Let $V$ be a normed, real vector space and let $B = \{v \in V : ||v|| \leq 1\}$. Consider all functionals on $B$ which are bounded by $[-1,1]$, where $[-1,1]$ has the subspace topology inherited from $\mathbb{R}$: $[-1,1]^B$. Let $B^*$ be the subspace of all linear functionals in $[-1,1]^B$; that is, for every $v,u \in B$ such that $v + u \in B$, the projection of $f \in B^*$ from along each vector is additive: $\pi_{v+u}(f) = \pi_v(f) + \pi_u(f)$. Furthermore, we also have the additional property of linearity that if $u \in B$ and $\alpha \in \mathbb{R}$ such that $\alpha u \in B$, then we have $\pi_{\alpha u}(f) = \alpha \pi_u(f)$. From Tychonoff's Theorem, we know that $[-1,1]^B$ is a compact space. We would like to prove that $B^*$ is also compact.
The strategy is to show that $B^*$ is a closed subspace of $[-1,1]^B$ and hence is compact. I have been struggling to show this. My approach is to show that $B^*$ contains all its boundary points, but that is proving difficult. I have so far managed to show that $B^*$ is not an open set, if that helps.
Best Answer
For fixed $u,v \in V$ consider the set
$L(u,v) = \{ f \in \mathbb{R}^V: \pi_u(f) + \pi_v(f) = \pi_{u+v}(f)\}$. Letting $p: \mathbb{R}^V \to \mathbb{R}^3$ be the map sending $f$ to $(f(u), f(v), f(u+v))$, which is continuous as a "triple projection", note that $L(u,v) = p^{-1}[H]$ where $H$ is the hyperplane $\{((x,y,z) \in \mathbb{R}^3: x+ y -z= 0\}$, which is closed in $\mathbb{R}^3$, so $L(u,v)$ is always closed.
Similarly, for each scalar $\alpha \in \mathbb{R}$ and $u \in V$ the set $M(\alpha, u) = \{f \in \mathbb{R}^V: \pi_{\alpha u}(f) = \alpha \pi_u(f)\}$ can be seen as $p'^{-1}[L]$ where $p': \mathbb{R}^V \to \mathbb{R}^2$ is given by $p'(f) = (f(\alpha u), f(u))$ and $L = \{(x,y) \in \mathbb{R}^2: x = \alpha y \}$ is closed. So all $M(\alpha, u)$ are closed too.
Now, denoting by $\pi_B$ the restriction map from $\mathbb{R}^V$ to $\mathbb{R}^B$, $$B^\ast = \pi_B\left[ [0,1]^V \cap \bigcap\{ L(u,v): u,v \in B, u+v \in B\} \cap \bigcap \{M(\alpha,u): u \in B, \alpha u \in B\} \right]$$ which is closed in $[0,1]^B$, as the set inside the $[]$ is compact (closed in $[0,1]^V$); intersections of closed sets are closed after all.
The crux is seeing "being a functional" as having to obey a lot of "closed conditions".