Given: Let $S$ be an infinite set, and let $\lbrace s_n \rbrace_{n=1}^{\infty}$ be a sequence of distinct points in $S$. Let $X$ be a Banach space of bounded functions on $S$, supplied with the supremum-norm. Suppose that, for all choices $s_{1}^{'}, \cdots , s_{k}^{'}$ of finitely many different elements of $\{ s_n : n = 0, 1, 2, \cdots \}$, and for all choices $\alpha_1, \cdots , \alpha_k$ of scalars such that $|\alpha_i| = 1$$(1 \leq i \leq k)$, there exists an element $f$ of $X$ such that $||f||_{\infty} \leq 1$ and $f(s_{i}^{'}) = \alpha_i (1 \leq i \leq k)$.
I need some help with the following questions;
- If $x = (x_1, x_2, \cdots) \in \ell^{1}$, then show that $\phi_x (f):= \sum_{n=1}^{\infty} x_n f(s_n)$ is a bounded linear functional on $X$.
- Show that the map $x \rightarrow \phi_x$ is an isometric embedding of $\ell^{1}$ into $X^{'}$.
- Show that $X$ is not reflexive.
- Show that $C[a, b]$ is not reflexive $(a < b)$.
I can answer the first two points for you as a hint, and I'll give a direction for the third point.
The linearity of $\phi_x$ and of the map $x\to\phi_x$ both follow directly from their definitions. For boundedness of $\phi_x$: $$\|\phi_x(f)\|\le|\sum_n x_n|\cdot\|f\|\le\sum_n |x_n|\cdot\|f\|=\|x\|_1\cdot\|f\|$$ via the triangle inequality. Note that this implies $\|\phi_x\|\le\|x\|_1$. So we only need to show $\|\phi_x\|\ge\|x\|_1$.
We can use the supposedly existing function from your question; for each $k$ we choose $s'_i=s_i$ and $\alpha_i=1/\text{sgn}(x_i)$, get an $f_k\in X$ as you described and observe:
$\sum_{i=1}^k x_i f_k(s_i)=\sum_{i=1}^k x_i\alpha_i=\sum_{i=1}^k|x_i|$.
This expression goes to $\|x\|_1$ as $k\to\infty$, and the set $\{|\sum_{i=1}^k x_i f_k(s_i)|:k\in\mathbb{N}\}$ is a subset of $\{|\sum_{i=1}^\infty x_i f(s_i)|:\|f\|\le1\}$, so the other inequality follows.
For the third point, if we denote $x\mapsto\phi_x$ with $\Phi$, we can use that $X$ reflexive $\iff$ $X'$ reflexive, and that if $\Phi(\ell^1)$ is a closed subspace of $X'$, and not reflexive, then $X'$ is not reflexive.