Re: concentration / compactness

On Mon, 30 Jul 2007 08:49:18 EDT, craigt <ctcowan1972@xxxxxxxxx>
wrote:

On Mon, 30 Jul 2007 03:03:14 EDT, craigt
<ctcowan1972@xxxxxxxxx>
wrote:

On Sat, 28 Jul 2007 17:52:56 EDT, craigt
<ctcowan1972@xxxxxxxxx>
wrote:
[...]

I have a new question:

Take C to be the uniformly continuous functions on
Omega and C* its dual.


Take Omega open in R^n and assume that f_m -> f a.e.
in Omega and that

f_m -> \mu (signed measure) weak* in C*.

Now suppose one can show \mu is absolutely
continuous with respect to n-dimeansional Lebesgue
measure.

I would be tempted to say that f_m -> f in L^1. An
example shows this is false.

Example: Take f_m to be the difference of two
delta function approximations at the same point. So
one can show that f_m -> 0 weak* in C* and one can
also show that f_m does not converge weakly in L^1.


So here is the question:

Suppose 0 \le f_m -> f a.e. in Omega and

f_m -> \mu weak* in C* where \mu is absolutely
continuous with respect to n-dimeansional Lebeqgue
measure.

Does f_m -> f in L^1. ((I can prove it does but my

proof might be flawed.))


The answer to this question is no. It seems possible
that this
question is not exactly what you meant to ask: Are we
also
supposed to assume that d mu = f dx?

Nope.

I also should of added we are on a bounded set but your example works in that case...


I agree your example shows what I want is false. But I still can't see the flaw in my proof.

If you wouldn't mind could you point to where the flaw is in my proof?


I will sketch it here. Lets work in say [0,1].


So assume 0 <=f_m -> f a.e. and f_m dx -> g dx weak*

where g is L^1. Let a < b. Let 0 <t (small), 0 <= h_t <=1 denote a continouos function with h_t =1 on [a,b] and which is supported in [a-1/t, b+ 1/t].

That last looked wrong, but I see it was a typo that you've already
corrected.

Then by taking a limit in m and then t one has

limsup_m \int_a^b f_m dx <= \int_a^b g dx.

Seems right - a fairly standard thing, iirc.
(Also not contradicted by the counterexample I gave.)

Now one sees we can replace (a,b) with any open set.


Let e >0 but small. By Egoroff there is some closed set E with |E^c| < e such that f_m -> f uniformly on E.


So we have

limsup_m \int f_m <= \int_E f + \int_{E^c} g

This is the first assertion that I noticed is simply false
given the counterexample. So the error must be somewhere
above. Took me a while to find it:

The error is where you say "Now one sees we can replace (a,b)
with any open set." That didn't seem so clear when I thought
about the example, but it obviously follows from the inequality
for (a,b).

Except that it doesn't follow. To get from (a,b) to an
arbitrary open set you're using the obvious inequality
limsup(sum) <= sum(limsup). And that inequality is false
for _infinite_ sums. For example, let a_{j,n] = 1 when
j = n, 0 otherwise. Then limsup_j(sum_n a_{j,n]) = 1
while sum_n(limsup_j a_{j,n]) = 0.

and where the second integral goes to zero as e-> 0.


So we have

limsup_m \int f_m <= \int f and now standard methods

show f_m -> f in L^1.







If that was
supposed to
be part of the question then I'm not sure yet.

Counterexample for the question as you asked it:

Choose a_m > 0 with sum a_m < infinity.
Let E_m be the union of the m intervals

[j/m, (j+a_m)/m],

for 1 <= j <= m. So |E_m| = a_m.

Let f_m = 1/a_m on E_m, 0 elswhere. Then
f_m -> 0 almost everywhere since sum |E_n| <
infinity.
But f_m -> the characteristic function of [0,1]
weak*.

thanks for the help.


craig









In any case I have an extra condition on the
$u_m$
which I did'nt tell you about gives me the desired
negative result.


I supect the answer in the full generality that I
asked is it can't ((or at least thats what scaling
suggests))




thanks

craig







(Or something like that...)

thanks


craig


************************

David C. Ullrich


************************

David C. Ullrich


************************

David C. Ullrich


************************

David C. Ullrich
.

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