The space of continuous function on $[0,1]$ in dense in $L^{\infty}$ respect to the weak$-*$ topology

Question

I am reading Rudins' book on Functional Analysis for self study. I stumbled upon an exercise and I would like someone to revise my solution: exercise 7 chapter $3$ second edition. The exercise asks to prove that the space of continuous functions $C([0,1])$ is dense in $L^{\infty}([0,1])$ respect to the weak-$*$ topology. I tried to prove that $$ N_{f_{1}}(g_{1},\epsilon_{1}) \, \cap C([0,1])\neq \emptyset \quad \forall g_{1} \in L^{\infty}([0,1]),\quad \forall f_{1} \in L^{1}([0,1]),\,\quad \forall \epsilon_{1}>0. $$ where $N_{f_{1}}(g_{1},\epsilon_{1})$ is defined as $$ N_{f_{1}}(g_{1},\,\epsilon_{1})=\left\{ m \in L^{\infty}: \left \vert \int_{0}^{1}f_{1}(x)\left(g_{1}(x)-m(x) \right)\,dx \right\vert<\epsilon_1 \right\}. $$ I thought that a continuous function of $[0,1]$ that might fall into $N_{f_{1}}(g_{1})$ could be one of the form $g_{1}*\rho_{\epsilon}$ where $\{\rho_{\epsilon}\}_{\epsilon >0}$ is a family of mollifiers. Firstly, I choose $M$ such that $$ \int_{0}^{1} \left\vert f_{1}(y)-f_{1}(y)\cdot \mathbb{1}_{\left\{ \vert f_{1} \vert \leq M \right\}}\right\vert \, dy \leq \dfrac{\epsilon_{1}}{2\left( 1+2\,\Vert g_1 \Vert_{\infty} \right)}. $$ Secondly, I split the main integral in the sum of two integral and use triangular inequality $$ \begin{gather*} \left\vert \int_{0}^{1}f_{1}(x)\left(g_{1}(x)-g_{1}*\rho_{r}(x) \right)\,dx \right\vert\\ \leq \left\vert \int_{0}^{1} f_{1}(y)\cdot \mathbb{1}_{\left\{ \left\vert f_{1} \right\vert > M \right\}} \, \left( g_{1}(y)-g_{1}*\rho_{r}(y) \right) \,dy \right\vert + \left\vert \int_{0}^{1}f_{1}(y)\cdot \mathbb{1}_{\left\{ \left\vert f_{1} \right\vert \leq M \right\}} \, \left( g_{1}(y)-g_{1}*\rho_{r}(y) \right) \,dy \right\vert \\ \leq 2\Vert g \Vert_{\infty}\cdot \int_{0}^{1}\vert f_{1}(y) \vert \, \cdot \mathbb{1}_{\left\{ \left\vert f_{1}\right\vert>M \right\}}(y)\,dy+ M\cdot \int_{0}^{1}\vert g_{1}(y)-g_{1}(y)*\rho_{r}(y) \vert\,dy \\ \leq \frac{\epsilon_{1}}{2}+ M\cdot \int_{0}^{1}\vert g_{1}(y)-g_{1}(y)*\rho_{r}(y) \vert\,dy. \end{gather*} $$
Finally, by the mollifiers properties, I can choose $r$ small enough so that $$ M\cdot \int_{0}^{1}\vert g_{1}(y)-g_{1}(y)*\rho_{r}(y) \vert\,dy<\frac{\epsilon_{1}}{2}, $$ and for such an $r$ we have $g_{1}*\rho_{r} \in N_{f_{1}}(g_{1},\,\epsilon_{1});$ since the sets of the form $N_{f_{1}}(g_{1},\epsilon_{1})$ are a pre-base for the weak-$*$ topology we conclude that $C([0,1])$ is dense in $L^{\infty}.$

Answer 1

Your solution is mostly correct. However, for a set $A$ to be dense in a topological space $X$, it is not sufficient that $A$ intersects a pre-base of the topology $(*)$. You must verify that $A$ intersects a base, and this follows from the argument you present. The proof can be shortened a bit if you use the fact (which you essentially reprove) that continuous functions are dense in $L^2$.

$(*)$ consider $\mathbb R^2$ with the pre-base of all products $$\{(a,b) \times \mathbb R : \, a<b \} \, \cup \, \{ \mathbb R \times (c,d)\, : \, c<d \} \,. $$ Then the union of the $x$ and $y$ axes intersects every set in the pre-base but is not dense in $\mathbb R^2$.

Answer 2

A relative simple proof can be obtained by application of Hahn-Banach's theorem.

Denote by $m$ the Lebesgue measure on $([0,1],\mathscr{B}([0,1])$. Let $M=\overline{C[0,1])}$ where the closure is taken with respect the weak topology $\sigma(L_\infty(m),L_1(m))$. Suppose $x^*\in L_1(m)$ such that $\langle x^*,x\rangle=\int^1_0 x(t)x^*(t)\,dt=0$ for all $x\in M$. In particular $$\int^1_0 \phi(t)x^*(t)\,dt=0,\qquad \phi\in C([0,1])$$ The Riesz representation theorem yields a unique finite Radon signed measure $\mu$ on $[0,1]$ such that $\langle x^*,\phi\rangle=\int \phi\,d\mu=0$ for all $\phi\in \mathcal{C}([0,1])$. Hence $\mu=x^*(t)\,dt\equiv0$ and so, $x^*=0$ $m$-almost surely . The conclusion follows from the Hahn-Banach theorem: If $y\in L_\infty\setminus M\neq\emptyset$ then there would be $x^*\in L_1$ such that $\langle x^*, x\rangle=0$ for all $x\in M$ and $\langle x^*,y\rangle =1$.