Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 106 / 5

Let $p_C(x)=\inf \{t>0:x\in tC\}$ be the Minkowski functional of $C$. Openness, convexity, and $0\in C$ make $p_C$ sublinear, with $p_C(x)<1$ for $x\in C$ and $p_C(x_0)\ge 1$. Define a linear functional on $\mathbb{R}{x}_0$ by $f(t{x}_0)=t{p}_C(x_0)$. The real Hahn-Banach theorem extends it to $X$ with $f\le p_C$. Hence

Apply this separation to the open ball of radius $\Vert x_0\Vert$ and rescale to obtain $f\in S_{X^{\ast }}$ with $f(x_0)=\Vert x_0\Vert$. For a closed subspace $Y$, apply it toThe separator must vanish on $Y$ because $C$ contains every translate along $Y$, and normalization gives

Let $F\subseteq X^{\ast }$ be finite-dimensional and $\Phi \in B_{X^{\ast \ast }}$. ConsiderIf $\Phi {|}_F$ did not belong to $S((1+\epsilon )B_X)$, finite-dimensional strict separation would produce some $f\in F$ withcontradicting $\Vert \Phi \Vert \le 1$. Thus there is $x\in X$ with $\Vert x\Vert <1+\epsilon$ and $f(x)=\Phi (f)$ for every $f\in F$.

Given a basic weak-star neighbourhood of $\Phi$ in $B_{X^{\ast \ast }}$, apply this result to the finite-dimensional span of its defining functionals and then replace $x$ by $x/(1+\epsilon )$. As $\epsilon \downarrow 0$, the resulting points of $J(B_X)$ enter that neighbourhood. Hence $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$, proving Goldstine theorem.

Identify $X$ with its canonical image in $X^{\ast \ast }$ and putFor $x\in S_X$ and $\lambda \in \mathbb{R}$, if $|\lambda |\le 1/(d+1)$ thenIf $|\lambda |\ge 1/(d+1)$, thenTherefore $d(x,\mathrm{span}\{\Phi \})\ge c$.

The restriction of $x\in X^{\ast \ast }$ to $\ker \Phi \subseteq X^{\ast }$ has normwhere the first equality is the Hahn-Banach distance formula and $(\ker \Phi {)}^{\perp }=\mathrm{span}\{\Phi \}$. Scaling from $S_X$ givesfor every $x\in X$. Hence $\ker \Phi$ is $c$-norming for $X$.