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 ?

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- Aug 25th 2007, 04:38 AMmazaheriP norm and essential supremum

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 ?

- Aug 25th 2007, 05:16 PMJakeD
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

- Aug 26th 2007, 01:18 PMRebesquesQuote:

is there a simpler one?

Not that I know of. :confused:

I only have the same question every time: Why are the limits for epsilon and for p interchangeable? :rolleyes::eek: - Aug 26th 2007, 07:36 PMJakeD
- Aug 27th 2007, 02:46 PMJakeD
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. - Aug 29th 2007, 12:40 AMRebesques
I also think the solution is simple (maybe we have the same textbook? :p)

I e-mailed a friend (who happens to be a better mathematician :o:() 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 :( - Aug 29th 2007, 03:23 PMJakeD
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. :(