"Dirac delta" as an element of $L^\infty(\mathbb{R})$

Question

It is known that if $(X, \Sigma, \mu)$ is a measure space,

$$L^1(d\mu) \hookrightarrow [L^\infty(d\mu)]^* \\ \quad \quad g \, \, \, \mapsto \int \_ \, \, g \, d\mu $$

is isometric. To show that the inclusion is not surjective, consider $\mu$ to be the Lebesgue measure on $X = \mathbb{R}$ and consider the following functional on the space $C_b(\mathbb{R})$ of bounded continuous functions on $\mathbb{R}$:

$$\varphi: C_b(\mathbb{R}) \to \mathbb{C}\\ \quad \quad \quad \quad \, f \, \, \, \mapsto f(0) \, .$$

By the Hahn-Banach theorem, $\varphi$ has an extension to $[L^\infty(\mathbb{R})]^*$, and one can show that $\varphi$ cannot be seen as $\int \_ \, g \, d\mu$ for any $g \in L^1(\mathbb{R})$.

My question is: how is $\varphi$ well defined on $[L^\infty(\mathbb{R})]^*$? The subset $\{ 0\} \subset \mathbb{R}$ has measure zero, so what is the meaning of mapping $f \mapsto f(0)$ from the point of view of $L^\infty(\mathbb{R}) $?

Answer 1

You are invoking Hahn-Banach, so the map is not $f\mapsto f(0)$ on the whole space $L^{\infty}(d\mu)$.

Instead, $\varphi$ is a bounded linear map on $L^{\infty}(d\mu)$ such that, if $f \in L^{\infty}(d\mu)$ has a continuous representative $\tilde f$, then $\varphi(f) = \tilde f(0)$. If $f$ has no continuous representative, then $\varphi(f)$ depends on the extension obtained using Hahn-Banach.

Answer 2

The Hahn-Banach extension is not unique so, for $f$ in $L^\infty ({\mathbb R})$, it is impossible to tell what is $\varphi (f)$ in general. Moreover in the proof of Hahn-Banach's Theorem the extension is obtained as an application of Zorn's Lemma, a process that is not constructive.

Nevertheless in the present case it is possible to exhibit an extension in a slightly more concrete form, as follows: for each natural number $n$, consider the linear functional $\varphi _n$ on $L^\infty ({\mathbb R})$ given by $$ \varphi _n(f) = \frac n{2}\int_{-1/n}^{1/n} f(t)\, dt. $$ Clearly $\|\varphi _n\|\leq 1$, so $\varphi _n$ belongs to the unit ball of $L^\infty ({\mathbb R})^*$, which is compact by Alaoglu's Theorem. One may then pick (here is the least constructive step of this process) an accumulation point $\varphi $ of the $\varphi _n$, which will have the desired properties.

In particular, if $f$ is such that $$ \lim_{n\to \infty }\frac n{2}\int_{-1/n}^{1/n} f(t)\, dt $$ exists, then $\varphi (f)$ will necessarily be this limit. Also notice that the set of such functions is substantially bigger than $C^b({\mathbb R})$!