Functional Analysis: Folland Problem Folland problem 5.36 (b)
Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_{n}\}_{n=1}^{\infty}$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T:L^{1}(\mu)\to \mathcal{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_{n} $.
Show that $T$ is surjective.
My approach: Consider any $x\in\mathcal{X}$ and let $x_{n}$ be a sequence such that $x_{n}\to x$ (since $\mathcal{X}$ is separable this can be done). Now consider $f^{k}$ such that $f^{k}(k) = 1$ and $f^{k}(j)= 0$ if $j\neq k$. Clearly $Tf^{k} = x_{k}\to x$. But now the problem is that $f^{k}$ doesn't converge.
 A: One problem you're going to have is that all of the $x_n$ only live in the unit ball, so there's no way you're going to be able to get any $x \in X$ as the limit point of a sequence of the $x_n$. But you're on the right track, since one thing you can do is realize any $x \in B$ (the unit ball) as a limit point of the $x_n$. And obviously, if $T$ surjects onto the unit ball, it will surject globally.
One construction is the following. For convenience, let $S$ denote your countable dense set. We'll build up a sequence of $f_n$ such that $T f_n \to x$. Given $x \in B$, pick some $x_{n_1} \in S$ such that $||x-x_{n_1}|| \le 1/2$, and set $f_1$ to be the sequence with a single $1$ in the $n_1$ spot. Then pick some $x_2$ such that $||x - x_{n_1} - x_{n_2}/2|| \le 1/4$. You can do this, because we picked $x_{n_1}$ so that $x-x_{n_1}$ lives in $B/2$. Define $f_2$ to be $f_1$ plus the sequence with a single $1/2$ in the $n_2$ spot. Continue. Now your sequence $f_n$ will converge, as it's just a reordering of a geometric series.
A: Here is a complete proof of 36
Problem 36 - Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{1}^{\infty}$ is a countable dense subset of the unit ball of $\mathcal{X}$, and define $T:L^{1}(\mu)\rightarrow \mathcal{X}$ by $Tf = \sum_{1}^{\infty}f(n)x_n$.
a.) $T$ is bounded.
b.) $T$ is surjective.
c.) $\mathcal{X}$ is isomorphic to a quotient space of $L^{1}(\mu)$ (Use Exercise 35).
Proof a.) - We have that $Tf = \sum_{1}^{\infty}f(n)x_n$, hence  $$\| Tf\|  = \| \sum_{1}^{\infty}f(n)x_n\| \leq \sum_{1}^{\infty}|f(n)|\| x_n\| \leq \sum_{1}^{\infty}|f(n)| = \| f\|$$
Thus $T$ is bounded.
Proof b.) - Since $T$ is linear, it's enough to show that every $x\in X$ with $||x||\leq 1$ is in the image of $T$. To do this, we proceed inductively: there exists $x_{n_1}$ such that $||x-x_{n_1}||<\frac{1}{2}$. If $y=2(x-x_{n_1})$ then $||y||<1$, so there exists $x_{n_2}\neq x_{n_1}$ such that $||y-x_{n_2}||<\frac{1}{2}$, hence $||x-x_{n_1}-\frac{1}{2}x_{n_2}||<\frac{1}{4}$. In general, if $x_{n_1},\dots,x_{n_k}$ have been chosen such that
$$ \Big|\Big|x-\sum_{j=1}^k2^{1-j}x_{n_j}\Big|\Big|<2^{-k}$$
then $y=2^k(x-\sum_{j=1}^k2^{1-j}x_{n_j})$ is in the unit ball, hence there exists $x_{n_{k+1}}\not\in\{x_{n_1},\dots,x_{n_k}\}$ such that $||y-x_{n_{k+1}}||<\frac{1}{2}$, i.e.
$$ \Big|\Big|x-\sum_{j=1}^{k+1}2^{1-j}x_{n_j}\Big|\Big|<2^{-k-1}$$
Finally, define $f\in \ell^1(\mathbb{N})$ by $f(n)=2^{1-k}$ if $n=n_k$, $f(n)=0$ otherwise. Then $Tf=x$.
proof c.) - We have that  $L^{1}(\mu)$ and $\mathcal{X}$  are Banach spaces and $T\in L(L^{1}(\mu),\mathcal{X})$. Moreover,
$T$ is surjective, so $\textrm{range}(T)= \mathcal{X}$ and $\mathcal{X}$ is closed in $\mathcal{X}$. So by exercise 35.
$L^{1}(\mu)/\mathcal{N}(T)$ is isomorphic to $\mathcal{X}$.
