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Can there be a continuous linear map, with a continuous inverse, from $l^{1}$ to $L^{1}(m)$ where $m$ is the Lebesgue measure on the unit interval $\left[0,1\right]?$

My thinking to this should be No. In $l^{1}$, we have a special property that weak convergence is actually equivalent to norm convergence; proven using a "gliding hump argument". This is certainly impossible in $L^{1}(m)$. A continuous linear map with continuous inverse is essentially a homeomorphism between the two spaces; so it should preserve norm convergence. I'm just wondering if my reasoning is correct and also if there are any resources out there that I can understand these ideas better.

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    See also [this thread](http://math.stackexchange.com/q/97126/5363) for a discussion of $\ell^p$ and $L^p$.2012-01-07

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The property you are referring to is called Schur's property, and it's preserved by isomorphisms (and this can be used to distinguish between $\ell^1$ and $L^1$).

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    Done, following your suggestion. Oh, well. Everyone goofs from time to time.2011-08-22
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NO, they are not isomorphic. My favorite off-beat reason: $l^1$ has the Radon-Nikodym Property and $L^1$ doesn't.
reference: Diestel & Uhl, Vector Measures

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Here's one more answer.

Every closed subspace of $\ell_1$ contains a copy of $\ell_1$ but $L_1$ contains a copy of $\ell_2$ - for example take the closed linear span of the Rademacher system and apply Khinchine's inequality.