigr\rvert\cdot\bigl(\lVert P_k(x_n)\rVert + \lVert P_k(x_m)\rVert\bigr).$$

Now $P_k$ never increases the norm, i.e. $\lVert P_k(x)\rVert \leqslant \lVert x\rVert$ for all $x\in \ell^2$, and since $\lVert x_n\rVert$ is a Cauchy sequence, it follows that $\bigl(\lVert x_n\rVert\bigr)$ is bounded, say $\lVert x_n\rVert \leqslant K$ for all $n$. Thus from the above we obtain

$$\lvert f_n(k) - f_m(k)\rvert \leqslant 2K\bigl\lvert \lVert P_k(x_n)\rVert - \lVert P_k(x_m)\rVert\bigr\rvert.$$

By the reverse triangle inequality it follows that

$$\lvert f_n(k) - f_m(k)\rvert \leqslant 2K\lVert P_k(x_n) - P_k(x_m)\rVert.$$

But $P_k$ is linear, and it never increases norm, so

$$\lvert f_n(k) - f_m(k)\rvert \leqslant 2K\lVert P_k(x_n - x_m)\rVert \leqslant 2K\lVert x_n - x_m\rVert.$$

This bound is independent of $k$, and hence

$$\lVert f_n - f_m\rVert_{\infty} := \sup \{ \lvert f_n(k) - f_m(k)\rvert : k \in \mathbf{N}\} \leqslant 2K\lVert x_n - x_m\rVert.$$

And since $(x_n)$ is a Cauchy sequence, it follows that $(f_n)$ is a Cauchy sequence with respect to the uniform norm, thus it converges uniformly to its pointwise limit $f$.

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  • $\begingroup$ +1 Very insightful and exceptionally clear. This makes it very clear why the convergence I was seeking is uniform without unnecessary fluff. I applaud this answer! $\endgroup$ – Alex Ortiz Dec 24 '16 at 23:29
  • $\begingroup$ Just to be clear, we get that $f_n\to f$ uniformly with Rudin's theorem 7.8? $\endgroup$ – Alex Ortiz Dec 25 '16 at 0:32
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    $\begingroup$ Yes, @AOrtiz. Although, since we already know that the sequence is pointwise convergent, we only need half of one direction of that theorem. $\endgroup$ – Daniel Fischer Dec 26 '16 at 20:25
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Your prerequisites that $\|\mathbf x_n\|$ is bounded and converges to $y$ pointwise is not enough for uniform convergence. In fact, it gives only weak convergence. A simple counterexample is $\mathbf x_n$ with $x_{n,j}=0$ for $j\ne n$ and $x_{n,n}=1$.

You really need to make more use of the fact that $\mathbf x_n$ is Cauchy. Denote by $a^{[k]}$ the truncation of the sequence $a$ $$ a^{[k]}=\{a_1,a_2,a_3,\ldots,a_k,0,0,\ldots\}. $$ Then for $m,n\ge N$ we have for every fixed $k$ $$ \left|\|\mathbf x_n^{[k]}\|-\|\mathbf x_m^{[k]}\|\right|\le\|\mathbf x_n^{[k]}-\mathbf x_m^{[k]}\|\le\|\mathbf x_n-\mathbf x_m\|\le\epsilon. $$ Taking the limit for $m\to+\infty$ we get for $n\ge N$ $$ \left|\|\mathbf x_n^{[k]}\|-\|\mathbf y^{[k]}\|\right|\le\epsilon,\ \forall k, $$ that is $\|\mathbf x_n^{[k]}\|$ converges to $\|\mathbf y^{[k]}\|$ uniformly.

This is all you need if you exchange the limits for $$ \phi_n(k)=\|\mathbf x_n^{[k]}\|=\sqrt{f_n(k)},\qquad \phi(k)=\|\mathbf y^{[k]}\|=\sqrt{f(k)}. $$

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  • $\begingroup$ Your answer is very much in the same spirit as Daniel Fischer's, and both answers are very good. Thank you so much for helping put this to rest. $\endgroup$ – Alex Ortiz Dec 24 '16 at 23:28

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