5

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.

t.b.
  • 78,116
wade23
  • 111
  • 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 07 '12 at 15:36

3 Answers3

6

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 $L^1$).

t.b.
  • 78,116
Angelo Lucia
  • 852
  • 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 08:02
  • 1
    Done, following your suggestion. Oh, well. Everyone goofs from time to time. – t.b. Aug 22 '11 at 08:07
3

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

GEdgar
  • 111,679
0

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.

Tomasz Kania
  • 16,361