If ||f ||p= {òx |f|p dm}1/p is p norm of measurable complex function
' f ' , m denoting
the measure on set X ,what is the relationship between lim|| f ||p (p®¥) (if exists)
and || f ||¥ the essential supremum of f ?
I've never seen this stated, but I think when
Is this proof correct or is there a simpler one?
Let
Define for
Since
and
by definition of
(Note: getting a positive expression here so the next limit equals 1 is the whole reason for introducing the function This would not necessarily hold for )
Therefore
and letting
Thanks for questioning the proof. I think the limits are not interchangeable and the proof is flawed.
The goal of the proof is to show
If , then almost everywhere and
.
Then the desired limit has the indeterminate form
My argument using to try to get around this is invalid.
I don't see the solution and it seems to me to be a non-trivial complication. I find this odd because the proof (or a simpler version of it) is left as an exercise for the reader in a textbook I have.
I also think the solution is simple (maybe we have the same textbook? )
I e-mailed a friend (who happens to be a better mathematician ) and he says you are justified to interchange the limits, as is monotonic for .
Other than this, the expert says the limits can switch, as long as the quantity is nonnegative; If one of the double limits is infinity, the other will be the same. This last one, I must admit, has got me thinking
I think your friend is not looking at the problem in enough detail.
The basic goal is to prove
This means given , there exists such that implies
We can try using as my "proof" does that for fixed
and
but it does not help.
To see this, write using the triangle inequality that
Now guarantees that for each we can find such that
But there is no corresponding guarantee that
For that we would need for given that
Instead we have for given that
Oops.