Prev: Completing two statistics problems
Next: Question about Sarkovskii Ordering
From: jane on 30 Nov 2007 23:58
Let U be the unit ball in L^2[0,1] with Lebesgue measure and consider L^2[0,1] as a subspace of L^1[0,1]. Is it possible to describe the
1). The closure of U in the norm topology of L^1[0,1]
2). In the weak-topology of L^1[0,1] ?

Thanks.
From: omega on 2 Dec 2007 17:34
1) The closure is U itself, due to the reflexivity of L^2.
2) Now, being a convex and closed subset of L^1, U is also weakly closed (Mazur).
From: TCL on 3 Dec 2007 13:09
> 1) The closure is U itself, due to the reflexivity of
> L^2.

Let me elaborate; correct me if I am wrong. By the reflexivity of L^2, U is weakly compact in the weak topology T_2 of L^2. Since the weak topology T_1 of L^1 induces a weaker topology than T_2 on L^2, U is also weakly compact in T_1 topology, and hence weakly closed in L^1. Since U is convex, it is also norm closed in L^1.

But I am not quite sure if the statement "the weak topology T_1 of L^1 induces a weaker topology than T_2 on L^2" is correct.

> 2) Now, being a convex and closed subset of L^1, U is
> also weakly closed (Mazur).
From: TCL on 3 Dec 2007 13:15
> > 1) The closure is U itself, due to the reflexivity
> of
> > L^2.
>
> Let me elaborate; correct me if I am wrong. By the
> reflexivity of L^2, U is weakly compact in the weak
> topology T_2 of L^2. Since the weak topology T_1 of
> L^1 induces a weaker topology than T_2 on L^2, U is
> also weakly compact in T_1 topology, and hence weakly
> closed in L^1. Since U is convex, it is also norm
> closed in L^1.
>
> But I am not quite sure if the statement "the weak
> topology T_1 of L^1 induces a weaker topology than
> T_2 on L^2" is correct.

I think it is correct since the dual of L^1 (L^\infty) is a subspace of the dual of L^2.

TCL
From: TCL on 3 Dec 2007 22:32
> > > 1) The closure is U itself, due to the
> > >reflexivity
> > >of
> > > L^2.
> >
> > Let me elaborate; correct me if I am wrong. By the
> > reflexivity of L^2, U is weakly compact in the
> > weak
> > topology T_2 of L^2. Since the weak topology T_1
> >of
> > L^1 induces a weaker topology than T_2 on L^2, U
> >is
> > also weakly compact in T_1 topology, and hence
> >weakly
> > closed in L^1. Since U is convex, it is also norm
> > closed in L^1.

I was dumb. A weakly closed set is of course norm closed; convexity is not needed.

> >
> > But I am not quite sure if the statement "the weak
> > topology T_1 of L^1 induces a weaker topology
> >than
> > T_2 on L^2" is correct.

> I think it is correct since the dual of L^1
> (L^\infty) is a subspace of the dual of L^2.
>

In fact, the topology induced by T_1 on U is identical with T_2 on U, because a compact Hausdorff topology is minimal Hausdorff.

TCL
 | 
Pages: 1
Prev: Completing two statistics problems
Next: Question about Sarkovskii Ordering