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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.

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up vote 4 down vote accepted

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.

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