Originally Posted by

**nirax** observe one thing : you can write $\displaystyle |u|$ as a sum of smooth functions each of which is non negative.

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 [tex] L [/math ] in R. consider the square function with value 1 on this interval. call this function $\displaystyle h$. take a sequence $\displaystyle \{f_n\}$ of smooth positive functions which increases and converges to this square function pointwise. convince yourself that it can always be done.

2. now since the above mentioned property of your distribution, observe that the sequence of real numbers $\displaystyle \{<u, f_n>\}$ is an 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 [tex] L [/math ] and take a smooth positive function $\displaystyle g$ which is strictly larger than $\displaystyle h$. convince yourself that it can always be done. observe that $\displaystyle <u, g>$ is a positive number which acts as an upper bound for our sequence.

4. call the limit of the sequence as $\displaystyle C_L$.

5. now suppose you given a smooth test function $\displaystyle \phi$ in $\displaystyle L$, convince yourself that $\displaystyle |<u, \phi>| \leq C_L\max{\{|\phi(x)| : x \in L\}}$ by breaking $\displaystyle |\phi|$ 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 $\displaystyle R^n$

I hope you agree that your problem is solved.