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Let $1\le p<\infty$, and let $U$ be the open unit ball in $\ell^p$. Let $\{x_k\}\subseteq\overline U$ be a sequence such that for every $n$ the limit $x(n):=\lim_{k\to\infty}x_k(n)$ exists. Can I then conclude that $x\in\overline U$?

Now I've managed to show using the proof here that $x_n$ converges weakly to $x$.

Then theorem 3.12 in Rudin's Functional Analsyis tells me that the norm closure of $U$ is equal to the weak closure of $U$, so it follows that $x\in\overline U$.

But using that proof feels like overkill so I wonder if there's a simpler way.

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    For fixed $N$, consider $$\sum_{n = 0}^N \lvert x_k(n)\rvert^p.$$ – Daniel Fischer Nov 01 '15 at 21:17
  • Well it's easy to see that $\lim_{k\to\infty}\sum_{n=0}^N\vert x_k(n)\vert^p=\sum_{n=0}^N\vert x(n)\vert^p$, but that's all I can see.

    I'm really not getting it.

     –  Nov 01 '15 at 23:31
  • Can you characterise $\overline{U}$ in terms of a similar-looking sum? – Daniel Fischer Nov 01 '15 at 23:33
  • Well for $y\in\overline U$ we have $\sum\vert y(n)\vert^p\le 1$, so I think I see what you're getting at: if we let $N\to\infty$ then $\sum_{n=0}^N|x(n)|^p$ is strictly increasing and from above we see that it's always $\le 1$ so it follows the limit as we let $N\to\infty$ exists and is $\le 1$. Is that what you mean or have I just written down some nonsense? –  Nov 01 '15 at 23:41
  • The "strictly" is wrong. We can have $x(n) = 0$ for some or even all $n$. Apart from that, it's what I was getting at. – Daniel Fischer Nov 01 '15 at 23:43
  • Ah nice. Thank you for both your help and your patience. –  Nov 01 '15 at 23:44

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A short proof is to apply Fatou's lemma to $\mathbb{N}$ with counting measure. It yields $$\sum_n|x(n)|^p \le \liminf_{k\to\infty} \sum_n|x_k(n)|^p \le 1$$ as desired.

For a self-contained proof one does what Daniel Fischer suggested. To prove $\|x\|_p\le 1$, it suffices to show that $\sum_{n=1}^N|x(n)|^p\le 1$ for every $N$; and the latter follows from coordinate-wise convergence.