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consider the Banach space $\ell^1$, and let $e_i$ be the sequence $(0,\dots,1,0\dots)$, with $1$ in the $i$-th position. show that $\{e_i\}$ converges weakly to $0$ in $\ell^1$ but not strongly.

My attempt : I found the answer here but i don't understand the answer

A sequence $(x_n)$ in a normed space $X$ is said to be weakly convergent if there is an $x \in X$ such that $$\lim_{n \to \infty}f(x_n)=f(x)$$ where $ f \in X'$( dual space of $X$)

Take $e^{(n)}\in\ell_1$ given by $e^{(n)}_i:=\begin{cases}1&\mbox{ if }n=i;\\\ 0&\mbox{otherwise.} \end{cases}$ Then $d\left(e^{(m)},e^{(n)}\right)=2$ if $m\neq n$

This implies sequence $\{e_i\}$ has no convergent subsequence

Thus $\{e_i\}$ is not strongly convergent

$l^1$ has infinitely linearly independent set, so we can write $x=a_1e_1+ a_2e_2+.....$

now $f : \ell^1 \to \mathbb{F}$ define by $$f(x) = f(a_1e_1+a_2e_2 +...)= af(e_1)+a_2f(a_2).....=\sum_{i=1}^{\infty}a_if(e_i)$$

since $\ell^{\infty}$ is the dual space of $\ell^1 $ so $f(x) \in \ell^{\infty}$

My confusion is that how to show $\lim_{n \to \infty}f(x_n)=f(x)?$

Arctic Char
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jasmine
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1 Answers1

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Your sequence does not converge to $0$ weakly. For example, let $f\in( \ell^1)^*$ be given by

$$f((a_i)_{i=1}^\infty)=\sum_{i=1}^\infty a_i.$$

Then $f(e^{(n)})=1$ for all $n$, but $f(0)=0$.

Arctic Char
  • 16,007