Spring 2019 - Problem 5

Solution.

1. Since step functions are dense in $L^p\p{\br{0,1}}$ for $1 \leq p < \infty$, it suffices to show that step functions can be approximated by a countable set. Consider

   $$V = \set{\sum_{j=1}^n c_j \chi_{\br{a_j, b_j}} \st n \in \N,\ a_j, b_j \in \Q \cap \br{0,1},\ c_j \in \Q}.$$

   By linearity, it suffices to show that we can approximate functions of the form $c\chi_{\br{a, b}}$, but this is easy by density of $\Q$ in $\R$ and the dominated convergence theorem, so $V$ is dense in $L^p\p{\br{0,1}}$, so it is separable.

   When $p = \infty$, consider $e_n = \chi_{\p{\frac{1}{n+1}, \frac{1}{n}}}$ so that each $e_n$ have disjoint supports. Let

   $$A = \set{\sum_{j=1}^\infty \sigma_j e_j \st \sigma_j \in \set{0, 1}},$$

   which is in bijection with $\set{0, 1}^\N$, i.e., $A$ is uncountable. Notice that if $\set{\sigma_n}_n \neq \set{\tau_n}_n$ are distinct elements of $A$, then there exists $n$ such that $\sigma_n \neq \tau_n$, and so $\sum_n \sigma_n e_n$ and $\sum_n \tau_n e_n$ differ on $\p{\frac{1}{n+1}, \frac{1}{n}}$, so distinct elements have distance at least $1$. Thus, because $A$ is uncountable, we see that $L^\infty\p{\br{0,1}}$ is not separable.

2. Suppose otherwise, and that there exists such a $T$. We have the transpose $\func{T^\dagger}{\p{L^1\p{\br{0,1}}}^*}{\p{L^p\p{\br{0,1}}}^*}$, which is given by $T^\dagger\p{\lambda} = \lambda \circ T$. Notice that $T^\dagger$ is injective: indeed, suppose $T^\dagger\p{\lambda} = 0$, and let $g \in L^1\p{\br{0,1}}$. Since $T$ is surjective, there exists $h \in L^p\p{\br{0,1}}$ such that $Th = g$, so

   $$0 = \p{T^\dagger \lambda}\p{h} = \lambda\p{Th} = \lambda\p{g}.$$

   Thus, $\lambda = 0$, so $T^\dagger$ has trivial kernel. But by Riesz representation, we see that

   $$L^\infty\p{\br{0,1}} \subseteq L^{p/\p{p-1}}\p{\br{0,1}}.$$

   But this contradicts the fact that $L^\infty\p{\br{0,1}}$ is not separable.