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I am trying to prove that the map stated in the title is continuous by showing it is Lipschitz. Not sure what should go in the $\dots$

$$\|\ {J(\left\{ x_n \right\} )} \|_{\ell^{1}}\ = \left\|\ \dfrac{x_n}{2^{n}} \right\|_{\ell^{1}}\ = \sum_{n=0}^{\infty} \left| \dfrac{x_n}{2^{n}} \right| = \dots \leq \sup_{n} |x_n| = \|\ \left\{x_n\right\} \|_{\ell^{\infty}}\ $$

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    $$\sum_{n = 0}^{\infty} \biggl\lvert \frac{x_n}{2^n}\biggr\rvert \leqslant \sum_{n = 0}^{\infty} \frac{\sup \{\lvert x_k\rvert : k \in \mathbb{N}\}}{2^n} = \lVert x\rVert_{\ell^{\infty}}\sum_{n = 0}^{\infty} \frac{1}{2^n},$$ and you need a factor of $2$ on the right hand side. – 2017-01-21
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    @Daniel Fischer Thank you. – 2017-01-21
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    @DanielFischer Does this also show that $\|\ x \|_{\ell^{\infty}} \not\leq c\|\ Jx \|_{\ell^{1}}$ for $c>0$? – 2017-01-21
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    Not by itself. Together with some theory, it does; for example one can argue that if there were a $c$ such that $\lVert x\rVert_{\ell^{\infty}} \leqslant c\lVert J x\rVert_{\ell^1}$ for all $x$, then $J$ would be an embedding, but since $\ell^1$ is separable and $\ell^{\infty}$ isn't separable, that is impossible. But it's easier to just note $\lVert e_n\rVert_{\ell^{\infty}} = 1$ and $\lVert Je_n\rVert_{\ell^1} = 2^{-n}$ for all $n$. – 2017-01-21

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$$\|\ J(\left\{ x_n \right\} ) \|_{\ell^{1}} = \left\|\ \dfrac{x_n}{2^{n}} \right\|_{\ell^{1}} = \sum_{n = 0}^{\infty} \biggl\lvert \frac{x_n}{2^n}\biggr\rvert \leqslant \sum_{n = 0}^{\infty} \frac{\sup \{\lvert x_k\rvert \space\ \big| \space\ k \in \mathbb{N}\}}{2^n} = \lVert \left\{ x_n \right\} \rVert_{\ell^{\infty}}\sum_{n = 0}^{\infty} \frac{1}{2^n} = 2 \|\ \left\{x_n\right\} \|_{\ell^{\infty}}$$