Spring 2019 - Problem 4

Hahn-Banach, operator theory

Let $\mathcal{V}$ be the subspace of $L^\infty\p{\br{0, 1}, \mu}$ (where $\mu$ is the Lebesgue measure on $\br{0, 1}$ ) defined by

\mathcal{V} = \set{f \in L^\infty\p{\br{0, 1}, \mu} \st \lim_{n\to\infty} n\int_{\br{0,1/n}} f \,\diff\mu \text{ exists}}.

Prove that there exists $\phi \in \p{L^\infty\p{\br{0, 1}}}^*$ such that $\phi\p{f} = \lim_{n\to\infty} n \int_{\br{0,1/n}} f \,\diff\mu$ for every $f \in \mathcal{V}$ .
Show that, given any $\phi \in \p{L^\infty\p{\br{0, 1}}}^*$ satisfying the condition in (a) above, there exists no $g \in L^1\p{\br{0,1}}$ such that $\phi\p{f} = \int fg \,\diff\mu$ for all $f \in L^\infty\p{\br{0, 1}}$ .

Let $\func{\phi}{\mathcal{V}}{\R}$ be given by
$\phi\p{f} = \lim_{n\to\infty} n \int_{\br{0,1/n}} f \,\diff\mu,$
which is well-defined by definition of $\mathcal{V}$ . It's clear that this is linear, and because
$\abs{n \int_{\br{0,1/n}} f \,\diff\mu} \leq \norm{f}_{L^\infty},$
we see that $\norm{\phi} \leq 1$ , so $\phi \in \mathcal{V}^*$ . Thus, by Hahn-Banach, $\phi$ extends to a function in $\p{L^\infty\p{\br{0,1}}}^*$ , which was what we wanted to show.
Suppose otherwise, and let $g \in L^1\p{\br{0,1}}$ represent $\phi$ . Notice that if $f = \chi_{\br{a,b}}$ with $a > 0$ , then $f \in L^\infty\p{\br{0,1}}$ and $\phi\p{f} = 0$ . Thus, if $x \in \poc{0, 1}$ is a Lebesgue point of $g$ , then for $0 < r < x$ , we have
$0 = \frac{\phi\p{\chi_{\br{x-r,x+r}}}}{2r} = \frac{1}{2r} \int_{\br{x - r, x + r}} g \,\diff\mu \xrightarrow{r\to0} g\p{x}.$
Thus, $g = 0$ a.e., but this is impossible since, for example,
$1 = \phi\p{1} = \int g \,\diff\mu = 0.$