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.