Hilbert transform of the line Poisson kernel

Let $q(u):=e^{-2\pi a\mathrm{\mid }u\mathrm{\mid }}$ for $u\in \mathbb{R}$. Then $q$ is continuous and real-valued: $u\mapsto \mathrm{\mid }u\mathrm{\mid }$ is continuous, so is $u\mapsto -2\pi a\mathrm{\mid }u\mathrm{\mid }$, and the composite with the continuous exponential is continuous [F13]; in particular $q$ is Borel measurable [F11]. For $R>0$ the second fundamental theorem [F6] applied on $[0,R]$ to the antiderivative $u\mapsto -e^{-2\pi au}\mathrm{/}(2\pi a)$ gives ${\int }_0^R{e}^{-2\pi au}du=\frac{1-e^{-2\pi aR}}{2\pi a}$, and $e^{-2\pi aR}\le \frac{1}{1+2\pi aR}\to 0$ as $R\to \mathrm{\infty }$ by [F13]; hence the improper Riemann integral of the nonnegative continuous function $q$ over $[0,\mathrm{\infty })$ converges to $\frac{1}{2\pi a}$, and [F6] makes $q$ Lebesgue integrable on $(0,\mathrm{\infty })$ with ${\int }_{(0,\mathrm{\infty })}qd\lambda =\frac{1}{2\pi a}$. The function $q{\chi }_{[0,\mathrm{\infty })}$ is already integrable by [F6]. Apply [F12] to this function and the reflection $T(u)=-u$, whose Jacobian has absolute value one: its pullback $q{\chi }_{(-\mathrm{\infty },0]}$ is integrable and has the same integral $1\mathrm{/}(2\pi a)$ (the singleton $\{0\}$ has measure zero). Thus $q$ is the sum of two known integrable functions $q{\chi }_{(-\mathrm{\infty },0]}$ and $q{\chi }_{(0,\mathrm{\infty })}$, so $q\in L^1(\mathbb{R})$ before applying additivity [F11], which gives ${\int }_{\mathbb{R}}qd\lambda =\frac{1}{\pi a}<\mathrm{\infty }$, that is, $q\in L^1(\mathbb{R})$. For every ball $B$ monotonicity [F12] gives ${\int }_{B}qd\lambda \le {\int }_{\mathbb{R}}qd\lambda <\mathrm{\infty }$, so $q\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$ as well.

For the integrability of $P_a$ used repeatedly below, note that $0<P_a(y)\le 1\mathrm{/}(\pi a)$ for all $y$, while $0<P_a(y)\le a\mathrm{/}(\pi y^2)$ for $y\mathrm{\ne }0$, because $y^2\le a^2+y^2$; hence $P_a\le \min (1\mathrm{/}(\pi a),a\mathrm{/}(\pi y^2))$ pointwise. By [F6] the continuous bounded function $P_a$ is integrable over $[-a,a]$, and the improper integrals ${\int }_a^{\mathrm{\infty }}a\mathrm{/}(\pi y^2)dy$ and ${\int }_a^{\mathrm{\infty }}{a}^2\mathrm{/}({\pi }^2{y}^4)dy$ converge by the second fundamental theorem applied to the antiderivatives $-a\mathrm{/}(\pi y)$ and $-a^2\mathrm{/}(3{\pi }^2{y}^3)$ with vanishing limits at infinity. Reflecting the already integrable positive-tail majorants by [F12] gives the corresponding negative-tail bounds. Together with integrability on $[-a,a]$, these bounds give $P_a\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, with ${\int }_{\mathbb{R}}\mathrm{\mid }{P}_a\mathrm{\mid }\le 4\mathrm{/}\pi$ and ${\int }_{\mathbb{R}}{P}_a^2\le 8\mathrm{/}(3{\pi }^{2}a)$.

Substituting $y=x-t$ in the displayed truncation of [F1] shows that for every $x$, every $0<\epsilon <R$ and every $f\in L^1(\mathbb{R})$, $\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }t\mathrm{\mid }<R}\frac{f(x-t)}{t}dt=\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R}\frac{f(y)}{x-y}dy$. Subtracting the constant $f(x)$, whose integral against $1\mathrm{/}(x-y)$ vanishes over the symmetric domain $\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R$ (the substitution $u=y-x$ makes the integrand odd), gives the identity $\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R}\frac{f(y)}{x-y}dy=\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R}\frac{f(y)-f(x)}{x-y}dy$, valid when $f$ is bounded near $x$; the subtraction changes no value.

For $\xi \in \mathbb{R}$ put $z:=2\pi (a+i\xi )$, so that $\mathrm{Re}z=2\pi a>0$ and $\mathrm{\mid }z\mathrm{\mid }\ge 2\pi a$, and let $u(t):=-z^{-1}{e}^{-zt}$ for $t\in \mathbb{R}$. By [F10], $e^{-zt}=e^{-2\pi at}(\cos (2\pi \xi t)-i\sin (2\pi \xi t))$, and differentiating the two real components with the product, chain, trigonometric and exponential derivative rules of [F10] gives $\frac{d}{dt}{e}^{-zt}=-z{e}^{-zt}$, so $u$ is complex $C^1$ on $\mathbb{R}$ with $u^{\mathrm{\prime }}(t)=e^{-zt}$; the complex fundamental theorem of calculus [F10] on $[0,R]$ then gives ${\int }_0^R{e}^{-zt}dt=u(R)-u(0)=\frac{1-e^{-zR}}{z}$, while $\mathrm{\mid }{e}^{-zR}\mathrm{\mid }=e^{-2\pi aR}\le \frac{1}{1+2\pi aR}\to 0$ by [F10] and [F13], so the truncated integrals converge to $1\mathrm{/}z$. Moreover $\mathrm{\mid }{e}^{-zt}\mathrm{\mid }=q(t)$ for $t\ge 0$ and $q$ is Lebesgue integrable on $(0,\mathrm{\infty })$ with ${\int }_{(0,\mathrm{\infty })}qd\lambda =\frac{1}{2\pi a}$ by step 1.1, so dominated convergence [F8] applied to the functions $1_{[0,R]}{e}^{-z\cdot }$, which converge pointwise to $e^{-z\cdot }$ and are dominated by $q$, gives

Fix $x\in \mathbb{R}$. Since $q$ is continuous at $x$ [F13], for every $\epsilon >0$ there is $\delta >0$ such that $\mathrm{\mid }q(y)-q(x)\mathrm{\mid }<\epsilon$ whenever $\mathrm{\mid }y-x\mathrm{\mid }<\delta$; for $0<r<\delta$ the pointwise bound $\mathrm{\mid }q-q(x)\mathrm{\mid }{\chi }_{B(x,r)}\le \epsilon {\chi }_{B(x,r)}$, the monotonicity and homogeneity of the nonnegative integral, and the value ${\int }_{\mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}}\epsilon {\chi }_{B(x,r)}d\lambda =\epsilon \lambda (B(x,r))$ of the simple integral [F12] give, since $\lambda (B(x,r))$ is positive and finite [F11] and $q\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$ by step 1.1, that the ball average $A_r(\mathrm{\mid }q-q(x)\mathrm{\mid })(x)$ of [F11] satisfies $A_r(\mathrm{\mid }q-q(x)\mathrm{\mid })(x)\le \epsilon$ for every $0<r<\delta$

Applying step 1.3 to $f=P_a$ and using

For the $L^2$ assertion, note that $\mathrm{\mid }{Q}_a(y)\mathrm{\mid }=\frac{\mathrm{\mid }y\mathrm{\mid }}{\pi (a^2+y^2)}\le \frac{1}{2\pi a}$ for all $y$ and $\mathrm{\mid }{Q}_a(y)\mathrm{\mid }\le \frac{1}{\pi \mathrm{\mid }y\mathrm{\mid }}$ for $\mathrm{\mid }y\mathrm{\mid }\ge a$; the same elementary integration as in step 1.2, by [F6], gives $Q_a\in L^2(\mathbb{R})$.

For $j\ge 1$ define ${\psi }_j(y):=P_a(y)\chi (y\mathrm{/}j)$ with $\chi$ as in [F3]. Each ${\psi }_j$ lies in $C_c^{\mathrm{\infty }}(\mathbb{R})$ and hence in $\mathcal{S}(\mathbb{R})$ by [F5], with $0\le {\psi }_j\le P_a$; and ${\psi }_j(y)=P_a(y)$ as soon as $j\ge \mathrm{\mid }y\mathrm{\mid }$, so ${\psi }_j\to P_a$ pointwise everywhere. Since $\mathrm{\mid }{\psi }_j-P_a\mathrm{\mid }\le 2P_a$ with $P_a\in L^1\cap L^2$ by step 1.2, dominated convergence [F8] gives $\mathrm{\parallel }{\psi }_j-P_a{\mathrm{\parallel }}_1\to 0$ and $\mathrm{\parallel }{\psi }_j-P_a{\mathrm{\parallel }}_2\to 0$.

By the definition of the transform [F4], $\widehat{q}(\xi )={\int }_{\mathbb{R}}q(t)e^{-2\pi i\xi t}d\lambda (t)$ for every $\xi$; the integrand $h_{\xi }:=q{e}^{-2\pi i\xi \cdot }$ satisfies $\mathrm{\mid }{h}_{\xi }\mathrm{\mid }=q\in L^1(\mathbb{R})$ by step 1.1, so $h_{\xi }$ is integrable and its indefinite integral is countably additive on pairwise disjoint measurable families [F11]. Splitting over the disjoint measurable sets $(-\mathrm{\infty },0]$ and $(0,\mathrm{\infty })$, which cover $\mathbb{R}$, and applying the reflection change of variables [F12] to the integrable function $h_{\xi }{\chi }_{(-\mathrm{\infty },0]}$, whose reflection is $q{\chi }_{(0,\mathrm{\infty })}{e}^{2\pi i\xi \cdot }$ because $q$ is even, gives, using step 2.1 on each half-line and step 2.1 again with $\xi$ replaced by $-\xi$,

Put $G(y):=\frac{x}{a}\arctan \frac{y}{a}+\frac{1}{2}\log (a^2+y^2)$. By the chain rule, the arctangent and logarithm derivatives of [F7], and [F9], $G$ is differentiable on $\mathbb{R}$ with $G^{\mathrm{\prime }}(y)=\frac{x}{a^2+y^2}+\frac{y}{a^2+y^2}=\frac{x+y}{a^2+y^2}$. Since the domain $\{\epsilon <\mathrm{\mid }x-y\mathrm{\mid }<R\}$ is the disjoint union of the intervals $(x-R,x-\epsilon )$ and $(x+\epsilon ,x+R)$ on which $y\mapsto (x+y)\mathrm{/}(a^2+y^2)$ is continuous, [F6] and the right-hand integral of step 2.3 give

Fix $x\in \mathbb{R}$ and $j>\mathrm{\mid }x\mathrm{\mid }$, and use the functions ${\psi }_j$ of step 2.5. For Schwartz ${\psi }_j$, [F2] represents the principal value at $x$ by the two-piece pairing, and the oddness cancellation of step 1.3 identifies it with $(W\ast {\psi }_j)(x)=\frac{1}{\pi }{\int }_{\mathrm{\mid }t\mathrm{\mid }>1}\frac{{\psi }_j(x-t)}{t}dt+\frac{1}{\pi }{\int }_{\mathrm{\mid }t\mathrm{\mid }<1}\frac{{\psi }_j(x-t)-{\psi }_j(x)}{t}dt$; combining this with the same identity for $P_a$ in step 1.3, and abbreviating ${\delta }_j:={\psi }_j-P_a$, gives for every $0<\epsilon <1$

By [F2] the isometry $H$ is defined on $L^2$ and is linear, so $\mathrm{\parallel }H{\psi }_j-H{P}_a{\mathrm{\parallel }}_2=\mathrm{\parallel }{\psi }_j-P_a{\mathrm{\parallel }}_2\to 0$ by step 2.5; that is, $H{\psi }_j\to H{P}_a$ in $L^2(\mathbb{R})$.

Both $q\in L^1(\mathbb{R})$ (step 1.1) and $\widehat{q}=P_a\in L^1(\mathbb{R})$ (step 1.2, step 3.1) are integrable, so the inversion theorem [F4] applied to $f:=q$ gives a bounded continuous function $g(x)={\int }_{\mathbb{R}}{P}_a(\xi )e^{2\pi ix\xi }d\lambda (\xi )$ that agrees with $q$ almost everywhere and agrees with $q(x)$ at every Lebesgue point $x$ of $q$; every real $x$ is such a point by step 2.2, so $g(x)=e^{-2\pi a\mathrm{\mid }x\mathrm{\mid }}$ everywhere, and writing the defining integral of $\widehat{P_a}$ [F4] at the frequency $-x$ identifies $g(x)=\widehat{P_a}(-x)$, so $\widehat{P_a}(-x)=e^{-2\pi a\mathrm{\mid }x\mathrm{\mid }}$ for every $x$, and replacing $x$ by $-\xi$ gives $\widehat{P_a}(\xi )=e^{-2\pi a\mathrm{\mid }\xi \mathrm{\mid }}$ for every $\xi$; this proves assertion 1.

Since $P_a\in L^1$ by step 1.2, for each fixed $\epsilon >0$ the full integral $H_{\epsilon }{P}_a(x)=\frac{1}{\pi }{\int }_{\mathrm{\mid }t\mathrm{\mid }>\epsilon }{P}_a(x-t)\mathrm{/}tdt$ converges absolutely and is the limit of its truncations at $R\to \mathrm{\infty }$; hence passing to the limit $R\to \mathrm{\infty }$ in step 3.2 is legitimate. As $R\to \mathrm{\infty }$, $\arctan \frac{x+R}{a}\to \frac{\pi }{2}$ and $\arctan \frac{x-R}{a}\to -\frac{\pi }{2}$ because $\arctan$ is increasing with supremum $\pi \mathrm{/}2$ and infimum $-\pi \mathrm{/}2$ on its range $(-\pi \mathrm{/}2,\pi \mathrm{/}2)$; the logarithmic argument tends to $\frac{a^2+(x-\epsilon {)}^2}{a^2+(x+\epsilon {)}^2}>0$, and log is continuous there by [F7]. Therefore

Letting $\epsilon \downarrow 0$ in step 4.2, continuity of $\arctan$ and $\log$ [F7] gives $\arctan \frac{x-\epsilon }{a}-\arctan \frac{x+\epsilon }{a}\to 0$ and $\log \frac{a^2+(x-\epsilon {)}^2}{a^2+(x+\epsilon {)}^2}\to \log 1=0$; hence

In the situation of step 3.3 one has ${\delta }_j(x)=0$, and ${\sup }_j\mathrm{\parallel }{\delta }_j^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$: indeed ${\delta }_j^{\mathrm{\prime }}=P_a^{\mathrm{\prime }}(\chi (\cdot \mathrm{/}j)-1)+P_a{\chi }^{\mathrm{\prime }}(\cdot \mathrm{/}j)\mathrm{/}j$ with $\mathrm{\parallel }\chi (\cdot \mathrm{/}j)-1{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and $\mathrm{\mid }{\chi }^{\mathrm{\prime }}(\cdot \mathrm{/}j)\mathrm{/}j\mathrm{\mid }\le \mathrm{\parallel }{\chi }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}j$, while $P_a$ and $P_a^{\mathrm{\prime }}$ are bounded. Hence the mean value theorem [F7] bounds the difference quotient of ${\delta }_j$ by a constant $C(a)$ uniformly in $j$ on $\mathrm{\mid }t\mathrm{\mid }<1$, and the integrand of the first term of step 3.3 is dominated by the integrable constant $C(a)$ on $0<\mathrm{\mid }t\mathrm{\mid }<1$; letting $\epsilon \downarrow 0$ by dominated convergence [F8], and using that $H_{\epsilon }{\psi }_j(x)\to H{\psi }_j(x)$ and $H_{\epsilon }{P}_a(x)\to Q_a(x)$ by [F2] and step 5.1,

The first term of step 6.1 tends to $0$ as $j\to \mathrm{\infty }$ by dominated convergence [F8]: for each fixed $t\mathrm{\ne }0$ the integrand tends to $0$ because ${\delta }_j\to 0$ pointwise and ${\delta }_j(x)=0$ for $j>\mathrm{\mid }x\mathrm{\mid }$, and it is dominated by $C(a)$ on the finite-measure set $0<\mathrm{\mid }t\mathrm{\mid }<1$; the second term tends to $0$ because $\mid {\int }_{\mathrm{\mid }t\mathrm{\mid }>1}{\delta }_j(x-t)\mathrm{/}tdt\mid \le \mathrm{\parallel }{\delta }_j{\mathrm{\parallel }}_1\to 0$ by step 2.5. Therefore $H{\psi }_j(x)\to Q_a(x)$ for every fixed $x$.

By step 3.4 the sequence $H{\psi }_j$ converges in $L^2$ to a representative of the class $H{P}_a$; by [F8] it has a subsequence converging almost everywhere to a representative of $H{P}_a$, while step 7.1 makes that same subsequence converge to $Q_a$ at every point. Hence $Q_a=H{P}_a$ almost everywhere, i.e. $Q_a=H{P}_a$ in $L^2(\mathbb{R};\mathbb{C})$, which is assertion 3.