I'm trying to understand a majorization that the professor has given of the Cauchy principal value.
The Cauchy principal value is a distribution defined as $$ \left<v_p\left(\frac{1}{x}\right), \phi \right> = \lim_{\epsilon \to 0} \int_{|x|\geq \epsilon} \frac{\phi(x)}{x}dx $$
By the mean value theorem, we have
for $|x| \leq 1$: $$\left|\frac{\phi(x)-\phi(0)}{x}\right| \leq \left| \frac{\phi'(\xi)(x-0)}{x} \right| \leq \sup_{|x|\leq 1}{|\phi'(x)|}$$ and for $|x| \geq 1$: $$\left|\frac{\phi(x)}{x}\right| \leq \sup_{|x| \geq1}|\phi(x) | $$
Thus (and this is the part I don't understand) $$ \left|\int_{|x|\geq \epsilon} \frac{\phi(x)}{x}\,dx\right| \leq \int_{1 \geq |x| \geq \epsilon} \left|\frac{\phi(x) - \phi(0)}{x}\right| \,dx + \int_{R \geq |x| \geq 1} \left|\frac{\phi(x)}{x}\right|\,dx \leq 2 R \sup_{|x| \leq R}{\left(|\phi(x)| + |\phi'(x)|\right)} $$ I don't understand why we have to multiply by $2R$. Thanks!
This is a rough bound.
$$ \int_{R \geq |x| \geq 1} |\frac{\phi(x)}{x}|dx \leq \int_{R \geq |x| \geq 1} \sup |\phi| dx \leq \sup |\phi|\int_{R \geq |x|} dx = 2R \sup |\phi|. $$