Proof
Lemma
For all
and
there exists an
such that
for all
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))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67905b3e55c7fe87bc5cec5e0b84d174e8cad017)
it is possible to find
with
for
Now let
![{\displaystyle Y:=\bigcap _{i}\ker \varphi _{i}=\ker \Phi .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0f601f008f5c91a9e4b783cfe7baf420a467ca4)
Every element of
satisfies
and
so it suffices to show that the intersection is nonempty.
Assume for contradiction that it is empty. Then
and by the Hahn–Banach theorem there exists a linear form
such that
and
Then
[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,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/306f9c2dfca117c21d810c47efe7a94acb318879)
which is a contradiction.
Proof of theorem
Fix
and
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\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e957a4552e574728eca5331f77eea028680a9aa5)
Let
be the embedding defined by
where
is the evaluation at
map. Sets of the form
form a base for the weak* topology,[2] so density follows once it is shown
for all such
The lemma above says that for any
there exists a
such that
and in particular
Since
we have
We can scale to get
The goal is to show that for a sufficiently small
we have
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)|.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab146ff38f917cea1e9bf659df946071fe91bfc3)
Note that one can choose
sufficiently large so that
for
[3] Note as well that
If one chooses
so that
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 .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35cdad5825a72a4f55c49bc2f1f93d664552a058)
Hence one gets
as desired.