Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
4  
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). –  t.b. Aug 21 '11 at 19:26
    
See also this thread for a discussion of $\ell^p$ and $L^p$. –  t.b. Jan 7 '12 at 15:36

2 Answers 2

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 $L^1$).

share|improve this answer
2  
oh, boy. I still manage to mix up weak and weak$^{\ast}$ convergence... –  t.b. Aug 21 '11 at 22:35
    
@Theo Buehler : your comment indeed seemed off-topic to me, but since it was a "nitpick", I didn't dare to question it... maybe you could edit the question title, setting it to something more meaningful and clear of "Basic Functional Analysis Question". Probably "Is $\ell^1$ isomorphic to $L^1$" would fit, or something similar... –  Angelo Lucia Aug 22 '11 at 8:02
1  
Done, following your suggestion. Oh, well. Everyone goofs from time to time. –  t.b. Aug 22 '11 at 8:07

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.