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 $\sum\limits_{n=1}^\infty f(n)x_n$?
Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks.
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