# Principal Valued Distributions

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.

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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

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