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.

  • $\begingroup$ 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

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 $f$.

  • $\begingroup$ +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
  • 1
    $\begingroup$ 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

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 $$ \phi_n(k)=\|\mathbf x_n^{[k]}\|=\sqrt{f_n(k)},\qquad \phi(k)=\|\mathbf y^{[k]}\|=\sqrt{f(k)}. $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.