Consider the measurable space $([0,1],\mathcal{B}([0,1]),l)$, where $\mathcal{B}([0,1])$ denotes the Borel algebra in $[0,1]$ and $l$ is the Lebesgue measure.
I need to prove that exists a bounded linear functional $\Gamma\neq 0$ over $L_{\infty}$ which vanishes over $\mathcal{C}([0,1]).$ Conclude that $(L_\infty)^{*}\neq L_{1}.$ The suggestion is to consider the subspace of bounded functions which admits limit in the origin, and apply the Hahn-Banach theorem.
Hahn-Banach Theorem: Let $V$ a real vector space and $p:V\rightarrow\Bbb{R}$ an application satisfying $p(cv)=cp(v)$ and $p(u+v)\leq p(u)+p(v)$ for any $c>0$ and $u,v\in V.$ Suppose that $\gamma:W\rightarrow \Bbb{R}$ is a linear functional over an vector subspace $W\subset V$ satisfying $\gamma(w)\leq p(w)$ for all $w\in W.$ So, $\gamma$ extends to a linear functional $\Gamma$ defined over all $V$ satisfying $\Gamma(v)\leq p(v)$ for any $v\in V.$
Looks like the question is not that hard, but I'm confused about how to apply the theorem for this case.
My function $p$ needs to be a function from $[0,1]$ to $\Bbb{R}$, right? And it needs to satisfies that $\gamma(x)\leq p(x)$ for all $x\in[0,1]$ for some linear functional $\gamma:W\rightarrow\Bbb{R}$. My first question is, what is $W$ in my example? Doesn't make sense $W$ be the set of all bounded functions such that exists limit in $0$, since all the functions in $\mathcal{C}([0,1])$ satisfies this. I need to take $W$ as a subspace of $\mathcal{C}[(0,1)]$ or $L_{\infty}?$
You can see that I'm really confused, so if someone could help me patiently, would help me.
Best Answer
Pick an $f_{0}\in L^{\infty}-C[0,1]$ and let $W=\text{span}(\{f_{0}\}\cup C[0,1])$. Consider $\Gamma(\alpha f_{0}+g)=\alpha d(f_{0},C[0,1])$ for any scalar $\alpha$, then $\Gamma$ vanishes on $C[0,1]$, certainly $\Gamma$ is a non-zero bounded linear functional on $W$.
Hahn-Banach Theorem gives an extension of $\Gamma$ to the whole $L^{\infty}$.