Uniform convergence and exchange of limits (for series)

Question

While reading the proof of the result that the sequence space $\ell^2$ is complete, I couldn't justify the exchange of limits that occured.

For any Cauchy sequence $\{x_n\}$ in $\ell^2$ we obtain a candidate limit by taking the pointwise limit of each coordinate, which we call $\tilde{x}$.

$$ \sum_{k=1}^{\infty}|\tilde{x}(k)|^2 = \sum_{k=1}^{\infty} |\lim_{n \to \infty} x_n(k)|^2 = \lim_{n \to \infty} \sum_{k=1}^{\infty} |x_n(k)|^2 = \lim_{n \to \infty} || x_n ||^2. $$ The exchange of limits is justified, since the convergence of $\lim_{n \to \infty} x_n(k)$ is uniform over $k$.

I can't find the result which justifies the exchange of limits. I know of three similar results:

1. Given a double sequence $f_{n,m}$ such that $\lim_n f_{m,n}\to f_m$ uniformly on $m$, then the iterated limit $\lim_n \lim_n f_{m,n}$ exists and equals $\lim_{m}\lim_n f_{m,n}$.

2. Discrete version of dominated convergence thm

3. A result on dominated convergence and uniform convergence

Problem with using 1): We know that $x_n\to\tilde x$, but it seems to me we would need to prove that $\sum_{i=1}^k \vert x_n(i)\vert^2$ converges to $\sum_{i=1}^k \vert \tilde x(i)\vert^2$ uniformly in $k$. Convergence is clear by additivity of limits, but why this convergence would be uniform in $k$ is not clear to me.

Problem with using 2): I would have to find a sequence $y(k)$ such that $\sum \vert y(k)\vert^2<\infty$ and such that $\vert x_n(k)\vert\leq \vert y(k)\vert$. Not sure how to go about that either.

Problem with using 3): The space $\mathbb N$ doesn't have finite measure.

Answer 1

Ok, sorry for my nebulous answer earlier. Another try:

Let us first prove that $\tilde{x}$ is in $\ell^2$. I assume that you know that (and why) $\lim_{n \rightarrow \infty} \lVert x_n - \tilde{x} \rVert_{\ell^2} = 0$. Indeed, for every $n \in \mathbb{N}$: $$ \lVert \tilde{x} \rVert_{\ell^2} \leq \lVert x_n - \tilde{x}\rVert_{\ell^2} + \lVert x_n \rVert_{\ell^2} $$ If we choose $\bar n$ so large that $\lVert x_{\bar n} - \tilde{x} \rVert_{\ell^2} < 1$, we get: $$ \lVert \tilde{x} \rVert_{\ell^2} \leq 1 + \lVert x_{\bar n} \rVert_{\ell^2} $$ So indeed, $\tilde{x} \in \ell^2$.

Hence, we can use the triangle inequality to get $$ \lvert \lVert x_n \rVert_{\ell^2} - \lVert \tilde{x} \rVert \rvert \leq \lVert x_n - \tilde{x} \rVert_{\ell^2} \rightarrow 0\text{ as }n \rightarrow \infty. $$ So $\lVert x_n \rVert_{\ell^2} \rightarrow \lVert \tilde{x} \rVert_{\ell^2}$ as $n \rightarrow \infty$. So exchanging limits as you want to do it is not really necessary.

Still, if you want to exchange limits, dominated convergence would probably be the theorem of choice. But as you already stated, there is no reason why such bound $y$ should exist.

As it pertains to uniform convergence, this is probably not enough. At least it is only clear, that $\ell^2$-convergence gives uniform convergence but not the other way round.

Answer 2

Thanks to the post linked to by Michh I see now that the statement

If $\lim_n f_n(m)=f(m)$ converges uniformly in $m$, then $\lim_n \sum_m f_n(m)=\sum_m f(m).$

is false. This means that the proof I was reading was not valid (the exchange of limits is correct, but the justification for it is not). That is okay, because an alternative proof is possible, and even quite straightforward, as pointed out by Meowdog.

I will write out the counter-exmple to the claim above (by MathematicsStudent1122):

Let $f_n(m)=1/n$ if $1\leq m\leq n$ and $f_n(m)=0$ if $m>n$. Then $$ \lim_n f_n(m)=\lim_n 1/n=0, $$ and this convergence is uniform in $m$, as can be seen from the inequality $\vert f_n(m)-0\vert\leq 1/n$ which holds for all $m,n$. Now note that $$ \sum_m\lim_n f_n(m)=\sum_m 0=0 $$ while $$ \lim_n \sum_m f_n(m)=\lim_n\sum_{m=1}^n 1/n=\sum_n 1/n=\infty. $$ Therefore $\lim_n \sum_m f_n(m)\neq \sum_m f(m)$.