Let $X$ be a Banach space (not necessarily reflexive) and let $X^{*}$ denote the continuous dual of $X$. Let $\psi:X \to X^{**}$ denote the canonical embedding of $X$ into its double dual $X^{**}$. We know that the weak$^*$ topology on $X^*$ is induced by the family of continuous linear functionals $\{\psi(x): x\in X\}$.
From this, can we say that every weak$^*$ continuous linear functional on $X^*$ is of the form $\psi(x)$ for some $x\in X$? If it is, I would like to know an explanation of the answer. If not then a counter example would help. I am a beginner on this domain and know very little. So, I do not know, if this question is trivial.
Thanks in advance.
Let $F$ be a continuous linear functional on $(X^{*}, weak*)$. There exists a basic neighborhood of $0$ in weak* topology, say $\{f\in X^{*}: |f(x_i)| <r_i, 1 \leq i \leq N\}$ such that $|F(f)| <1$ whenever $f$ lies in this neighborhood. In particular, this implies that if $f(x_i)=0$ for each $i$ then ($nf$ is in the neghborhod for each $n \in \mathbb N$ so) $|F(nf)| <1$ for all $n$. This proves that $\bigcap_i ker (F_{x_i}) \subseteq ker F$, where $F_x(f)\equiv f(x)$. An elementary lemma in linear algebra now says that $F$ is a linear combination of $F_{x_1},F_{x_2},,...,F_{x_N}$ so $F=F_x$ for some $x$ of the form $\sum c_ix_i$and this is exactly what we are trying to prove.