Spring 2014 - Problem 5

Solution.

1. Let $D \subseteq \C$ be a countable, dense set (e.g., rational linear combinations of $1$ and $i$). Then

   $$A = \set{\set{a_n}_n \mid a_n \in D \text{ and } a_n \neq 0 \text{ for only finitely many terms}}$$

   is a countable union of countable sets, hence countable itself. We will show that $A$ is dense in both $\ell^1\p{\N}$ and $\ell^2\p{\N}$: let $p \in \set{1, 2}$, $\set{x_n}_n \in \ell^p\p{\N}$, and $\epsilon > 0$. Then there exists $N \in \N$ such that

   $$\sum_{n=N+1}^\infty \abs{x_n}^p < \epsilon.$$

   For each $1 \leq n \leq N$, there exists $q_n$ such that $\abs{x_n - q_n}^p < \frac{\epsilon}{N}$, by density. Set $q_n = 0$ for $n > N$ which means $\set{q_n}_n \in A$. Then

   $$\begin{aligned} \sum_{n=1}^\infty \abs{x_n - q_n}^p &= \sum_{n=1}^N \abs{x_n - q_n}^p + \sum_{n=N+1}^\infty \abs{x_n - q_n}^p \\ &\leq \sum_{n=1}^N \frac{\epsilon}{N} + \sum_{n=N+1}^\infty \abs{x_n}^p \\ &\leq 2\epsilon, \end{aligned}$$

   so $A$ is dense in both $\ell^1\p{\N}$ and $\ell^2\p{\N}$.

   To show that $\ell^\infty\p{\N}$ is not separable, consider the set $A$ of infinite binary strings, i.e., $\set{x_n}_n$ so that $x_n \in \set{0, 1}$ for all $n \geq 1$. Certainly $A \subseteq \ell^\infty\p{\N}$, and if $x \neq y$ are elements of $A$, there exists some $n_0$ so that $x_{n_0} \neq y_{n_0}$. Hence, $\norm{x_{n_0} - y_{n_0}}_{\ell^\infty} = 1$.

   Suppose $\set{a_n}_n$ is dense in $\ell^\infty\p{\N}$. Then for each $x \in A$, there exists $a_{n\p{x}}$ such that $\norm{a_{n\p{x}} - x}_{\ell^\infty} < \frac{1}{2}$. But $x \mapsto n\p{x}$ is an injective function from an uncountable set to a countable one, which is impossible. Thus, $\ell^\infty$ is not separable.

2. We first prove the following lemma: if $\func{T}{X}{Y}$ is a continuous linear map between Banach spaces, then $T^\dagger$ is injective if and only if the range of $T$ is dense in $Y$, where $T^\dagger$ is the transpose of $T$.

   Suppose $T^\dagger$ is injective, but the range of $T$ is not dense. Let $M = \cl{\range{T}}$, which is not all of $Y$ by assumption. Hence, there exists some $y \in Y \setminus M$ with distance $\delta > 0$ away from $M$. By Hahn-Banach, there exists $f \in Y^*$ such that $f\p{x} = 0$ on $M$ and $f\p{y} = \delta$. Thus, $T^\dagger\p{f} = f \circ T \in X^*$, but by construction $f\p{T\p{x}} = 0$ for all $x \in X$. Hence, $T^\dagger\p{f} = 0$, but $f \neq 0$, a contradiction.

   Conversely, suppose $\range{T}$ is dense in $Y$, and suppose $f \in Y^*$ is such that $T^\dagger\p{f} = 0$. Since $f \circ T = T^\dagger\p{f}$ and $\range{T}$ is dense, it follows that $f\p{y} = 0$ for any $y \in Y$, by continuity, so $f = 0$ and hence $T^\dagger$ is injective.

   In our problem, our $T$ is surjective, in particular dense in $\ell^1\p{\N}$. Thus, $\func{T^\dagger}{\ell^\infty\p{\N}}{\ell^2\p{\N}}$ is an injective bounded linear map, so $T^\dagger$ is an isomorphism $\ell^\infty\p{\N} \to T^\dagger\p{\ell^2\p{\N}}$, and so we have a continuous linear map $T^\dagger\p{\ell^2\p{\N}} \to \ell^\infty\p{\N}$. But $\ell^2\p{\N}$ is separable, which would imply by continuity that $\ell^\infty\p{\N}$ is also separable, a contradiction. Hence, no continuous surjection could have existed to begin with, which completes the proof.