I have the follow problem:
Let u a distribution on have de property that for all real valued nonnegative . Show that is of order 0.
--edited!.. sorry about math formulas.
first let me rewrite your problem with correct formatting. Let u a distribution on R^n have de property that for all real valued nonnegative . Show that is of order 0.
i will write my answer in a new reply.
the following analysis becomes simpler to express if I assume that n=1, but the idea is valid for all n.
1. take any connected closed interval in R. consider the square function with value 1 on this interval. call this function . take a sequence of smooth positive functions which increases and converges to this square function pointwise. convince yourself that it can always be done.
2. now because of the above mentioned property of your distribution, observe that the sequence of real numbers is a positive increasing one. if this sequence is bounded, it must have a limit.
3. to show that it is bounded, construct a larger interval, which completely encompasses your chosen interval and take a smooth positive function which is strictly larger than . convince yourself that it can always be done. observe that is a positive number which acts as an upper bound for our sequence.
4. call the limit of the sequence as .
5. now suppose you r given a smooth test function in , convince yourself that by breaking into a finite sum of nonnegative smooth functions as pointed out at the begining of this reply.
6. Now generalize this process to an arbitrary compact subset of R.
7. generalize this to arbitrary
I hope you agree that your problem is solved.
in case you havent noticed the statement
isn't quite correct. the sum may not be finite. but all the pieces have supports which intersect only at the boundaries if at all. so the argument remains unchanged by replacing the finite sum with infinite sum if required.observe one thing : you can write as a finite sum of smooth functions each of which is non negative.