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For $a\in \mathbb{D}$, this holomorphic function has H1 space on the unit disk norm one: its boundary modulus is the unit-mass Poisson kernel on the circle. The radial integrals are bounded by that boundary integral by subharmonic comparison and tend to it by continuity. Choosing $a$ at depth comparable to an arc's length makes $|f_a|$ bounded below by the reciprocal length throughout its hyperbolic geodesic cap. It tests necessity in the Carleson embedding theorem on the disk.

For continuous boundary data on the counterclockwise unit circle and its Cauchy transform $\Phi$, $\Phi (r\omega )-\Phi (r^{-1}\omega )=P\varphi (r\omega )$ for $0<r<1$. This follows by subtracting the two Cauchy kernels and obtaining the Poisson kernel on the circle. Consequently the paired interior-minus-exterior difference tends to $\varphi (\omega )$ as $r\uparrow 1$. For $r\downarrow 1$ from above, the same written difference has the opposite sign. This paired limit does not require each individual boundary value of the transform to exist.

// Destination: analysis.bigb

The Harnack inequality on the unit disk states that a nonnegative harmonic function satisfiesOne proof applies the Poisson integral on the unit disk on a circle of radius $R>|z|$: the Poisson kernel on the circle lies between $(R-|z|)/(R+|z|)$ and its reciprocal. The mean value property for harmonic functions identifies the mean boundary value with $u(0)$; let $R\uparrow 1$.

A harmonic conjugate on the simply connected unit disc makes $H=u+iv$ holomorphic. The Carathéodory derivative bound for the right half-plane gives $|H^{\prime }(0)|\le 2u(0)$. The Cauchy-Riemann equations give $|H^{\prime }|=|\nabla u|$, so $|\nabla u(0)|\le 2u(0)$. For a fixed $z$, use the disk automorphismThe chain rule and conformality multiply the gradient length by $|{\psi }^{\prime }(0)|$. Applying the estimate at zero to $u\circ \psi$ proves the sharp gradient bound for positive harmonic functions:Sharpness follows from $u(\zeta )=\mathrm{Re}[(1+\zeta )/(1-\zeta )]$, which has equality at zero, and then composing with ${\psi }^{-1}$. These extremizers are positive multiples of the Poisson kernel on the circle with its boundary pole fixed at one point.

We prove necessity of the geometric condition from the embedding condition. Use normalized angular measure $m$ and the H1 space on the unit disk norm $\Vert f{\Vert }_1={\sup }_{r<1}\int |f(r\xi )|dm(\xi )$. Testing $f=1$ first gives $\mu (\mathbb{D})\le N(\mu )$.

For a short arc $I$ of normalized length $\ell =m(I)$ centered at ${\xi }_0$, put $a=(1-\ell ){\xi }_0$ and use the normalized squared Cauchy kernelIts boundary modulus is the Poisson kernel on the circle, so $\Vert f_a{\Vert }_1=1$. The geodesic-cap and Carleson-box comparison gives $|z-{\xi }_0|\le C\ell$ for $z\in Q(I)$. Therefore $|1-\overline{a}z|\le C^{\prime }\ell$, whereas $1-|a{|}^2=2\ell -{\ell }^2\ge \ell$. Thus $|f_a(z)|\ge c/\ell$ throughout the hyperbolic geodesic cap. The embedding bound givesFor arcs longer than a fixed small threshold, the total-mass estimate gives the same form after changing the absolute constant. HenceThis is precisely the Carleson measure condition, in its geodesic-cap version.

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 $|\zeta |=1$ 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 $r\downarrow 1$.

Identify the circle with $\mathbb{T}=\mathbb{R}/(2\pi \mathbb{Z})$ and use normalized Fourier coefficients and convolution:For $0\le r<1$, summing two geometric series gives the Poisson kernel on the circle:This Fourier series converges absolutely and uniformly for each fixed $r<1$. Consequently we can integrate it term by term in the convolution; the substitution $u=t-s$ givesThe resulting series is itself absolutely and uniformly convergent, since $|\hat{f}(n)|\le \Vert f{\Vert }_{\infty }$ and ${\sum }_n{r}^{|n|}<\infty$.

The Poisson kernel on the circle is nonnegative and has normalized integral one, as its constant Fourier coefficient is one. Its mass concentrates near zero. Indeed, for $r\ge 1/2$ and $\delta \le |s|\le \pi$,Hence its normalized integral outside $(-\delta ,\delta )$ tends to zero as $r\uparrow 1$. Given $\eta >0$, uniform continuity of $f$ gives a $\delta >0$ such that $|f(t-s)-f(t)|<\eta$ whenever the circular distance of $s$ from zero is less than $\delta$. Splitting the convolution error into this arc and its complement yieldsThe second term tends to zero, and $\eta$ is arbitrary. Thus the Abel sums converge uniformly to $f$, or $\boxed{\Vert f\ast P_r-f{\Vert }_{\infty }\to 0}$. This is uniform Poisson summability of continuous circle functions: positivity, unit mass and concentration supply the required approximate identity argument.

The precise conclusion is that finite one-sided limits must agree, so a jump discontinuity is impossible. We prove this using explicit kernels and derive every conjugate-series identity used. The distinction from a removable discontinuity is addressed at the end.

Write the Fourier coefficients asTranslation multiplies each coefficient by a unimodular factor and preserves the hypothesis, so place the proposed discontinuity at zero. Suppose the finite limits are $A=f(0^{-})$ and $B=f(0^{+})$. For $0<r<1$, geometric summation gives the Poisson kernel on the circle and the conjugate Poisson kernel on the circle:These series converge uniformly and absolutely for each $r<1$. Therefore they may be integrated term by term against the Lebesgue integrable function $f$. With normalized convolution, $P_r$ has multiplier $r^{|n|}$, while $Q_r$ has multiplier $-i\mathrm{sgn}(n)r^{|n|}$ for $n\ne 0$ and zero at $n=0$. More explicitly, at zero,The assumed vanishing of negative Fourier coefficients thus proves, rather than presupposes, the conjugate relation

We next show that the right-hand side is bounded as $r\uparrow 1$. The kernel $P_r$ is positive and even, and its normalized integral is one by its series. Its mass outside any fixed $(-\delta ,\delta )$ tends to zero; in fact its supremum there is $O_{\delta }(1-r)$. On $(-\delta ,0)$, $f$ is uniformly close to $A$ for sufficiently small $\delta$, and on $(0,\delta )$ it is uniformly close to $B$. Evenness places asymptotically half the mass on each side. The outside contribution tends to zero because $f$ is integrable and the outside kernel tends uniformly to zero. Bounding the inside errors by the one-sided approximation errors givesHence $(Q_r\ast f)(0)$ is bounded by the proved conjugate relation.

On the other hand, oddness of $Q_r$ givesFor $0<\delta <\pi$, differentiating $1-2r\cos t+r^2$ proves the exact identityAlso $Q_r\ge 0$ on $(0,\pi )$ and is uniformly bounded on $[\delta ,\pi ]$ as $r\uparrow 1$. Write $f(-t)-f(t)=A-B+e(t)$ with $e(t)\to 0$ as $t\downarrow 0$. The integral over $[\delta ,\pi ]$ is $O_{\delta }(1)$ by integrability. Given any $\eta >0$, choose $\delta$ so that $|e(t)|\le \eta$ on $(0,\delta )$. Positivity and the preceding identity show that the near-zero error, divided by $\log (1/(1-r))$, has limiting upper bound at most $\eta /\pi$. Letting $\eta \downarrow 0$ provesBoundedness of the numerator makes the left side tend to zero. Therefore $A=B$. This proves one-sided Fourier spectrum excludes jumps, and supplies the required conjugate-sum argument in full.

There is a point-value qualification to the phrase “first kind.” If it means finite unequal one-sided limits, the proof establishes exactly the stated result. If the definition also includes removable discontinuities, the literal statement is false for an arbitrary integrable representative: the function that equals zero everywhere except $f(0)=1$ has every Fourier coefficient zero but a removable discontinuity at zero. Integral hypotheses cannot detect a change at one point. Thus no jump is possible; equal one-sided limits need not equal an arbitrarily assigned point value.

The Poisson kernel on the circle is nonnegative, has normalized integral one, and its mass outside any fixed arc about zero tends to zero as $r\uparrow 1$. Split the convolution error into a short arc, controlled by the modulus of continuity, and its complement, controlled by the vanishing kernel mass. This proves uniform convergence for every continuous periodic function. Its Fourier multiplier is $r^{|n|}$, so the convolution is the Abel sum of the Fourier series.