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 lemma

By the surjectivity of

$${\displaystyle {\begin{cases}\Phi :X\to \mathbb {C} ^{n},\\x\mapsto \left(\varphi _{1}(x),\cdots ,\varphi _{n}(x)\right)\end{cases}}}$$

it is possible to find ${\displaystyle x\in X}$ with ${\displaystyle \varphi _{i}(x)=x^{\prime \prime }(\varphi _{i})}$ for ${\displaystyle 1\leq i\leq n.}$

Now let

$${\displaystyle Y:=\bigcap _{i}\ker \varphi _{i}=\ker \Phi .}$$

Every element of ${\displaystyle z\in (x+Y)\cap (1+\delta )B}$ satisfies ${\displaystyle z\in (1+\delta )B}$ and ${\displaystyle \varphi _{i}(z)=\varphi _{i}(x)=x^{\prime \prime }(\varphi _{i}),}$ so it suffices to show that the intersection is nonempty.

Assume for contradiction that it is empty. Then ${\displaystyle \operatorname {dist} (x,Y)\geq 1+\delta }$ and by the Hahn–Banach theorem there exists a linear form ${\displaystyle \varphi \in X^{\prime }}$ such that ${\displaystyle \varphi {\big \vert }_{Y}=0,\varphi (x)\geq 1+\delta }$ and ${\displaystyle \|\varphi \|_{X^{\prime }}=1.}$ Then ${\displaystyle \varphi \in \operatorname {span} \left\{\varphi _{1},\ldots ,\varphi _{n}\right\}}$^[1] and therefore

$${\displaystyle 1+\delta \leq \varphi (x)=x^{\prime \prime }(\varphi )\leq \|\varphi \|_{X^{\prime }}\left\|x^{\prime \prime }\right\|_{X^{\prime \prime }}\leq 1,}$$

which is a contradiction.

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,^[2] 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 ${\displaystyle 1\leq i\leq n.}$^[3] 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.