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Let $\{x^k\}$ be a weak convergent sequence in $\ell_1$, and its weak limit is 0. Is the following property true:

For $\forall \epsilon >0$ and $\forall n>0$, there exists a K, s.t. $$\sum_{i=1}^{n}|x^K_i|<\epsilon.$$

This comes from a proof I read that tries to prove the equivalence of weak and strong convergence in $\ell_1$. It uses this property which I don't know why.

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2 Answers 2

up vote 1 down vote accepted

Fix $\varepsilon$ and $n$. For $k$ integer, the map $l_k\colon\{x_j\}_j\mapsto x_k$ is a linear continuous functional, so by weak convergence you can find $N_k$ such that for $j\geq N_K$, $|l_k(x^j)|=|x_k^{j}|\leq \varepsilon/n$. Now take for example $K:=\max\{N_k,1\leq k\leq n$.

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Thank you so much! –  henryforever14 May 1 '12 at 20:17
    
You are welcome! –  Davide Giraudo May 1 '12 at 20:25

For each positive integer $i$, since $(x_j)$ converges weakly to $0$, we have $\lim\limits_{j\rightarrow\infty} e_i(x_j)=0$, where $e_i$ is the standard $i^{\rm th}$ unit vector in $\ell_\infty$. That is, the coordinates of $x_j$ converge to $0$. So select $K$ so large that each of the first $n$ coordinates of $x_K$ are less than $\epsilon/n$ in absolute value.

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Thanks! This does seem like a silly question. –  henryforever14 May 1 '12 at 20:16

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