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I'll preface the examples I am trying to understand with the definitions of weak and weak-* convergence in a Banach space $X$:

"A sequence $(x_n)_{n=1}^\infty$ converges weakly to $x$ in $X$ if for all $f\in X'$, $f(x_n)\to f(x)$ in $\mathbb F$ as $n\to\infty.$

A sequence $(f_n)_{n=1}^\infty\subset X'$ converges weak-* to $f$ in $X'$ if for all $x\in X$, $f_n(x)\to f(x)$ in $\mathbb F$ as $n\to\infty.$"

In my notes I am given the following two examples, although, I am having trouble following them completely. In all examples $(e_n)_{n=1}^\infty$ denotes the infinite sequence with zero in every entry apart from its $n$-th entry, which is one.

Example $1$:

Let $(e_n)_{n=1}^\infty\subset\ell^2$. Then $(e_n)_{n=1}^\infty$ does not converge strongly anywhere. But since $(\ell ^2)'$ is isometrically isomorphic to $\ell^2$ we can represent every $f\in(\ell^2)'$ as the action of some $a\in \ell^2$ on another $x\in\ell^2$. Letting $f:=f_a$ we have that then $f_a(x)=\sum_{k=1}^\infty a_kx_k$. We then have that $f(e_n)=a_n$. My notes then go on to say that if $a\in\ell^2$ then $a_n\to0$ as $n\to\infty$ and $x_n\rightharpoonup x$ in $\ell^2$.

It is at this last sentence that I am lost. I understand that $f(e_n)=a_n$, but how does the fact that $a\in\ell^2$ allow us to deduce that $a_n\to0$ as $n\to\infty$ and then in turn allow us to establish the weak convergence?

Example $2$:

Take $(e_n)_{n=1}^\infty\subset\ell^1$. Then for every $f\in(\ell^1)'$ we can define similarly to what we done above $f:=f_a$ for some $a\in\ell^\infty$. But then my notes say that $f_a(e_n)=a_n$ does not have to go to zero as $n\to\infty$. From which point it then goes on to say that it is not true that $e_n\rightharpoonup 0$ in $\ell^1$.

Again, similar to the last example, I am completely lost on the latter part of the example. Why in this case for $a\in\ell^\infty$ does $f_a(e_n)=a_n$ not have to converge strongly to zero as $n$ increases?

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As for 1. note that every sequence in $\ell_p$ ($p\in [1,\infty)$) converges to 0, hence the claim follows.

As for 2. take $f=((-1)^n))_{n=1}^\infty\in\ell_\infty$. Then $(f(e_n))_{n=1}^\infty$ fails to have a limit.

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  • $\begingroup$ Perhaps this is a fundamental misunderstanding on my part, but I thought that $(x_n)_{n=1}^\infty\in\ell^p\,\,\text{if}\,\,\left(\sum_{i=1}^\infty|x_i|^p\right)^{1/p}\lt\infty$. How does that necessarily mean that $(x_n)_{n=1}^\infty\in\ell^p$ converges to $0$? $\endgroup$ – Jeremy Jeffrey James Nov 22 '16 at 11:31
  • $\begingroup$ @Jeremy Jeffrey James, otherwise there would be $\delta > 0$ and an infinite set $A\subset \mathbb{N}$ such that $|x_n|>\delta$. In particular $\|(x_n)\|^p \geqslant \sum_{n\in A} |x_n|^p \geqslant \sum_{n\in A}\delta^p = \infty$. $\endgroup$ – Tomek Kania Nov 22 '16 at 12:08
  • $\begingroup$ Yes it is, thanks for clarifying! $\endgroup$ – Jeremy Jeffrey James Nov 22 '16 at 21:19

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