Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently studying applied functional analysis and I see a proof about principal valued distributions. It is easy to prove that $x \times P/x =1$, where $P/x$ is the principal valued distribution. Interestingly, the other direction is very similar as well. If $f$ is a distribution and $xf=1$, then $f(x) = P/x + C\delta(x)$. The starting proof strategy is intuitive, and this is how it goes:

Choose a test function $\Theta$ such that $\Theta(0)=1$. Then for any test function $\phi$, $\phi(x) -\phi(0) \Theta(x) \rightarrow 0$ as $x\rightarrow 0$, and we know that $k(x)=[\phi(x) - \phi(0)\Theta(0)]/x$ is a test function and doesn't blow up at $x=0$. Then,

\begin{align}\int f(x) \phi(x) dx &= \int f(x) [xk(x) + \phi(0)\Theta(0)] dx\\\ &= \int k(x)dx + \phi(0)\int f(x) \Theta(x) dx\\\ &=\int P/x [\phi(x) - \phi(0)\Theta(x)]dx + \phi(0) \int f(x) \Theta(x) dx\\\ &= \int P/x \phi(x)dx + \phi(0) \int [f(x) - P/x] \Theta(x) dx\\\ &=\int [P/x + C\delta(x)]\phi(x) dx \end{align}

I am having trouble understanding the ultimate step here. I see the importance of the fact that $\Theta(0)=1$, because I think the proof is showing $[f(x) - P/x] = C\delta(x)$, in which case

$$\phi(0) \int [f(x) - P/x] \Theta(x) dx = \phi(0) \Theta(0) = \phi(0),$$ but I do not see it.

share|improve this question
    
You simply need to set $C:= \int [f(x) - P/x] \theta(x) dx$ and you are done. –  Vobo May 5 '12 at 6:25
    
Gratzi! very simple –  Thiagarajan May 5 '12 at 7:03
    
So then I posted it as an answer. –  Vobo May 5 '12 at 10:46
    
And please keep the tag "distribution theory". –  Vobo May 5 '12 at 10:51
add comment

1 Answer

You simply need to set $C:= \int [f(x) - P/x] \Theta(x) dx$ and you are done.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.