## Tag Archives: Hilbert transform

### Various proofs of Hilbert transform: Hardy’s inequality, Minkowski integral inequality

$\mathbf{Problem:}$ For $x,y>0$, define

$\displaystyle Tf(x)=\int_0^\infty \frac{f(y)}{x+y}dy$

prove that $T:L^p\to L^p$ is a bounded operator.

Proof by Hardy’s inequality

Hardy’s inequality: Let $1\leq p<\infty$ and $f\geq 0$

1. $\forall\, \alpha<-1$

$\displaystyle \int_0^\infty\left(\int_0^x f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+|\alpha|}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt$

2. $\forall\, \alpha>-1$

$\displaystyle \int_0^\infty\left(\int_x^\infty f(t)dt\right)^px^\alpha dx\leq \left(\frac{p}{1+\alpha}\right)^p\int_0^\infty f^p(t)t^{\alpha+p}dt$

$\mathbf{Proof:}$ Split $Tf(x)$ as $\displaystyle Tf(x)=\int_0^x \frac{f(y)}{x+y}dy+\int_{x}^\infty\frac{f(y)}{x+y}dy$. Then

$\displaystyle \left\|\int_0^x \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_0^x \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\frac{1}{x}\int_0^x |f(y)|dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy$

Here we used Hardy’s inequality case (1) with $\alpha=-p<-1$.

$\displaystyle \left\|\int_x^\infty \frac{f(y)}{x+y}dy\right\|_p^p=\int_0^\infty \left|\int_x^\infty \frac{f(y)}{x+y}dy\right|^pdx\leq \int_0^\infty \left(\int_x^\infty\frac{|f(y)|}{y}dy\right)^pdx\leq C_p\int_0^\infty|f(y)|^pdy$

Here we used Hardy’s inequality case (2) with $\alpha=0$.

So $\displaystyle \|Tf\|_p\leq \left\|\int_0^x\frac{f(y)}{x+y}dy\right\|_p+\left\|\int_x^\infty\frac{f(y)}{x+y}dy\right\|_p\leq C\|f\|_p$.

$\text{Q.E.D}\hfill \square$

Proof by Minkowski integral inequality

$\mathbf{Proof:}$ Changing variable by $y=tx$

$\displaystyle \int_0^\infty \frac{f(y)}{x+y}dy=\int_0^\infty\frac{f(tx)}{1+t}dt$

So by Minkowski Integral Inequality

$\displaystyle \|Tf\|_p=\left( \int_0^\infty \left(\int_0^\infty \frac{f(tx)}{1+t}dt\right)^pdx\right)^\frac{1}{p}\leq \int_0^\infty\left(\int_0^\infty \left|\frac{f(tx)}{1+t}\right|^pdx\right)^\frac{1}{p}dt\\=\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}\left(\int_0^\infty|f(y)|^p\right)^\frac{1}{p}dt=\|f\|_p\int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt$

Since $p>1$, $\displaystyle \int_0^\infty \frac{1}{(1+t)t^\frac{1}{p}}dt<\infty$. We are done.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$

Pay attention to the tricks using here. The proof by Hardy’s inequality is hinted from Prof. Chanillo’s notes. For Minkowski method, see stein’s book.