The Hilbert transform is bounded on Lp

Size and smoothness of the kernel. $\mathrm{\mid }k(x)\mathrm{\mid }=1\mathrm{/}(\pi \mathrm{\mid }x\mathrm{\mid })$ for $x\mathrm{\ne }0$, which is the pointwise size bound with constant $1\mathrm{/}\pi$; and for $\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }>0$ the difference is $\mathrm{\mid }k(x-y)-k(x)\mathrm{\mid }=\frac{1}{\pi }\mid \frac{1}{x-y}-\frac{1}{x}\mid =\frac{\mathrm{\mid }y\mathrm{\mid }}{\pi \mathrm{\mid }x\mathrm{\mid }\mathrm{\mid }x-y\mathrm{\mid }}\le \frac{2}{\pi }\frac{\mathrm{\mid }y\mathrm{\mid }}{\mathrm{\mid }x{\mathrm{\mid }}^2}$, because $\mathrm{\mid }x-y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }-\mathrm{\mid }y\mathrm{\mid }\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$. Hence $k$ satisfies the standard $1$-Hölder condition with constant $A_2^{\mathrm{\prime }}=2\mathrm{/}\pi$.

The smooth kernel is measurable and locally integrable away from zero. Its oddness gives $k(r)+k(-r)=0$, and [F5] gives ${\int }_{R\le \mathrm{\mid }x\mathrm{\mid }\le 2R}\mathrm{\mid }k(x)\mathrm{\mid }dx=(2\mathrm{/}\pi ){\int }_R^{2R}dr\mathrm{/}r=2\log 2\mathrm{/}\pi$. For each $y\mathrm{\ne }0$, directly integrating the estimate of step 1.1 gives ${\int }_{\mathrm{\mid }x\mathrm{\mid }\ge 2\mathrm{\mid }y\mathrm{\mid }}\mathrm{\mid }k(x-y)-k(x)\mathrm{\mid }dx\le (2\mathrm{/}\pi )\mathrm{\mid }y\mathrm{\mid }2{\int }_{2\mathrm{\mid }y\mathrm{\mid }}^{\mathrm{\infty }}{r}^{-2}dr=2\mathrm{/}\pi$. Thus the base kernel conditions hold with $A_1=2\log 2\mathrm{/}\pi$ and $A_2=2\mathrm{/}\pi$; together with step 1.1 they establish standard $1$-Hölder status with $A_2^{\mathrm{\prime }}=2\mathrm{/}\pi$.

$H$ is a Calderón–Zygmund operator with kernel $k$ and $L^2$ norm $B=1$. It is $L^2$-bounded with norm one by [F2], so it remains to prove the off-support representation. Let $f\in L^2$ be compactly supported, let $\phi \in C_c^{\mathrm{\infty }}$ be supported off $\mathrm{supp}f$, and put $d:=\mathrm{dist}(\mathrm{supp}f,\mathrm{supp}\phi )>0$. By [F2] and the reality of $k$, $⟨Hf,\phi ⟩=-\int f(y)\overline{H\phi (y)}dy=-\frac{1}{\pi }\iint \frac{f(y)\overline{\phi (x)}}{y-x}dxdy,$ where the inner limit defining $\overline{H\phi (y)}=\frac{1}{\pi }\mathrm{p}\mathrm{.}\mathrm{v}\mathrm{.}\int \frac{\overline{\phi (x)}}{y-x}dx$ is an absolutely convergent integral because $\mathrm{\mid }x-y\mathrm{\mid }\ge d>0$ on $\mathrm{supp}f\times \mathrm{supp}\phi$; the double integral is absolutely convergent over the bounded supports, so Fubini's theorem may be applied and the sign of the denominator changed: $⟨Hf,\phi ⟩=\frac{1}{\pi }\iint \frac{f(y)\overline{\phi (x)}}{x-y}dxdy=\int (\int k(x-y)f(y)dy)\overline{\phi (x)}dx,$ the last equality by the definition $k(x-y)=1\mathrm{/}(\pi (x-y))$ and Fubini. Since the pairing against every test function supported off $\mathrm{supp}f$ determines the $L^2$ class off that support, the $L^2$ function $Hf$ agrees almost everywhere off $\mathrm{supp}f$ with the locally integrable function $x\mapsto \int k(x-y)f(y)dy$, which is the representation (3) required of a Calderón–Zygmund operator.

Applying the strict-range theorem [F3] to the Calderón–Zygmund operator $H$ with constants $A_1=2\log 2\mathrm{/}\pi$, $A_2=2\mathrm{/}\pi$ and $B=1$ yields a unique bounded extension of $H$ to $L^p(\mathbb{R};\mathbb{C})$ for every $1<p<\mathrm{\infty }$ with $\mathrm{\parallel }Hg{\mathrm{\parallel }}_p\le C_{1,p}(1+2\mathrm{/}\pi )\max (p,(p-1{)}^{-1})\mathrm{\parallel }g{\mathrm{\parallel }}_p$; since the dimension is one, $C_{1,p}$ depends only on $p$. This is the assertion.