Say $\{x_n\}$ is Cauchy in $\ell^p$ and $x$ is its pointwise limit. To argue that $x \in \ell^p$ would the following be correct:

Let $\varepsilon > 0$ and let $N$ be s.t. $n,m > N$ $\Rightarrow$ $|x_n - x_m|_p < \varepsilon$. Then $\lim_{m \to \infty} |x_n - x_m|_p = |x-x_n|_p \le \varepsilon$.

I saw the following different argument: $|x_n - x_m|_p < \varepsilon$ implies $\left(\sum_{k=0}^M |(x_n - x_m)_k|^p\right)^{1/p} < \varepsilon$ for all $M$ therefore $\lim_{m \to \infty}\left(\sum_{k=0}^M |(x_n - x_m)_k|^p\right)^{1/p} = \left(\sum_{k=0}^M |(x_n - x)_k|^p\right)^{1/p} \le \varepsilon$ for all $M$ therefore $\lim_{M \to \infty} \left(\sum_{k=0}^M |(x_n - x)_k|^p\right)^{1/p} \le \varepsilon$.

The difference is to use a finite sum step in between. Is it correct to drop it? And if not: why not? Norm seems to be continuous so one should be able to exchange norm and limit. Thanks.


Edit: The second argument is correct, not the first one. But that's not the only thing you have to prove. There are three steps. See the usual elementary proof below.

Now if you really want to swap some limits, here is what you can do. Fix $\epsilon>0$ and take $N$ such that $\|x_n-x_m\|_p\leq \epsilon$ for all $n,m\geq N$. Now fix $n\geq N$. And consider the sequence $y_m=x_n-x_m$ which converges pointwise to $x_n-x$, so that $\liminf y_m(k)=x_n(k)-x(k)$ for all $k$. We have $$ \|y_m\|_p\leq \epsilon\qquad\forall m\geq N \qquad\Rightarrow\qquad \liminf_m\|y_m\|_p\leq \epsilon. $$ Now by Fatou's lemma (which is easily proved in $\ell^p$), you have $$ \|x_n-x\|_p=\|\liminf_m y_m\|_p\leq \liminf_m \|y_m\|_p\leq \epsilon $$ for all $n\geq N$. So this proves everything at the same time.

Usual elementary proof:

  1. Find a pointwise limit $x$ by pointwise completeness. You've done that already.

  2. Prove $x$ belongs to $\ell^p$. You've done that also.

  3. Check that $x_n$ tends to $x$ for the $\ell^p$ norm.

ALthough the technique is essentially the same, one needs to treat 2 and 3 separately.

About 2. There is actually an $\epsilon$ free argument at this point, see the note at the end. Now to prove this the way you did, one usually picks $\epsilon=1$. Then there is $N$ such that for all $n,m\geq N$, $\|x_n-x_m\|_p\leq 1$. In particular, for all $K$ and all $m\geq N$: $$ \left(\sum_{k=1}^K|x_m(k)|^p\right)^{1/p}\leq \left(\sum_{k=1}^K|x_N(k)|^p\right)^{1/p}+\left(\sum_{k=1}^K|x_m(k)-x_N(k)|^p\right)^{1/p} $$ $$ \leq \|x_N\|_p+\|x_m-x_N\|_p\leq \|x_N\|_p+1. $$ Now let $m$ tend to $+\infty$. And finally $K$ to $+\infty$. You get $$ \|x\|_p\leq \|x_N\|_p+1 $$ so $x\in\ell^p$.

For 3., now. Fix $\epsilon>0$. Take $N$ such that $\|x_n-x_m\|_p\leq \epsilon$ for all $n,m\geq N$. Then for $n,m\geq N$ and all $K$: $$ \left(\sum_{k=1}^K|x_m(k)-x_n(k)|^p\right)^{1/p}\leq \|x_m-x_n\|_p\leq \epsilon. $$ Let $m$ tends to $+\infty$. Then $K$. You get $\|x-x_n\|_p\leq \epsilon$ for all $n\geq N$.

Note: Proof of 2 without $\epsilon$. Since $x_n$ is Cauchy, it is bounded, say by $M$. Then for all $K$ and all $n$: $$ \left(\sum_{k=1}^K|x_n(k)|^p\right)^{1/p}\leq \|x_n\|_p\leq M. $$ Now let $n$ tend to $=\infty$. And then $K$. You get $\|x\|_p\leq M$. So $x\in\ell^p$.

  • $\begingroup$ thank you! i know about the other steps... i had done the other steps and then realized i was not sure about this step. $\endgroup$ – user66372 Mar 12 '13 at 15:56
  • $\begingroup$ could you explain why the first argument doesn't work? it's really all i'm asking since i've done the other proofs. norm seems to be continuous so that i should be able to exchange limit and norm. $\endgroup$ – user66372 Mar 12 '13 at 16:01
  • $\begingroup$ @user66372 Your first argument says: $\lim_m \|x_n-x_m\|_p=\|x_n-x\|_p$. Yes the norm is continuous. But to use continuity, you need to know already that $x_m$ tends to $x$ for the norm. This requires to prove that $x$ is indeed in $\ell^p$ (point 2), and that $x_m$ tends to $x$ (point 3). $\endgroup$ – Julien Mar 12 '13 at 16:27
  • $\begingroup$ @user66372 If you really want to swap sum and limit, you need to consider limninf. I'll add such an argument. $\endgroup$ – Julien Mar 12 '13 at 16:32
  • $\begingroup$ @user66372 Note also that proving point 2 the way you did it afterwards, it is really more "elegant" to take $\epsilon=1$. Note also that the edit I gave below is an interesting alternative, if you know that every Cauchy sequence in a metric space is bounded. ANd that"s not difficult to prove. $\endgroup$ – Julien Mar 12 '13 at 16:39

It's not correct to drop the finite sum step. Your equation $$\lim_{m\to\infty} |x_n-x_m|_p = |x-x_n|_p$$ assumes already that $x_m\to x$ in $\ell^p$. The finite sum step is a trick to allow you to take a limit. (It's always OK to take a termwise limit of a finite sum, but it's not always OK for an infinite sum.)

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    $\begingroup$ wait could you say a wee bit more about why it already assumes $x_m \to x$? i thought for continuous $f$, $\lim_{n\to\infty} f(x_n) = f(\lim_{n \to \infty} x_n)$ and $|\cdot|_p$ is continuous. $\endgroup$ – user66372 Mar 12 '13 at 15:31
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    $\begingroup$ @user66372 Indeed that's true, and the only question is whether $\lim x_n = x$, and that's exactly what you're trying to prove! (Note $\lim$ here refers to the topology of $\ell^p$, not the pointwise topology.) $\endgroup$ – Sean Eberhard Mar 12 '13 at 16:17

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