Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

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 $K:=\max\{N_k,1\leq k\leq n$.

share|improve this answer
Thank you so much! –  henryforever14 May 1 '12 at 20:17
You are welcome! –  Davide Giraudo May 1 '12 at 20:25
add comment

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.

share|improve this answer
Thanks! This does seem like a silly question. –  henryforever14 May 1 '12 at 20:16
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.