distribution property

**yemino** · August 26th 2009, 12:58 PM

Hi,

I have the follow problem:

Let u a distribution on $\mathbb{R}^n$ have de property that $<u,\phi> \geq 0$ for all real valued nonnegative $\phi \in C_c^{\infty}(\mathbb{R}^n)$ . Show that $u$ is of order 0.

--edited!.. sorry about math formulas.

**nirax** · August 26th 2009, 08:25 PM

first let me rewrite your problem with correct formatting. Let u a distribution on R^n have de property that $<u,\phi> \geq 0$ for all real valued nonnegative $\phi \in C_c^{loc}(\mathbb{R}^n)$ . Show that $u$ is of order 0.

i will write my answer in a new reply.

**nirax** · August 26th 2009, 08:44 PM

Originally Posted by nirax

first let me rewrite your problem with correct formatting. Let u a distribution on R^n have de property that $<u,\phi> \geq 0$ for all real valued nonnegative $\phi \in C_c^{loc}(\mathbb{R}^n)$ . Show that $u$ is of order 0.

i will write my answer in a new reply.

observe one thing : you can write $|\phi|$ as a finite 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 $L$ in R. consider the square function with value 1 on this interval. call this function $h$ . take a sequence $\{f_n\}$ of smooth positive functions which increases and converges to this square function pointwise. convince yourself that it can always be done.

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

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

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

I hope you agree that your problem is solved.

**nirax** · August 26th 2009, 08:52 PM

Originally Posted by nirax

observe one thing : you can write $|u|$ 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 $h$ . take a sequence $\{f_n\}$ 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 $\{<u, f_n>\}$ 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 $g$ which is strictly larger than $h$ . convince yourself that it can always be done. observe that $<u, g>$ is a positive number which acts as an upper bound for our sequence.

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

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

I hope you agree that your problem is solved.

Can anybody point out why the formatting is not coming out proper ?
Thanks a lot.

**Opalg** · August 27th 2009, 12:08 AM

Originally Posted by nirax

Can anybody point out why the formatting is not coming out proper ?
Thanks a lot.

In writing "[tex] L [/math ]", you have inserted a space between "/math" and "]". Delete that space, and the compiler will recognise this as a TeX formula.

Originally Posted by nirax

first let me rewrite your problem with correct formatting. Let u a distribution on R^n have de property that $<u,\phi> \geq 0$ for all real valued nonnegative $\phi\inC_c^{loc}(\mathbb{R}^n)$ . Show that $u$ is of order 0.

The reason that this formula ([tex]\phi\inC_c^{loc}(\mathbb{R}^n)[/tex]) does not compile is that there should be a space between "\in" and "C". In this case, the compiler interprets "\inC" as a single (unknown) control sequence, which it regards as suspicious and therefore refuses to compile.

**nirax** · August 27th 2009, 12:56 AM

Thanks a lot !

**yemino** · August 27th 2009, 02:15 PM

Thanks nirax,

but I have a question:

Originally Posted by nirax

2. now because of the above mentioned property of your distribution, observe that the sequence of real numbers $\{<u, f_n>\}$ is a positive increasing one. if this sequence is bounded, it must have a limit.

How can I prove that $\{<u, f_n>\}_{n\in\mathbb{N}}$ is increasing?

thanks

**nirax** · August 27th 2009, 08:45 PM

Originally Posted by yemino

Thanks nirax,

but I have a question:

How can I prove that $\{<u, f_n>\}_{n\in\mathbb{N}}$ is increasing?

thanks

since $\{f_n\}_{n\in\mathbb{N}}$ is a pointwise increasing sequence of functions so the successive differences are always nonnegative. if you apply $u$ to this successive difference it will be nonnegative (why?). hence $\{<u, f_n>\}_{n\in\mathbb{N}}$ is a increaisng sequence of +ve real numbers.

**nirax** · August 28th 2009, 07:51 PM

in case you havent noticed the statement

observe one thing : you can write $|\phi|$ as a finite sum of smooth functions each of which is non negative.

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.

**yemino** · August 30th 2009, 05:34 AM

Originally Posted by nirax

in case you havent noticed the statement..

I was just thinking about it.

Sorry, can you explain me again all the idea?

I can't understand why you take a increasing function secuence.

Thanks..

**yemino** · September 13th 2009, 09:50 AM

Thanks, I fully understand yor hint.

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