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See the suggested proof here. The thing I don't understand is that the functional $f\in(\ell^1)^*$, corresponding to a sequence $y\in\ell^\infty$ depends on the index $j$ in its definition. So let $f_j$. Then David Bowman shows that $(|f_j(x^{N_j})|)_{j=1}^{\infty}$ is strictly bounded away from zero for each $j\in\mathbb{N}$. Why does this contradict that $(x^{N_j})_{j=1}^{\infty}$ convergence weakly to $0$?

Edit: It looks like I just misread the definition. It does not depend on $j$.

sum_math
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  • If $(x^{N_j})_{j\in\mathbb{N}}$ converges weakly to $0$ in $\ell^1$, then $f(x^{N_j})$ converges to $f(0) = 0$ for every $f\in(\ell^1)^\ast$ by definition of weak convergence. Does this help? – stange Oct 31 '23 at 14:18
  • The issue was resolved when I realized that $f$ indeed is not dependent on $j$. – sum_math Oct 31 '23 at 16:49

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