Let $k$ be a discrete field and $V$ an infinite dimensional $k$-vector space equipped with the discrete topology. Let $V'$ denote the continuous dual of $V$, and $V''$ its continuous double dual (both equipped with the corresponding weak-*-topology).
I believe that $V$ is isomorphic to $V''$. Does someone know a simple direct proof? I guess one could use the map $T_{\cdot} \colon V \to V''$, $v \mapsto T_v$, where $T_v \colon V' \to k$, $T_v(\phi) := \phi(v)$. Since $T_{\cdot} \colon V \to V''$ is linear and injective, one would only have to show that
i) $V''$ is discrete (which would turn $T_{\cdot}^{-1} \colon V'' \to V$ continuous), and
ii) $T_{\cdot} \colon V \to V''$ is surjective.
Can someone help me?
For $S\subseteq V$ denote $S^\perp=\{\alpha\in V':\alpha(s)=0\text{ for all }s\in S\}$.
For $\mathscr{S}\subseteq V'$ denote $\mathscr{S}^\perp=\{s\in V:\alpha(s)=0\text{ for all }\alpha\in \mathscr{S}\}$.
Step 1: If $S_1\subseteq \operatorname{span}(S_2)$ then $S_2^\perp\subseteq S_1^\perp$.
Proof: If $\alpha\in S_2^\perp$ then $\alpha(s)=0$ for all $s\in S_2$. Every $t\in S_1$ is a linear combination of the elements of $S_2$, so $\alpha(t)=0$. This shows $\alpha\in S_1^\perp$.
Step 2: The collection $\{F^\perp:F\subseteq V, F\text{ finite}\}$ is a neighborhood base at zero for the topology of $V'$.
Proof: The definition of the topology of $V'$ is that a neighborhood base at zero has the form $N_{F,U}=\bigcap_{s\in F}\{\alpha\in V':\alpha(s)\in U\}$ for some fixed finite $F\subseteq V$ and $U\subseteq k$ such that $0\in U$. Taking $U=\{0\}$ we get that $F^\perp=N_{F,\{0\}}$ is an open neighborhood of zero in $V'$. Moreoever for any such $U$, $0\in F^\perp\subseteq N_{F,U}$, so the $F^\perp$ with $F$ ranging over finite subsets of $V$ form a neighborhood base at zero for $V'$.
Let $\mathscr{B}$ be a Hamel basis for $V$ and $\mathscr{B}^*=\{e^*:e\in\mathscr{B}\}$ be the dual set. This is defined by $e^*(e)=1$ and $e^*(f)=0$ for $f\in\mathscr{B}\setminus\{e\}$.
Step 3: The collection $\{F^\perp:F\subseteq \mathscr{B}, F\text{ finite}\}$ is a neighborhood base at zero for the topology of $V'$.
Proof: If $F_1\subseteq V$ is finite, there is some finite $F_2\subseteq\mathscr{B}$ such that $F_1\subseteq\operatorname{span}(F_2)$. By Step 1, $0\in F_2^\perp\subseteq F_1^\perp$.
Step 4: $T_{\cdot}\colon V\to V''$ is surjective.
Proof: Let $\gamma\in V''$. Since $\gamma$ is continuous, there is some finite $F\subseteq\mathscr{B}$ such that $\gamma(F^\perp)=\{0\}$. For $e\in F$ let $\lambda_e=\gamma(e^*)$. Let $s_\gamma=\sum_{e\in F}\lambda_e e\in V$. We'll show that $T_{\cdot}(s_\gamma)=\gamma$. Let $\alpha\in V'$. Let $\alpha'=\alpha-\sum_{e\in F}\alpha(e)e^*$. For $e\in F$, $\alpha'(e)=\alpha(e)-\sum_{f\in F}\alpha(f)f^*(e)=\alpha(e)-\alpha(e)=0$. This shows that $\alpha'\in F^\perp$. Therefore $\gamma(\alpha')=0$. Now $\gamma(\alpha)=\gamma(\alpha')+\sum_{e\in F}\alpha(e)\gamma(e^*)=\sum_{e\in F}\alpha(e)\lambda_e=T_{\cdot}(s_\gamma)(\alpha)$. Since $\alpha\in V'$ was arbitrary, it follows that $T_{\cdot}(s_\gamma)=\gamma$.
Step 5: The collection $\{\mathscr{F}^\perp:\mathscr{F}\subseteq V', \mathscr{F}\text{ finite}\}$ is a neighborhood base at zero for the topology of $V$ induced from $V''$ (i.e. the weak topology on $V$ from $V'$).
Proof: This is similar to Step 2, but in the other direction.
Step 6: The collection $\{W\subseteq V:W\text{ is a subspace of finite codimension}\}$ is a neighborhood base at zero for the topology of $V$ induced from $V''$.
Proof: $\mathscr{F}^\perp$ for $\mathscr{F}\subseteq V$, $\mathscr{F}$ finite is a subspace of $V$ of finite codimension. Exercise: prove that any subspace of $V$ of finite codimension has this form for some $\mathscr{F}$ a finite set of linearly independent functionals.