Goldstine theorem

Lemma

For all ${\displaystyle x^{\prime \prime }\in B^{\prime \prime },}$ ${\displaystyle \varphi _{1},\ldots ,\varphi _{n}\in X^{\prime }}$ and ${\displaystyle \delta >0,}$ there exists an ${\displaystyle x\in (1+\delta )B}$ such that ${\displaystyle \varphi _{i}(x)=x^{\prime \prime }(\varphi _{i})}$ for all ${\displaystyle 1\leq i\leq n.}$

Proof of theorem

Fix ${\displaystyle x^{\prime \prime }\in B^{\prime \prime },}$ ${\displaystyle \varphi _{1},\ldots ,\varphi _{n}\in X^{\prime }}$ and ${\displaystyle \epsilon >0.}$ Examine the set $${\displaystyle U:=\left\{y^{\prime \prime }\in X^{\prime \prime }:|(x^{\prime \prime }-y^{\prime \prime })(\varphi _{i})|<\epsilon ,1\leq i\leq n\right\}.}$$

Let ${\displaystyle J:X\rightarrow X^{\prime \prime }}$ be the embedding defined by ${\displaystyle J(x)={\text{Ev}}_{x},}$ where ${\displaystyle {\text{Ev}}_{x}(\varphi )=\varphi (x)}$ is the evaluation at ${\displaystyle x}$ map. Sets of the form ${\displaystyle U}$ form a base for the weak* topology, so density follows once it is shown ${\displaystyle J(B)\cap U\neq \varnothing }$ for all such ${\displaystyle U.}$ The lemma above says that for any ${\displaystyle \delta >0}$ there exists a ${\displaystyle x\in (1+\delta )B}$ such that ${\displaystyle x^{\prime \prime }(\varphi _{i})=\varphi _{i}(x),}$ ${\displaystyle 1\leq i\leq n,}$ and in particular ${\displaystyle {\text{Ev}}_{x}\in U.}$ Since ${\displaystyle J(B)\subset B^{\prime \prime },}$ we have ${\displaystyle {\text{Ev}}_{x}\in (1+\delta )J(B)\cap U.}$ We can scale to get ${\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B).}$ The goal is to show that for a sufficiently small ${\displaystyle \delta >0,}$ we have ${\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B)\cap U.}$

Directly checking, one has $${\displaystyle \left|\left[x^{\prime \prime }-{\frac {1}{1+\delta }}{\text{Ev}}_{x}\right](\varphi _{i})\right|=\left|\varphi _{i}(x)-{\frac {1}{1+\delta }}\varphi _{i}(x)\right|={\frac {\delta }{1+\delta }}|\varphi _{i}(x)|.}$$

Note that one can choose ${\displaystyle M}$ sufficiently large so that ${\displaystyle \|\varphi _{i}\|_{X^{\prime }}\leq M}$ for ${\displaystyle 1\leq i\leq n.}$ Note as well that ${\displaystyle \|x\|_{X}\leq (1+\delta ).}$ If one chooses ${\displaystyle \delta }$ so that ${\displaystyle \delta M<\epsilon ,}$ then $${\displaystyle {\frac {\delta }{1+\delta }}\left|\varphi _{i}(x)\right|\leq {\frac {\delta }{1+\delta }}\|\varphi _{i}\|_{X^{\prime }}\|x\|_{X}\leq \delta \|\varphi _{i}\|_{X^{\prime }}\leq \delta M<\epsilon .}$$

Hence one gets ${\displaystyle {\frac {1}{1+\delta }}{\text{Ev}}_{x}\in J(B)\cap U}$ as desired.