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Given the the sequence $(e_n)_n$ in $l^\infty$, I want to show that that $e_n$ converges weakly to $0$ in $l^\infty$, i.e. $$e_n\rightharpoonup 0 \text{ as } n\to \infty.$$ By $e_n\in l^\infty$, I mean the sequence $e_n^{(m)}=\delta_{m,n}$.

Should I try to show this by looking at the dual of $l^\infty$ which is not trivial, or is there another way?

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  • $\begingroup$ I think, it actually proves that it converges. See my answer. $\endgroup$
    – Peter
    Nov 17 '14 at 17:01
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Alternatively and directly:

Assume by contradiction that $(e_n)_{n}$ does not converge weakly to zero. Then there exists a $\epsilon >0$ and a functional $f\in l_\infty^\ast$ s.t. $|f(e_n)|\geq \epsilon$ for infinitely many $n\in\mathbb{N}$. By passing to that subsequence, we have that $|f(e_{n_k})|\geq \epsilon$ for all $k\in\mathbb{N}$. Let $\lambda_k :=\text{sign } f(e_{n_k})$ and set $x_N:=\sum_{k=1}^{N} \lambda_k e_{n_k}\in l_\infty$, we have that $\|x_N\|_\infty=1$ and $|f(x_N)|= \sum_{k=1}^N |f(e_{n_k})|\geq N\epsilon$.

Since we can do that for every $N\in\mathbb N$, we get a contradiction to the fact that $f\in l_\infty^\ast$.

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  • $\begingroup$ I corrected the typo. You are right it should have said $f\in \ell_\infty^\ast$ as opposed to $\ell_1^\ast$. Thanks! Yes, the unboundedness is the contradiction. $\endgroup$
    – Peter
    Dec 14 '17 at 6:24
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From here we know that $(e_n)_n$ converges weakly to $0$ iff it is bounded and every subsequence converges quasi-uniformly to $0$.

Clearly, $(e_n)_n$ is bounded and pointwise converging to $0$. Given a subsequence $(e_{n_k})_k$, $\epsilon>0$ and $n_0\in \mathbb{N}$. Set $\alpha_1 = {n_0+1}$ and $\alpha_2={n_0+2}$. Then for any $m\in\mathbb N$ we have $$\min_{i={1,2}}e_{n_{n_0+i}}(m)=\min_{i={1,2}} (\delta_{n_0+i,m})=0<\epsilon.$$

Hence $e_n \to 0$ weakly as $n\to\infty$.

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