I was trying to solve an exercise for my class, but then I have found somewhere a solution. I need to understand the meaning of a certain step. The exercise and the solution read as follow.
Exercise Consider the space $l^\infty:=\{x\in\mathbb{R}\ :\ \sup|x(n)|<\infty\}$ equipped with the usual $\sup$ norm. Let $(\delta_n)_n\subset (l^\infty)'$ be defined as $ \langle\delta_n,x\rangle:=x(n).$ Show that it does not exist any subsequence of $(\delta_n)_n$ weakly convergent in $(l^\infty)'$.
Solution Take a subsequence $(\delta_{h_k})_k$ of $(\delta_n)_n$, where $(h_k)$ is an increasing sequence of positive integers. By definition we have:
$$ \delta_{h_k} \rightharpoonup \delta\iff\langle f,\delta_{h_k}\rangle\rightarrow\langle f,\delta\rangle, \ \ \ \ \ \ \forall \ f\in(l^{\infty})'' \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
$(*)$ By duality the funcion $\langle\cdot,x\rangle:(l^\infty)'\rightarrow\mathbb{R}$ is a continuous and linear functional for any $x\in l^\infty$, thus the previous implications reads:
$$ \delta_{h_k} \rightharpoonup \delta\Rightarrow\langle \delta_{h_k},x\rangle\rightarrow\langle \delta,x\rangle, \ \ \ \ \ \ \forall \ x\in l^{\infty}. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$$
Now define the following sequence $$x(n)= \begin{cases} (-1)^k \ \ \ & \text{if} \ n=h_k\\ 0 & \text{otherwise}\end{cases}$$ Then $\langle \delta_{h_k},x\rangle=x(h_k)=(-1)^k$ does not converge, then the subsequence $\delta_{h_k}$ cannot weakly converge.
Question My problem is with the sentence labelled with $(*)$. Can someone help me to understand why we can pass from $(1)$ to $(2)$. Also, why we need $(h_k)$ to be an increasing sequence of positive integers? Thank you very much.
What needs to be shown is that given the subsequence $(\delta_{h_k})$ there is a functional $f\in X''$ such that $(f(\delta_{h_k}))$ does not converge. Now let $x$ be the sequence defined under $(2)$. Then $\hat x:X'\rightarrow \mathbb{K},x'\mapsto x'(x)$ is a bounded linear functional and $\hat x(\delta_{h_k})$ doesn't converge, so $(\delta_{h_k})$ doesn't converge weakly.