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If a finite complex Borel measure $\mu$ on the circle has Fourier coefficients $\hat{\mu }(n)=0$ for all $n<0$, then it is absolutely continuous with respect to arc length. The reflected Fourier convention gives the equivalent theorem with positive coefficients vanishing. The Poisson integral on the unit disk is analytic because of the one-sided Fourier spectrum. Subharmonic comparison for its square root and an $L^2$ maximal estimate produce an integrable radial maximum. This maximum dominates the weak-star boundary measure, forcing absolute continuity of measures. The same proof on the real line says that a finite measure with Fourier transform of a finite measure supported in $[0,\infty )$ has a Lebesgue density.

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.

For a nonnegative harmonic function on the unit disc, the displayed inequalities are sharp. Apply the Poisson integral on the unit disk on radius $R>|z|$, bound its positive kernel between $(R-|z|)/(R+|z|)$ and its reciprocal, and use the mean value property for harmonic functions. Letting $R\uparrow 1$ proves the assertion. Positive Poisson kernels with one boundary pole give equality in the appropriate radial directions.

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.

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.

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 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 $g$.

Apply this with $g=u{|}_{\partial \mathbb{D}}$. The maximum principle for harmonic functions applied to the harmonic difference $u-V$, which vanishes on the boundary, givesAt boundary points the formula is understood through its continuous interior limit.

Now suppose a continuous function has the mean value property and has a local maximum $M=u(z)$ at $z$. Shrink the neighborhood so that $u\le M$ there and all sufficiently small centered circles satisfy the mean value property. On each such circle the average is $M$. If one value were smaller than $M$, continuity would make it smaller on an arc of positive length, contradicting this average. Thus every sufficiently small circle, and hence an entire neighborhood of $z$, has value $M$.

A harmonic function has the mean value property. Indeed, if $m(r)$ is its average on a circle centered at $z$, the divergence theorem givesAs $r\downarrow 0$, $m(r)\to u(z)$, giving the asserted equality.

Conversely, suppose $u$ is continuous and has the mean value property. On an arbitrary closed disk compactly contained in its domain, form the harmonic Poisson integral $V$ with boundary values $u$. The difference $w=u-V$ is continuous, has the mean value property, and vanishes on the boundary. If its maximum were positive, it would be attained inside. The set attaining that maximum is relatively closed and, by the local argument already proved, relatively open. Connectedness of the disk makes $w$ constant, contradicting its boundary value. Thus $w\le 0$. Applying the same argument to $-w$ gives $w\ge 0$. Therefore $u=V$ on every such disk, proving

For the disc, put $u(r,\phi )=R_0(r)\Theta (\phi )$. The polar Laplace equation separates aswhere periodicity requires an integer $n\ge 0$. The radial ansatz $R_0=r^{\alpha }$ gives $\alpha =\pm n$; for $n=0$ the solutions are a constant and $\log r$. Regularity at the origin excludes negative powers and the logarithm.

Writing the boundary data as a Fourier serieswith $a_0=(2\pi {)}^{-1}{\int }_0^{2\pi }{u}_D$, $a_n={\pi }^{-1}{\int }_0^{2\pi }{u}_D\cos n\phi$, and the analogous formula for $b_n$, givesEquivalently it is the Poisson integral on the unit disk with kernelFor continuous boundary data this integral converges to the prescribed value at every boundary point, is smooth and harmonic inside, and uniqueness follows from the maximum principle for harmonic functions.

For the square, superpose four solutions, each carrying one side's data and zero data on the other three sides. Set $k_n=n\pi /a$ andSeparation of variables gives the explicit sumEach summand has zero Laplacian and the required zero values on three open sides. The Fourier sine series supplies the fourth side. For regular continuous data with matching values at the corners, the harmonic extension also takes the common corner values; termwise boundary evaluation of the sine series at the corners is not valid in general. For incompatible corner data there is no continuous solution on the closed square, but the displayed harmonic solution still attains the prescribed data on each open side.