Let $X$ be a separable Banach space and let $\mu$ be the counting measure on $\mathbb{N}$. If $\{x_n\}_{n=1}^{\infty}$ is a countable dense subset of the unit ball of $X$, and $T:L^1(\mu)\to X$ is defined by $Tf = \sum_{n=1}^{\infty}f(n)x_n$, then $T$ is bounded and surjective.
This came up as an exercise in a book I'm reading and it seems simple enough on the outside but I can't seem to reason why it is true. Does anyone have an easy way to see why this is true?
For the boundedness, note that $$ \|Tf\|_X = \|\sum_{n=1}^{\infty} f(n) x_n \|_X \leq \sum_{n=1}^{\infty} \|f(n) x_n\|_X = \sum_{n=1}^{\infty} |f(n)| = \|f\|_{L^1 (\mu)}.$$ For the surjectivity, note that by linearity, it is enough to prove that the unit ball is contained in the range of $T$. Clearly, $\text{ran}(T)$ contains $\{x_n: n \in \mathbb{N}\}$, so by taking the closure of this inclusion we obtain $$ \{x \in X: \|x\| \leq 1\} \subseteq \overline{\text{ran}(T)}.$$ We see that the assertion follows if we can show that $\text{ran}(T)$ is closed. It is easy to see that the adjoint of $T$ is given by \begin{align}T': X' &\rightarrow L^{\infty}(\mu) \\ x' &\mapsto \{x'(x_n)\}_{n=1}^{\infty}.\end{align} Observe that $\|T'x'\|_{L^{\infty}(\mu)} = \|x'\|_{X'}$, which implies that $\text{ran}(T')$ is closed. The closed range theorem now yields that $\text{ran}(T)$ is closed.
