Goldstine theorem

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

By the surjectivity of

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

we can find ${\displaystyle x\in X}$ with ${\displaystyle \varphi _{i}(x)=x''(\varphi _{i})}$ for ${\displaystyle 1\leq i\leq n}$.

Now let

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

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

Assume for contradiction that it is empty. Then dist(x, Y) ≥ 1 + δ and by the Hahn–Banach theorem there exists a linear form φ ∈ X ′ such that φ|_Y = 0, φ(x) ≥ 1 + δ and ||φ||_X ′ = 1. Then φ ∈ span{φ₁, ..., φ_n} [1] and therefore

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

which is a contradiction.

Fix ${\displaystyle x''\in B''}$, ${\displaystyle \varphi _{1},\ldots ,\varphi _{n}\in X'}$ and ${\displaystyle \epsilon >0}$. Examine the set

${\displaystyle U:=\{y''\in X'':|(x''-y'')(\varphi _{i})|<\epsilon ,1\leq i\leq n\}.}$

Let ${\displaystyle J:X\rightarrow X''}$ 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 if we can show ${\displaystyle J(B)\cap U\neq \varnothing }$ for all such ${\displaystyle U}$. The lemma above says that for all ${\displaystyle \delta >0}$ there exists an ${\displaystyle x\in (1+\delta )B}$ such that ${\displaystyle {\text{Ev}}_{x}\in U}$. Since ${\displaystyle J(B)\subset B''}$, 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, we have

${\displaystyle \left|\left[x''-{\frac {1}{1+\delta }}{\text{Ev}}_{x}\right](\varphi _{i})\right|=\left|\varphi _{i}(x)-{\frac {1}{1+\delta }}\varphi _{i}(x)\right|={\frac {\delta }{1+\delta }}|\varphi _{i}(x)|}$.

Note that we can choose ${\displaystyle M}$ sufficiently large so that ${\displaystyle \|\varphi _{i}\|_{X'}\leq M}$ for ${\displaystyle 1\leq i\leq n}$.[3] Note as well that ${\displaystyle \|x\|_{X}\leq (1+\delta )}$. If we choose ${\displaystyle \delta }$ so that ${\displaystyle \delta M<\epsilon }$, then we have that

${\displaystyle {\frac {\delta }{1+\delta }}|\varphi _{i}(x)|\leq {\frac {\delta }{1+\delta }}\|\varphi _{i}\|_{X'}\|x\|_{X}\leq \delta \|\varphi _{i}\|_{X'}\leq \delta M<\epsilon .}$

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