Hahn-Banach theorem in separable space

Question

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?

Answer 1

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 $\lim_k\varphi_k(x)=0$.