Principal Value of 1/x Proof

Question

How can we formally show that the Cauchy principal value of the function y(x)=1/x is a distribution, I understand that a distribution is a continuous linear functional on spaces of test functions, but I would like to know how to formally define the 'principal value' of the function as a distribution, and why is it a distribution perhaps in an intuitive sense, thank you!

Answer 1

Okay, so we define

$$ \langle PV \frac{1}{x},\varphi\rangle=\lim_{\varepsilon\to 0^+} \int_{\mathbb{R}\setminus (-\varepsilon,\varepsilon)} \frac{\varphi(x)}{x}\textrm{d}x $$ for $\varphi\in C_c^{\infty}$. I don't really know if there's much "intuition" about why this should define anything continuous beyond doing calculations. The calculations, however, are not too bad.

For any $\varepsilon>0$, we have

$$\int_{-1}^{-\varepsilon} \frac{\varphi(x)}{x}\textrm{d}x+\int_{\varepsilon}^{1} \frac{\varphi(x)}{x} \textrm{d}x=\int_{\varepsilon}^1 \frac{\varphi(x)-\varphi(-x)}{x}\textrm{d}x, $$ and applying the mean value theorem (and the axiom of choice I guess), there is some function $\xi:(0,1)\to (0,1)$ such that $\xi(x)<x$ and

$$ \frac{\varphi(x)-\varphi(-x)}{x}=2\varphi'(\xi(x)) $$

and thus, $\int_{0}^1 \left|\frac{\varphi(x)-\varphi(-x)}{x} \right|\textrm{d}x=\int_0^1 \left|2\varphi'(\xi(x))\right|\textrm{d}x\leq 2\|\varphi'\|_{\infty}<\infty$.

I guess at this point, you might say that the role of the Mean Value Theorem gives some sort of intuition: Basically the sign change of $\frac{1}{x}$ kills the integral around $0$ so the singularity does not cause any problems.

All in all, we can now apply the Dominated Convergence Theorem to get that

$$ \langle PV \frac{1}{x},\varphi\rangle=\int_{\mathbb{R}\setminus (-1,1)} \frac{\varphi(x)}{x}\textrm{d}x+\int_0^1 \frac{\varphi(x)-\varphi(-x)}{x}\textrm{d}x, $$ defines a linear functional of $\varphi$. Furthermore, we see that if $supp(\varphi)\subseteq [-K,K]$, then

$$ |\langle PV \frac{1}{x}, \varphi\rangle|\leq \|\varphi\|_{\infty} \| 1_{(K\geq|x|\geq 1)}\frac{1}{x}\|_{L^1}+ 2\|\varphi'\|_{\infty}, $$ so this defines a continuous functional on $C_c^{\infty}$.

Answer 2

Here's another approach.

Since $\ln|x| \in L^1_{\text{loc}}(\mathbb R)$ it defines a distribution through $$\langle \ln|x|, \varphi \rangle := \int \ln|x| \, \varphi(x) \, dx = \lim_{\epsilon \to 0}\int_{|x|>\epsilon} \ln|x| \, \varphi(x) \, dx.$$

Then we show that $\text{pv}\frac{1}{x}$ is the distributional derivative of $\ln|x|$: $$\begin{align} \int_{|x|>\epsilon} \frac{1}{x} \, \varphi(x) \, dx &= \int_{-\infty}^{-\epsilon} (\ln|x|)' \, \varphi(x) \, dx + \int_{\epsilon}^{\infty} (\ln|x|)' \, \varphi(x) \, dx \\ &= [\ln|x| \, \varphi(x)]_{-\infty}^{-\epsilon} - \int_{-\infty}^{-\epsilon} \ln|x| \, \varphi'(x) \, dx + [\ln|x| \, \varphi(x)]_{\epsilon}^{\infty} - \int_{\epsilon}^{\infty} \ln|x| \, \varphi'(x) \, dx \\ &= \ln\epsilon \, \varphi(-\epsilon) - \int_{-\infty}^{-\epsilon} \ln|x| \, \varphi'(x) \, dx - \ln\epsilon \, \varphi(\epsilon) - \int_{\epsilon}^{\infty} \ln|x| \, \varphi'(x) \, dx \\ &= - \epsilon\ln\epsilon \frac{\varphi(\epsilon) - \varphi(-\epsilon)}{\epsilon} - \int_{|x|>\epsilon} \ln|x| \, \varphi'(x) \, dx \\ &\to 0 \cdot 2\varphi'(0) - \int \ln|x| \, \varphi'(x) \, dx \\ &= - \langle \ln|x|, \varphi' \rangle \\ &= \langle (\ln|x|)', \varphi \rangle \end{align}$$

From this it follows that $\text{pv}\frac{1}{x}$ is a distribution.