Hello,
The claim should be okay for the almost everywhere. There's some thing dealing with equivalence classes.
Can't you use Lebesgue dominated theorem for your problem ?
Let be a sequence in .Suppose that and a.e. on .
Show that for all g .
is a Hilbert space,hence it is complete under the norm .So, if ,then .But I am not sure whether we can claim the same thing on almost everywhere converges sequence in Hilbert space.
Am I in the right way to prove the question?If I am wrong,can anyone please help me
Your question has been puzzling me...
Fatou's lemma gives , hence . (and not because of completeness)
I think the conclusion (what we want to prove) is equivalent to .
It would be tempting to apply the bounded convergence theorem, like Moo suggests, but there is no obvious dominating function. Sure we can write by Cauchy-Schwarz, but what we would need is for some integrable function .
If nobody can provide an answer on the forum, please tell us your teacher's solution when you get it.
I don't know if this is useful (I personally think I'm in the right direction, but there are some things missing) so I'm posting it here to see if anyone can do something with it:
Since it follows that there exists such that weakly i.e. for all . I'm stuck trying to prove and that in fact weakly.
I know it's not much, but at least it looks like what we're trying to prove. Hope it helps, I'll keep trying to figure it out.
wait, I just noticed that this is not valid, because what is bounded is the norm in of , and to use DCT you need to be bounded by an integrable function which doesn't follow from Hölder, the best you would get is (where is the norm in ). so DCT is not enough, at least not like that.
I also believe you are in the correct direction. The ball with radius M in is weakly compact, hence any "weaker" (less fine than the weak) Hausdorff topology agree with the weak topology in this ball. The point is showing that almost everywhere convergence implies a convergence in a topology with this characteristiscs, but unfortunately I don't see how to define such a topology. Perhaps it is trivial but at the moment I don't see.
Perhaps it is possible to answer the first question in your given argument. tends weakly to 0. Then I believe that taking charecteristic functions of bounded intervals implies almost everywhere convergence to zero of (using DCT in bounded intervals because the functions are essentially bounded in each interval by divided the lenghth of the interval) and then in . But at the moment I don't know how to deduce from this fact the convergence of the whole sequence.
This seems to be the right way indeed. More precisely, the fact that bounded sequences in have weakly convergent subsequences.
Let us prove that any weakly convergent subsequence of converges weakly toward (i.e. that is the only cluster point of the sequence for the weak topology). Then we will be able to conclude that converges weakly toward . Indeed, if not, then there would be and such that, for all , , which contradicts the existence of a subsequence of converging toward . This is a general feature: in a sequentially compact space (here, a closed -ball for weak topology), if a sequence has only one cluster point, then it converges (to the unique cluster point).
So, let's assume that a subsequence converges weakly toward some limit . In order to ease the writing, I shall take (or I could write and study ). We know that for all , and for almost-all .
Let us first reduce to a finite measure space (maybe there's a quicker way, I am biased by my acquaintance with probability). Let be a bounded interval of . We have of course for all measurable function on (taking outside ). Let . By Egoroff's theorem, there is a measurable subset such that (Lebesgue measure) and converges uniformly toward on . Then because of Lebesgue's bounded convergence theorem: (by uniform convergence) and (because the measure of is finite). Thus, we have both and (by applying our hypothesis to the function ), hence for all , and finally almost-everywhere in . This gives . Therefore, . In other words, for almost all in . Since is -finite (a big word for just writing ), we deduce finally from the finite measure case that for almost all . QED. If there is no mistake.