# Hahn-Banach theorem in separable space

Let $$E$$ be a normed separable space with infinite dimension.

a) Show that there exist a sequence $$(M_k)$$ of finite dimension subspaces such that $$M_k\subset M_{k+1}$$, for all $$k$$, and $$\bigcup_{k=1}^{\infty}M_k$$ is a dense subset of $$E$$.

b) Using Hahn-Banach Theorem, show that there exist a sequence of functions $$(\varphi_k)\subset E'$$ such that $$\Vert \varphi_k \Vert=1$$, for all $$k$$, and $$\displaystyle\lim_{k\rightarrow\infty}\varphi_k(x)=0$$ for all $$x\in E$$.

For item a), let $$D=\{x_k;k\in\mathbb{N}\}$$ a dense subset of $$E$$ which exists beacuse $$E$$ is separable. Then $$M_k=span\{x_1,...,x_k\}$$ solves.

My problem is with item b), can anyone help me with ideas?

• HB will provide a norm-1 functional that vanishes on $M_k$. Commented Dec 11, 2020 at 18:50
• can you show me this result? because I just know that one wich provides a norm-1 functional f such that f(x_0)=||x_0|| for each x_0 in E. Commented Dec 11, 2020 at 19:19
• See proposition 6.6 here. Commented Dec 11, 2020 at 19:23

Fix $$k$$. Choose $$y\in E\setminus M_k$$. On $$\operatorname{span}M_k\cup\{y\}$$, define a linear functional $$\varphi_k$$ with $$\varphi_k(y)=1$$, $$\varphi_k(x_j)=0$$, $$j=1,\ldots,k$$. This works because $$y$$ is linearly independent with $$M_k$$.
We need to show that $$\varphi_k$$ is bounded. We have \begin{align} \|\varphi_k\|&=\sup\{\frac{|\varphi_k(x+\lambda y)|}{\|x+\lambda y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm] &=\sup\{\frac{|\lambda|}{\|x+\lambda y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm] &=\sup\{\frac{1}{\|x/\lambda+ y\|}:\ \lambda\in\mathbb C\setminus\{0\},\ x\in M_k\}\\[0.3cm] &=\sup\{\frac{1}{\|y-x\|}:\ x\in M_k\}\\[0.3cm] &=\frac1{\inf\{\|y-x\|:\ x\in M_k\}}\\[0.3cm] &=\frac1{\operatorname{dist}(y,M_k)}<\infty \end{align} Knowing that $$\varphi_k$$ is bounded, we can extend it by Hahn-Banach to $$\varphi_k\in E'$$, that satisfies $$\varphi_k|_{M_k}=0$$.
Now, given $$x\in E$$ and $$\varepsilon>0$$, there exists $$k_0$$ such that $$\|x-x_{k_0}\|<\varepsilon$$. Then, for any $$k>k_0$$, $$|\varphi_k(x)|\leq|\varphi_k(x-x_{k_0})|+|\varphi_k(x_{k_0})| =|\varphi_k(x-x_{k_0})|\leq\|x-x_{k_0}\|<\varepsilon.$$ Thus $$\lim_k\varphi_k(x)=0$$.