$Tf=\sum\limits_{n=1}^\infty f(n)x_n$ is surjective from $\ell^1$ to a separable Banach space 
Let $X$ be a separable Banach space and let $\mu$ be a counting measure on $\mathbb{N}$. Suppose that $\{x_n\}_{n=1}^\infty$ is a countable dense subset of the unit ball of $X$ and define 
  $T: L^1(\mu)\rightarrow X$ by $$Tf=\sum\limits_{n=1}^\infty f(n)x_n.$$
Then $T$ is surjective.

This is an exercise from Folland's "Real Analysis", P164 Ex36(b).
How can an element of $X$ be expressed as $\sum\limits_{n=1}^\infty f(n)x_n$?
 A: Let $x\in B(X)$, the unit ball of $X$. 
Choose $x_{n_1}$ so that $$\Vert x-x_{n_1}\Vert\le 1/2^2.$$
Note $$\Vert 2(x-x_{n_1})\Vert\le 1/2.\tag{1}$$
From $(1)$, and the denseness of $(x_n)$ in $B(X)$, you can choose $x_{n_2}$ so that $$\Vert 2(x-x_{n_1})-x_{n_2}\Vert\le 1/2^3.$$
From this, it follows that 
 $$\Vert 4(x-x_{n_1})-2x_{n_2} \Vert\le 1/2^2\tag{2},$$
and
$$\Vert x-  (x_{n_1} +x_{n_2}/2 )\Vert\le1/2^4.\tag{$2'$}$$
Now, using $(2)$ and the denseness of $(x_n)$ in $B(X)$, choose $x_{n_3}$ so that  $$\Vert \bigl(4(x-x_{n_1})-2x_{n_2}\bigr)-x_{n_3}\Vert\le 1/2^4.$$
Note then that
$$
\Vert \bigl(8(x-x_{n_1})-4x_{n_2}\bigr)-2x_{n_3}\Vert\le 1/2^3.
$$
and
 $$\Vert x-(x_{n_1} +x_{n_2}/2 +x_{n_3}/4)\Vert\le 1/2^6.\tag{$3'$}$$

Continuing in this manner produces a 
$$y=e_{n_1}+e_{n_2}/2+e_{n_3}/2^2+\cdots\in\ell_1,$$
where $e_n$ is the standard $n$'th unit vector in $\ell_1$.
Since for each $n$, $Te_n=x_n$, it follows from $(2'), (3'),\cdots$ that
 $Ty=x$.
As far as I know, this argument is due to Banach and Mazur.
