# $Tf=\sum\limits_{n=1}^\infty f(n)x_n$ is surjective from $\ell^1$ to a separable Banach space [duplicate]

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 $\sum\limits_{n=1}^\infty f(n)x_n$?

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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.