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For the positive Laplacian convention $\Delta G=\delta$, the method of images in the first quadrant reflects an interior pole $(a,b)$ across each axis with negative sign and across both with positive sign:The Dirichlet Green function vanishes on both axes. Its outward normal derivative on the horizontal axis isand the vertical-axis kernel follows by exchange of coordinates. Green second identity converts these derivatives into the quadrant Poisson integral for boundary data.

A bounded holomorphic function on the unit disc has finite nontangential limits almost everywhere on its boundary. To outline a proof, bounded real and imaginary harmonic parts have $L^{\infty }$ boundary representations: apply the Banach-Alaoglu theorem to their circular traces and pass to the limit in the Poisson integral on the unit disk. Almost every boundary point is a Lebesgue point. The fact that a Poisson integral converges nontangentially at Lebesgue points gives the desired limits. Their $L^{\infty }$ norm equals the interior supremum norm, and the Poisson integral recovers the original function. Exceptional points can lack even radial limits; $\exp (i\mathrm{Log}(1-z))$ is a bounded example at the boundary point one.

On a relatively compact coordinate disk $V$ in a Riemann surface, replace a continuous subharmonic function $u$ inside $V$ by the Poisson integral of its boundary values, leaving it unchanged outside. The resulting function $L_{V}u$ is continuous and subharmonic, equals $u$ off $V$, is harmonic on $V$, and satisfies $L_{V}u\ge u$. The maximum principle for subharmonic functions proves the inequality. The pasting assertion follows from harmonic comparison: a harmonic comparison function dominating the pasted function on the boundary of a test domain first dominates $u$, then dominates the disk lift on its part inside $V$.

A harmonic function on one side of a hyperplane that is continuous up to an open portion of the hyperplane and zero there extends harmonically by odd reflection. One local proof takes a ball centered on the boundary and assigns odd continuous boundary data on its sphere. The Poisson integral on the ball is odd, so it vanishes on the equatorial disk. Uniqueness by the maximum principle for harmonic functions identifies it with the original function on the half-ball. Thus the odd extension is harmonic near each boundary point. A globally bounded zero-boundary harmonic function on a half-space is consequently zero by the Liouville theorem for harmonic functions.

For continuous subharmonic function $u$ on the unit disk, let $H_r$ be the Poisson integral of its values on $|z|=r$, evaluated inside that disk. Harmonic comparison gives $H_s\ge H_r\ge u$ for $|z|<r<s<1$. The Harnack inequality for harmonic functions makes their limit either finite and harmonic everywhere or everywhere infinite. In the finite case every harmonic majorant dominates every $H_r$, so their limit is the least harmonic majorant. The supremum must restrict $r>|z|$; the Poisson expression evaluated outside its circle does not have this interpretation.

For every holomorphic function and $p>0$, $|f{|}^p$ is subharmonic. Away from its zeros the displayed Laplacian calculation proves this. At the zeros, use the smooth approximations $(|f{|}^2+\epsilon {)}^{p/2}$, whose Laplacians are nonnegative, and let $\epsilon \downarrow 0$ in their mean inequalities. This permits Poisson integral comparison for fractional modulus powers even when a global analytic fractional power does not exist.

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

Choose the positive Green function conventionAway from its pole, $G$ is harmonic, and the maximum principle for harmonic functions makes it positive inside $D$. The sign of its logarithmic singularity is important: it is the convention for $-\Delta$, whereas a Green function for $+\Delta$ has the opposite sign.

Green's second identity, justified by removing a small disk about the pole and taking its radius to zero, givesSince $\Delta u=-f$, $\Delta G=-{\delta }_{\xi }$ and $G$ vanishes on the outer boundary, this rearranges toThe local logarithmic singularity is integrable, so continuous $f$ gives a well-defined volume integral.

In complex coordinates $z=x+iy$, $\zeta =\xi +i\eta$ for the unit disk, useThe numerator has no zeros inside the disk, so its logarithm is harmonic there. On $|z|=1$, the numerator and denominator have equal modulus, giving the required zero boundary value. Differentiating in the outward radial direction at $z=e^{i\theta }$ givesWith zero forcing the representation therefore becomes the Poisson integralFinally replace $\eta$ by $-\eta$ and change variable $\theta \mapsto 2\pi -\theta$. Cosine stays the same, sine changes sign and the assumed boundary symmetry leaves $g$ unchanged. The integral is therefore invariant, so $\boxed{u(\xi ,-\eta )=u(\xi ,\eta )}$. Equivalently, reflection produces a harmonic function with the same Dirichlet data, and uniqueness identifies it with the original solution.

For fixed source $(\xi ,\eta )\in D$, define the Dirichlet Green function by $-{\Delta }_{x,y}G={\delta }_{(\xi ,\eta )}$ in $D$, $G=0$ on $\partial D$. It is harmonic away from the source and has singularity $-(2\pi {)}^{-1}\log \sqrt{(x-\xi {)}^2+(y-\eta {)}^2}$. The outward normal vector derivative is ${\partial }_{n}G=\mathbf{n}\cdot {\nabla }_{x,y}G$ at a boundary point of arc-length parameter $s$.

Green's second identity gives ${\int }_D(u\Delta G-G\Delta u)={\int }_{\partial D}(u{\partial }_{n}G-G{\partial }_{n}u)ds$. Substituting the two PDEs and the boundary data yieldsThe singular version follows by excising a small source disk and taking its radius to zero, or directly by the distributional identity.

For the upper half-plane use the opposite-sign image source at $(\xi ,-\eta )$:The image lies outside the domain, so it is harmonic inside; the true source has the required unit singularity, and the two distances agree at $y=0$, giving zero boundary value. On that boundary $\mathbf{n}=-{\mathbf{e}}_y$, soThe assumed far-field decay makes the large semicircle contribution vanish. With $f=0$, the representation becomesFinally set $s=\xi +\eta \tan \theta$, $ds=\eta {\sec }^2\theta d\theta$. The Jacobian matrix cancels the denominator, givingIf $g$ is even, the substitution $s\mapsto -s$ in the Poisson integral gives $u(-\xi ,\eta )=u(\xi ,\eta )$. At $\eta =0$ this agrees with the continuous even boundary trace.

Let $f$ be a nonzero meromorphic function in a neighborhood of the closed disk $|z|\le R$, with no boundary zero or pole. List its interior zeros $a_j$ with multiplicities $m_j$ and its interior poles $b_k$ with orders $n_k$. Define the disk Green function and Poisson kernel byAt every interior point that is neither a zero nor a pole, the Poisson-Jensen formula isIn particular, at $z=0$ when $f(0)$ is finite and nonzero, this becomes Jensen's formula, with each zero contributing $-m_j\log (R/|a_j|)$ and each pole contributing $n_k\log (R/|b_k|)$.

Here is a proof that also explains the signs. The disk Blaschke factorshave a simple zero at $a$, no poles in the disk and boundary modulus one. Thereforeextends to a zero-free holomorphic function throughout the disk. Its logarithmic modulus is a harmonic function and agrees with $\log |f|$ on the boundary. The Poisson integral represents $\log |F(z)|$ by that boundary data. Since $\log |B_a|=-G_R(\cdot ,a)$, substitution gives the displayed Poisson-Jensen formula. Boundary zeros or poles can be handled by taking limiting radii: their logarithmic singularities are integrable. When zero is itself a zero or pole, factor out its signed power of $z$ before evaluating Jensen's formula there.

Use the Nevanlinna proximity function $m(r,f)=(2\pi {)}^{-1}\int {\log }^{+}|f(r{e}^{i\theta })|d\theta$ and the Nevanlinna integrated counting function $N(r,f)$ for poles, counting multiplicity. If the pole order at zero is ${\nu }_0$, thenBoundedness of the Nevanlinna characteristic refers to $r\uparrow 1$, equivalently to a uniform bound on $r_0\le r<1$ for any fixed $r_0>0$. This avoids the harmless origin normalization when ${\nu }_0>0$.

The Nevanlinna first main theorem states, for every fixed finite value $a$ and every nonconstant meromorphic function on the disk,For $a=\infty$, this is the definition of $T$. The error is bounded independently of $r$. Translation changes $T$ by $O(1)$, while Jensen's formula gives $T(r,h)-T(r,1/h)=\log |c|$ when $h(z)=c{z}^{\nu }(1+O(z))$ near zero; these observations explain the first theorem.

The Nevanlinna second main theorem in the unit disc states that for $q\ge 3$ distinct values $a_j$ on the Riemann sphere,as $r\uparrow 1$ outside an exceptional set $E$ satisfying ${\int }_{E}dr/(1-r)<\infty$. The truncated Nevanlinna counting function $\overline{N}$ counts each distinct preimage once. The boundary-distance error cannot simply be suppressed: on the disk it need not be $o(T)$. The disk error and exceptional-set estimates follow from the logarithmic-derivative estimates in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

Suppose first that $T(r,f)$ is bounded. Since $m(r,f)\ge 0$, its poles satisfy the Blaschke condition, counting their orders: let $r\uparrow 1$ in their integrated count and use $1-|b|\le -\log |b|$. Let $B$ be their Blaschke product, including the origin factor if necessary. Then $g=Bf$ extends to a holomorphic function on the whole disk, and $|B|\le 1$ givesHere the last bound follows from bounded $T$ and the uniform lower bound $N(r,f)\ge {\nu }_0\log r_0$ on $r\ge r_0$.

The nonnegative subharmonic function ${\log }^{+}|g|$ has a finite harmonic majorant. Indeed its Poisson integrals $H_R$ on expanding disks dominate it by the maximum principle for subharmonic functions, increase with $R$, and have $H_R(0)\le C$. The Harnack inequality for harmonic functions bounds them on every compact subdisk, so their increasing limit is a finite harmonic function $H\ge {\log }^{+}|g|$. Since the disk is simply connected, $H$ has a harmonic conjugate, giving a holomorphic function $A$ with $\mathrm{Re}A=H$. ConsequentlyThis proves the constructive half of the quotient characterization of bounded characteristic.

Conversely, if $f=u/v$ with $u,v$ bounded holomorphic functions and $v\not\equiv 0$, each pole of $f$ is a zero of $v$ of at least the same order. The elementary inequality ${\log }^{+}|u/v|\le {\log }^{+}|u|+{\log }^{+}|1/v|$ and the Nevanlinna first main theorem giveAny cancellation at an origin zero contributes only a bounded multiple of $\log r$ on $r\ge r_0$, which is included in this $O(1)$. The zero function is immediate; constant nonzero $v$ is handled directly. Thus bounded characteristic is exactly a quotient of two bounded analytic functions.

An unbounded example with bounded characteristic isIt is a ratio of the bounded holomorphic functions $1$ and $1-z$. More explicitly, Jensen's formula gives the mean of $\log |1-r{e}^{i\theta }|$ as zero, so $m(r,1/(1-z))=m(r,1-z)\le \log 2$, although $g(r)\to \infty$.

Finally, set $Y=(1+|g{|}^2{)}^{p/2}\ge 1$. Pointwise ${\log }^{+}|g|\le p^{-1}\log Y$. Jensen's inequality for the concave logarithm givesThere is no pole term because $g$ is holomorphic. Hence the stated integral bound implies bounded characteristic. This argument in fact works for every $p>0$.

Use the absolute value around the logarithm in the original PDF. Set $v(z)={\log }^{+}|f(z)|=\max (0,\log |f(z)|)$. This is a nonnegative subharmonic function, including at zeros by its continuous extension. Its radial means are nondecreasing: Poisson integral comparison on a larger concentric circle bounds the smaller mean by the larger one. The hypothesis givesConsequently $T(r)=0$ for every $r$, and continuity of $v$ implies $v=0$ throughout the unit disc. Thus $|f|\le 1$ everywhere, which is stronger than the requested conclusion. In particular $f$ has bounded characteristic, because ${\sup }_{r}T(r)<\infty$; it belongs to the Nevanlinna class and to H-infinity space. The hypothesis also excludes the identically zero function, whose absolute logarithmic integral is infinite.

A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

For a positive Borel measure $\nu$, the centered Hardy-Littlewood maximal function and the uncentered maximal function of a finite measure areWe use open balls and allow infinite values. For signed or complex measures the definition uses their total variation measure. Since a centered ball is also an admissible uncentered ball, and every ball $B(c,r)$ containing $x$ lies in $B(x,2r)$,For functions, $m(f)$ means the centered Hardy-Littlewood maximal function of $|f|dx$.

The set $E_{\alpha }$ is the union of the open balls $B$ with $\nu (B)>\alpha |B|$. In particular, each witnessing ball lies entirely inside $E_{\alpha }$. If $K\subset E_{\alpha }$ is compact, take a finite witnessing cover of $K$. Choose its largest-radius ball, discard all balls meeting it, and repeat. The selected balls $B_j$ are disjoint. Every discarded ball has no larger radius than the selected ball that discarded it and is contained in its threefold dilation. HenceTaking the supremum over compact $K$ by inner regularity of Lebesgue measure proves the stronger, localized estimateIf the right side is infinite there is nothing to prove. Positivity of the measure is important: a signed version uses $|\nu |(E_{\alpha })$, not $\nu (E_{\alpha })$.

For the radial-kernel inequality, integrability implies $\varphi (r)\to 0$. For $0<s<\varphi (0)$, let $r(s)$ be the unique radius with $\varphi (r(s))=s$. The layer cake representation and Tonelli theorem giveThe last integral equals the total mass of $\varphi (|\cdot |)$, namely one. Taking the absolute value of the original convolution therefore proves the claim with constant exactly one. This is radial decreasing kernel domination by the maximal function.

An application is boundary convergence of the Poisson integral. Its normalized radial profile is $c_d(1+r^2{)}^{-(d+1)/2}$, so ${\sup }_{t>0}|P_t\ast f|\le m(f)$. For a continuous compactly supported $g$, the approximate identity property gives $P_t\ast g\to g$ pointwise. For arbitrary $f\in L^1$, put $h=f-g$; thenThe preceding weak type (1,1) estimate and the elementary integral bound for $|h|$ show that the set where this exceeds $2a$ has measure at most $(3^d+1)\Vert h{\Vert }_1/a$. Continuous compactly supported functions are dense in $L^1$, so this measure is zero. Thus the Poisson integrals of an integrable function converge to it almost everywhere; the same approximate identity also gives convergence in $L^1$.

Put $M={\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy)|dx$. The goal is to obtain an integrable non-tangential maximal function without assuming in advance that $f$ has an integrable boundary density. We use a square root to place the boundary data in $L^2$, where the Hardy-Littlewood maximal function is strongly bounded.

First recall this strong bound directly from Q2. For a nonnegative $h\in L^2(\mathbb{R})$ and $\lambda >0$, split $h=h{1}_{\{h>\lambda /2\}}+h{1}_{\{h\le \lambda /2\}}$. The second summand has centered maximal function at most $\lambda /2$; the first is in $L^1$. Q2's weak estimate, in dimension one, therefore givesThe layer cake representation and Tonelli theorem then implyThis is the needed Strong Lp bound for the Hardy-Littlewood maximal function, with an explicit adequate constant.

For $\epsilon >0$, let $f_{\epsilon }(z)=f(z+i\epsilon )$ and $h_{\epsilon }(x)=|f(x+i\epsilon ){|}^{1/2}$. Q1's elementary harmonic pointwise estimate makes $f_{\epsilon }$ bounded on the closed upper half-plane. Its boundary value is continuous, and $\Vert h_{\epsilon }{\Vert }_2^2\le M$. The function $|f_{\epsilon }{|}^{1/2}$ is subharmonic: away from zeros, for any $q>0$,and regularization, or the submean inequality at zeros, extends this to the whole domain. Bounded subharmonic comparison with the Poisson integral givesHere $h_{\epsilon }$ need not be in $L^1$. The Poisson integral is nevertheless defined, since $h_{\epsilon }\in L^2\cap L^{\infty }$ and $P_y\in L^1\cap L^2$, and it tends to its continuous boundary values. To justify comparison on this unbounded domain, the difference between the two sides is bounded above, subharmonic, and zero on the real boundary. On an upper half-disk of radius $R$, use the positive harmonic barrierOn its semicircular arc, $b_R(R{e}^{i\theta })=2/\sin \theta \ge 2$, while $b_R(z)\to 0$ for every fixed interior $z$ as $R\to \infty$. The maximum principle for subharmonic functions, with a fixed bound for the difference, proves the comparison. The barrier's positive blow-up at the arc endpoints causes no difficulty.

For $|x-x^{\prime }|\le y$, the elementary inequality $y^2+|x-s{|}^2\le 3(y^2+|x^{\prime }-s{|}^2)$ givesQ2's radial decreasing kernel domination by the maximal function extends to nonnegative $L^2$ inputs by applying it to increasing bounded compactly supported truncations. Hence, taking the supremum over the cone,For any fixed cone point, $f_{\epsilon }(z)\to f(z)$ as $\epsilon \downarrow 0$. Therefore $f^{\ast }(x)\le {\mathrm{liminf}}_{\epsilon \downarrow 0}{f}_{\epsilon }^{\ast }(x)$; use a sequence of positive shifts and the Fatou lemma to concludeThe precise numerical constant is unimportant. Suprema over countable dense sets of cone points give the same values, so the maximal functions used here are measurable.

This yields boundary absolute continuity of measures rather than assuming it. The holomorphic function $f$ is a complex-valued harmonic function, so Q1 represents it as $f(x+iy)=P_y\ast \mu (x)$ for a finite complex measure $\mu$, with $f(\cdot +iy)dx\to \mu$ weak-star. Since $|f(x+iy)|\le f^{\ast }(x)$, every compactly supported continuous $\psi$ satisfiesThe dual characterization of the total variation measure implies $|\mu |\le f^{\ast }dx$, so the Radon-Nikodym theorem gives $\mu =Fdx$ with $F\in L^1$. The approximate identity now gives $f(\cdot +iy)\to F$ in the L1 norm and almost everywhere. Thus the analytic Hardy space has an integrable boundary function, unlike the unrestricted harmonic $h^1$ space.

In the real-line form of the F. and M. Riesz theorem, start with a finite complex measure whose Fourier transform of a finite measure vanishes for $\xi <0$. Its Poisson integral is holomorphic: the transformed operators ${\partial }_y$ and $i{\partial }_x$ act by $-|\xi |$ and $-\xi$, which agree on that spectrum. These identities can first be read as distributional identities and then classically, since the Poisson integral is smooth. Its harmonic $h^1$ norm is finite by Q1, so the preceding argument makes the measure absolutely continuous. A one-sided Fourier spectrum forces a finite boundary measure to have a Lebesgue density.

For the customary circle form, normalize arc length by $dm=d\theta /(2\pi )$. If $\hat{\mu }(n)=\int e^{-in\theta }d\mu (\theta )=0$ for $n<0$, thenis analytic and $F(r{e}^{i\theta })$ is the Poisson integral on the unit disk of $\mu$. Positivity and unit mass of the Poisson kernel on the circle imply ${\sup }_r\int |F(r{e}^{i\theta })|dm\le \Vert \mu {\Vert }_{\mathrm{TV}}$. The same square-root argument on the disk gives an integrable radial maximum, as follows. For $0<\rho <1$, put $h_{\rho }(\theta )=|F(\rho e^{i\theta }){|}^{1/2}$. The maximum principle for subharmonic functions on the disk givesThe supremum of these Poisson integrals is bounded by a constant times the circular Hardy-Littlewood maximal function. Indeed, for $r\ge 1/2$ the circular kernel is bounded by a constant times $(1-r)/((1-r{)}^2+{\theta }^2)$ for $|\theta |\le \pi$; splitting into arcs of radii $2^j(1-r)$ gives a summable geometric series of arc averages. For $r<1/2$ the kernel is uniformly bounded, and the whole-circle average suffices. The covering proof and the truncation argument already used on the line give the strong $L^2$ maximal bound for arcs on the circle as well. ThusLet $\rho \uparrow 1$ and apply Fatou lemma to obtain an integrable radial maximum $R(\theta )={\sup }_{0<r<1}|F(r{e}^{i\theta })|$. The measures $F(r{e}^{i\theta })dm$ tend weak-star to $\mu$ and are dominated in magnitude by $Rdm$. The same continuous-test-function argument gives $|\mu |\le Rdm$. This proves the F. and M. Riesz theorem: a finite complex measure on the circle with all negative Fourier coefficients zero is absolutely continuous with respect to arc length. Reversing the Fourier convention interchanges positive and negative indices.

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.

For $|\zeta |=1$ and $|z|<1$, the Poisson kernel on the circle satisfiesFor fixed $\zeta$, it is the real part of a holomorphic function of $z$, hence a harmonic function. On each compact subdisk, its derivatives are bounded uniformly in $\zeta$ because the denominator stays away from zero. Differentiation under the integral therefore proves that $P\varphi$ is harmonic, with the word interpreted componentwise when $\varphi$ is complex-valued.

The geometric seriesconverges uniformly in $\zeta$ for a fixed compact subdisk. With normalized Lebesgue measure, integration of its real part gives kernel mass one. Fix a boundary point $\omega$ and choose $\delta >0$ so that $|\varphi (\zeta )-\varphi (\omega )|<\epsilon$ when $|\zeta -\omega |<\delta$. ThenIf $z\to \omega$ from inside the unit disc, the remaining denominator is at least $(\delta /2{)}^2$ for sufficiently close $z$, while $1-|z{|}^2\to 0$. Hence $P\varphi (z)\to \varphi (\omega )$ along every interior approach, not just radial ones.

Orient the unit circle counterclockwise in the Cauchy transform. Its kernel and every $w$-derivative are uniformly bounded for $w$ in any compact subset of $\mathbb{C}\setminus \mathbb{T}$. Thus differentiation under the integral givesand proves that $\Phi$ is a holomorphic function on both components of its domain. Since $d\zeta =2\pi i\zeta dm(\zeta )$, for $0<r<1$ and $|\omega |=1$ the difference of its two kernels isConsequently the paired radial jump of a continuous Cauchy transform obeys the exact identityand the established Poisson integral boundary convergence givesThis proves the paired difference without assuming that the two individual boundary values exist for every continuous datum.

The printed direction $r\downarrow 1$ has the opposite sign. It is present in the original PDF, not merely in the converted TeX. With that direction $r>1$, so substituting $s=1/r<1$ in the exact identity givesThe simplest counterexample is $\varphi \equiv 1$: the Cauchy integral formula gives $\Phi (w)=1$ inside and $\Phi (w)=0$ outside, so the printed difference is constantly $-1$. The requested positive jump is corrected by taking $r\uparrow 1$, or by reversing the order of the difference when $r\downarrow 1$.

For a continuous boundary function $g(e^{i\theta })$, its Poisson integral on the unit disk isThe kernel is the real part of the holomorphic function $(e^{i\theta }+z)/(e^{i\theta }-z)$, so differentiation under the integral on compact subsets shows that $V$ is harmonic. The kernel is positive and has integral $2\pi$: expand the holomorphic fraction as $1+2{\sum }_{n\ge 1}{z}^n{e}^{-in\theta }$ and integrate. As $z$ approaches a boundary point, the mass outside any fixed arc around that point tends to zero. Uniform continuity of $g$ therefore shows that $V$ extends continuously with boundary values $g$.