Constructing a nonzero bounded linear functional on $L^\infty[0,1]$ that vanishes on $C[0,1]$

Question

I want to provide an example of a nonzero bounded linear functional on $L^\infty[0,1]$ that vanishes on $C[0,1]$. I know that the existence of such a functional is guaranteed by the Hahn-Banach Theorem. My issue is with actually constructing one. For example, I know $C[0,1]$ is a subspace of $L^\infty[0,1]$, so I tried defining a functional $f$ on $C[0,1]$ as $f(u)=\lim_{x\to\frac{1}{2}^+}u(x)-\lim_{x\to\frac{1}{2}^-}u(x)$. This vanishes on $C[0,1]$ but the one-sided limits are not necessarily well-defined on $L^\infty[0,1]$. I also thought of considering $L:=$ the span of $C[0,1]$ and some element in $L^\infty[0,1]\backslash C[0,1]$ like $\chi_{[0,\frac{1}{2}]}$. Then for $y=u+\lambda\chi_{[0,\frac{1}{2}]}\in L$, I can define $f(y)=\lambda$.

$f$ vanishes on $C[0,1]$ and $||f||\neq0$ and I know I can extend $f$ to $\phi$ on $L^\infty[0,1]$ but I want to be able to know what $\phi$ looks like explicitly. Maybe that is not possible for this $\phi$, but is there an example of some other $\phi$ that is explicitly defined for every element in $L^\infty[0,1]$?

Answer 1

The bounded linear functionals on $L^\infty[0,1]$ are precisely those given by integration against finitely additive complex measures that are absolutely continuous with respect to Lebesgue measure. So what you are looking for is an example of an additive but $\sigma$-additive measure. And the examples are not terribly explicit: here is the kind of thing to expect.

Answer 2

I don't think this is possible. Continuous linear functionals on $L^{\infty}$ are of the type $f \to \int f d\mu$ where $\mu$ is a finitely additive measure. If $\mu$ is countably additive then $\int f d\mu=0$ for all $f \in C[0,1]$ forces $\mu$ to be $0$. Hence we need a finitely additive measure which is not countably additive but there is no explicit construction of such a measure.

Answer 3

The dual of $E=L^1([0,1])$ is $E'=L^\infty([0,1])$. What you are asking for is a linear functional $\lambda\in E''$ that vanishes on the subspace $V=C([0,1]) \subset E'$. Now, any $g\in E$ gives rise to an element $\lambda_g\in E''$ by duality, and by standard integration theory it is a non-zero functional on $V$ iff $g$ is non-zero in $E$.

So you need a genuine element $\lambda\in E''\setminus E$. Such an element does not exists in a pure ZF system. This suggests strongly that you will not be able to describe explicitly the action of such an element on all of $E'$. AoC seems unavoidable in some form.