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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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    Two nitpicks: 1. weak convergence should be weak sequential convergence. 2. (more important) how do you *define* weak convergence intrinsically? The pre-dual of a Banach space is not uniquely determined. But if you're happy with arguing with pre-duals, notice that $l^1$ is a dual space, while $L^1$ isn't (the unit ball doesn't have *any* extremal points).2011-08-21
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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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