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Let $(u_n)_n\subset \ell^p(J)$ and $||u_n||_p<c$ for all $n \in \mathbb{N}$. Let $u\in \ell^p(J)$ so that $u_{n,j} \rightarrow u_j$ for all $j \in J$.

I want to show that $(u_n)_n$ converges weakly to u. I know that $(\ell^p(J))'\cong \ell^q(J)$ via $$f_v(u)=\sum\limits_{j\in J}v_j u_j$$ for $v\in \ell^q(J)$ where $\frac{1}p+\frac{1}{j}=1$. Now $\ell^p(J)$ should be reflexive and thus $B_c(0)\subset \ell^p(J)$ weakly compact. So i get a weakly convergent partial sequence $(u_{n_k})_k$. It's easy to so that $u_{n_k}$ weakly converges to $u$. Now I am stuck showing that this is true for the whole sequence.

Thank you

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Let $c_{00}$ denote the set of sequences with only finitely many non-zero terms. If $u_{n,j} \to u_j$, then show that for each $f \in c_{00} \subset \ell^q(J) \cong \ell^p(J)'$, $$ f(u_n) \to f(u) $$ Now use the fact that $c_{00}$ is norm-dense in $\ell^q(J)$ (and the fact that the $\{u_n\}$ are uniformly bounded) to conclude the same for all $f\in \ell^q(J)$.

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This can be handled with a well-known subsequence-subsequence lemma:

Let $\{x_n\} \subset X$ be a sequence in the topological space $X$. Suppose there is $x \in X$ with the property that every subsequence $\{x_{n_k}\}$ of $\{x_n\}$ has a subsequence $\{x_{n_{k_l}}\}$ which converges towards $x$. Then, $\{x_n\}$ converges towards $x$.

You can invoke this lemma with the weak topology in $\ell^p(J)$.

The proof of this lemma is quite simple, and I leave it to you.

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