Is there $f:\ell^1\to \ell^\infty$ so that
$f$ is surjective
$\forall x,y\in \ell ^1, \|x-y\|_1=\|f(x)-f(y)\|_\infty$
2026-04-01 06:53:17.1775026397
Isometry from $\ell^1$ to $\ell^\infty$
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Hint: you can show that $l^1$ is separable, and $l^\infty $ is not.