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Harmonic Hardy Classes and Fatou Boundary Limits

1 · Prerequisites

2 · Summary

The harmonic Hardy classes $h^{p} (D)$ collect the complex harmonic functions on the unit disc whose radial traces stay bounded in $L^{p} (T, m)$ . This page builds their boundary theory on top of harmonic-functions-and-the-poisson-integral, using the measure, duality and maximal-function machinery of its declared prerequisites: the Poisson integral is extended from continuous boundary data to finite regular complex Borel measures, and the density case is identified with integration against the corresponding $L^{1}$ function.

The boundary theory proceeds through maximal estimates. Centred circular arcs, the circle maximal function, and the nontangential regions $Γ_{A}$ are defined, and the weak-type $(1, 1)$ inequality for the circle maximal function is proved by a covering argument. The nontangential maximal function of a Poisson integral is then dominated by the circle maximal function of its boundary measure, with constant $(A + 1)^{2}$ . These two estimates yield the Fatou theorem: the Poisson extension of an $L^{1}$ datum converges to that datum $m$ -almost everywhere within every nontangential region, while tangential paths remain unconstrained.

The page then identifies the boundary behaviour of the Hardy classes themselves. Every $h^{1}$ function is the Poisson integral of a unique finite regular complex boundary measure, with equality of norms and weak-star convergence of the radial measures; the boundary measure need not have an $L^{1}$ density. For $1 < p \leq \infty$ the class $h^{p}$ is exactly the Poisson image of $L^{p}$ , isometrically, with $L^{p}$ radial convergence for finite $p$ and weak-star convergence at $p = \infty$ . Nonnegative harmonic functions are characterized as Poisson integrals of finite nonnegative measures with mass $u (0)$ , normalized positive families are shown to be compact, and bounded harmonic functions are recovered as Poisson integrals of $L^{\infty}$ data with nontangential limits almost everywhere.

The Axiom of Choice is stated where it is used, and each item identifies the step that spends it: the maximal-function and Fatou results use countable choice, while the measure-representation, representation, positivity and bounded-function results carry the full axiom.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-10-02Open item page →

The Poisson integral of a finite complex boundary measure

Definition

Assume countable choice. Identify the torus $T = R / Z$ with the Euclidean unit circle through $φ ([t]) = e^{2 π i t}$ and write $m$ for the normalized Haar measure of $T$ (The one-dimensional torus and its normalized Haar integral). For $z \in D$ and $ζ \in T$ put $P (z, ζ) : = P (z, φ (ζ)) = \frac{1 - ∣ z ∣^{2}}{∣ φ (ζ) - z ∣^{2}},$ where the second expression is the Poisson kernel of The Poisson kernel on the unit disc evaluated at the boundary point $φ (ζ)$ of the unit circle. For each fixed $z \in D$ the function $ζ \mapsto P (z, ζ)$ is continuous and positive on the compact space $T$ , because $φ$ is continuous and $∣ φ (ζ) - z ∣ \geq 1 - ∣ z ∣ > 0$ .

Integral against a finite complex measure. Let $μ$ be a finite regular complex Borel measure on $T$ (Regular complex Borel measures). Define the Poisson integral of $μ$ by $P [μ] (z) : = \int_{T} P (z, ζ) d μ (ζ) (z \in D),$ the integral being the one against a signed or complex measure (Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|)). This is well defined: for fixed $z$ the integrand is continuous on the compact space $T$ , hence bounded, so it belongs to $L^{1} (∣ μ ∣)$ , and by Integrals against signed or complex measures are bounded by total variation $∣ P [μ] (z) ∣ \leq (\sup_{ζ \in T} P (z, ζ)) ∣ μ ∣ (T) < + \infty .$ Linearity in $μ$ is the linearity of the integral in the measure.

Integral against an $L^{1}$ density. For $f \in L^{1} (T, m)$ (The class $L^{1} (μ)$ of integrable functions) let $f m$ denote the finite complex measure $(f m) (E) : = \int_{E} f d m (E \subseteq T Borel),$ which is a complex measure with $∣ f m ∣ (E) = \int_{E} ∣ f ∣ d m$ (A complex L^1 density defines a complex measure whose total variation is |h| dmu). Since $T$ is compact Hausdorff and second-countable, the finite positive Borel measure $∣ f m ∣$ is regular by Locally finite Borel measures on second-countable LCH spaces are regular under the assumed countable choice. Thus $f m$ is a finite regular complex measure. The displayed measurable-set formula supplies the density $f$ directly, uniquely up to $m$ -almost-everywhere equality by Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree. For a simple measurable $g = \sum_{j = 1}^{k} c_{j} 1_{E_{j}}$ in canonical disjoint form, each $f 1_{E_{j}}$ is integrable since $∣ f 1_{E_{j}} ∣ \leq ∣ f ∣$ , and $∣ f m ∣ (E_{j}) \leq ∥ f ∥_{1} < \infty$ . Thus The simple integral against a signed or complex measure and The Lebesgue integral is linear on $L^{1} (μ)$ give $\int_{T} g d (f m) = \sum_{j = 1}^{k} c_{j} (f m) (E_{j}) = \sum_{j = 1}^{k} c_{j} \int_{E_{j}} f d m = \int_{T} g f d m .$ For bounded measurable $g$ , the same identity follows as follows. Set $s_{n} : = 2^{- n} (⌊ 2^{n} Re g ⌋ + i ⌊ 2^{n} Im g ⌋) .$ Each $s_{n}$ is a measurable simple function with finite range, and $∣ s_{n} - g ∣ \leq \sqrt{2} 2^{- n}$ . Hence $\int_{T} ∣ s_{n} - g ∣ d ∣ f m ∣ \leq \sqrt{2} 2^{- n} ∥ f ∥_{1} ⟶ 0,$ so the definition of integration against a complex measure gives $\int s_{n} d (f m) \to \int g d (f m)$ . Also, by The Lebesgue integral is linear on $L^{1} (μ)$ and The modulus of an integral is bounded by the integral of the modulus, $∣ \int_{T} s_{n} f d m - \int_{T} g f d m ∣ \leq \sqrt{2} 2^{- n} ∥ f ∥_{1} ⟶ 0.$ Passing to the limit in the simple-function identity proves it for every bounded measurable $g$ . The density and approximants are supplied explicitly, so this argument uses no Radon-Nikodym existence theorem or additional choice assumption. In particular, for $z \in D$ with $g = P (z, \cdot)$ , $P [f] (z) : = P [f m] (z) = \int_{T} P (z, ζ) f (ζ) d m (ζ) .$ If $f = f^{'}$ $m$ -almost everywhere, then $(f m) (E) = \int_{E} f d m = \int_{E} f^{'} d m$ for every Borel $E$ (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree), so $f m = f^{'} m$ and $P [f] = P [f^{'}]$ : the Poisson integral of an $L^{1}$ class is independent of the chosen representative.

Radial functions. For $0 \leq r < 1$ and $f \in L^{1} (T, m)$ write $P_{r} * f : T \to C, (P_{r} * f) (ζ) : = \int_{T} P (r ζ, η) f (η) d m (η),$ so that $(P_{r} * f) (ζ) = P [f] (r ζ)$ by the preceding display. Equivalently, writing $P_{r} (u) : = (1 - r^{2}) / (1 - 2 r \cos (2 π u) + r^{2})$ for $u \in T$ , one has $(P_{r} * f) (ζ) = \int_{T} P_{r} (ζ - η) f (η) d m (η)$ , the torus convolution of $f$ with the kernel $P_{r}$ , in agreement with the second display of The Poisson kernel on the unit disc under the identification $ζ = e^{2 π i θ}$ .

Agreement with the published continuous-data integral. Let $ψ : \partial D \to R$ be continuous and let $F : = ψ \circ φ : T \to R$ , a continuous hence bounded function. By the definition of the torus integral, $\int_{T} P (z, ζ) F (ζ) d m (ζ) = \int_{[0, 1)} P (z, e^{2 π i t}) ψ (e^{2 π i t}) d t = \frac{1}{2 π} \int_{0}^{2 π} P (z, e^{i s}) ψ (e^{i s}) d s,$ the last step by the linear change of variables $s = 2 π t$ on $(0, 1)$ (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions). The right-hand side is exactly the Poisson integral of $ψ$ from The Poisson integral on the unit disc, so $P [ψ \circ φ] = P [ψ]$ on $D$ . The definition above therefore extends, and does not conflict with, the published real continuous-data definition.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Harmonic Hardy classes on the unit disc

Definition

Assume countable choice. Identify the torus $T = R / Z$ with the unit circle as in The one-dimensional torus and its normalized Haar integral, and write $m$ for its normalized Haar measure, a probability measure. For a function $u : D \to C$ and $0 \leq r < 1$ write $u_{r} : T \to C, u_{r} (ζ) : = u (r ζ) .$

A complex-valued function $u = U + i V$ on $D$ is called harmonic when both components $U$ and $V$ are real-valued plane harmonic functions in the sense of Plane harmonic functions; this is the componentwise convention of Complex Lp classes and Euclidean test-function conventions. Every such $u$ is continuous, because a $C^{2}$ real function is continuous and both components are of class $C^{2}$ .

The classes $h^{p} (D)$ . For $1 \leq p < \infty$ let $h^{p} (D) : = {u : D \to C harmonic : \sup_{0 \leq r < 1} ∥ u_{r} ∥_{L^{p} (T, m)} < + \infty},$ where $u_{r}$ is regarded as an element of the quotient space $L^{p} (T, m; C)$ of Complex Lp classes and Euclidean test-function conventions through its continuous representative, and $∥ \cdot ∥_{L^{p} (T, m)}$ is the norm of Complex Holder, Minkowski, and the quotient norm. For $p = \infty$ set $h^{\infty} (D) : = {u : D \to C harmonic : \sup_{0 \leq r < 1} \sup_{ζ \in T} ∣ u (r ζ) ∣ < + \infty} .$ Here the inner supremum may equivalently be read as the essential supremum of $u_{r}$ with respect to $m$ (The essential supremum of a measurable function with respect to a measure): a continuous function has the same supremum and essential supremum, because a nonempty open subset of $T$ contains the image $q ((a, b))$ of an open interval with $0 < b - a < 1$ (the map $q$ is open and its images of rational-endpoint intervals form a base, as proved in The one-dimensional torus and its normalized Haar integral), and such a set has $m$ -measure $b - a > 0$ ; hence a continuous function bounded by $M$ almost everywhere is bounded by $M$ everywhere. In particular $\sup_{0 \leq r < 1} \sup_{ζ \in T} ∣ u (r ζ) ∣ = \sup_{z \in D} ∣ u (z) ∣,$ because every $z \in D$ has the form $r ζ$ with $r = ∣ z ∣$ and $ζ \in T$ .

The Hardy norms. For $u \in h^{p} (D)$ put $∥ u ∥_{h^{p}} : = \sup_{0 \leq r < 1} ∥ u_{r} ∥_{L^{p} (T, m)} (1 \leq p < \infty), ∥ u ∥_{h^{\infty}} : = \sup_{z \in D} ∣ u (z) ∣ .$ Then $h^{p} (D)$ is a complex vector space: harmonicity and finiteness of the suprema are preserved by finite linear combinations, and $∥ u + v ∥_{h^{p}} \leq ∥ u ∥_{h^{p}} + ∥ v ∥_{h^{p}}$ , $∥ λ u ∥_{h^{p}} = ∣ λ ∣ ∥ u ∥_{h^{p}}$ for $λ \in C$ , by the corresponding statements at each radius in Complex Holder, Minkowski, and the quotient norm. The assignment is definite: if $∥ u ∥_{h^{p}} = 0$ , then $∥ u_{1 / 2} ∥_{L^{p}} = 0$ , so $u_{1 / 2} = 0$ almost everywhere, hence $u_{1 / 2} = 0$ everywhere by continuity and the preceding paragraph; the Poisson representation formula A harmonic function is recovered from its values on any containing circle by the Poisson formula then gives $u = 0$ on the disc $∣ z ∣ < \frac{1}{2}$ , and the identity principle A plane harmonic function that vanishes on a nonempty open set vanishes everywhere on the domain, applied to the two components on the domain $D$ , gives $u = 0$ . Thus $∥ \cdot ∥_{h^{p}}$ is a norm on $h^{p} (D)$ for every $1 \leq p \leq \infty$ . No containment among the classes $h^{r} (D) \subseteq h^{p} (D)$ for $p < r$ is asserted here; only the containment $h^{\infty} (D) \subseteq h^{1} (D)$ will be used, and it is proved where it is needed.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Poisson extension is an Lp contraction and converges in finite Lp

Statement

Assume countable choice. Let $1 \leq p \leq \infty$ and let $f \in L^{p} (T, m; C)$ . Then $P [f]$ is complex harmonic, lies in the Hardy class $h^{p} (D)$ of Harmonic Hardy classes on the unit disc, and $∥ P_{r} * f ∥_{p} \leq ∥ f ∥_{p} (0 \leq r < 1) .$ If $p < \infty$ , then $∥ P_{r} * f - f ∥_{p} \to 0$ as $r ↑ 1$ . No $L^{\infty}$ norm-convergence statement is made for arbitrary data.

Facts & Assumptions

Given: Countable choice, an exponent $1 \leq p \leq \infty$ , and a function $f \in L^{p} (T, m; C)$ .

[L1]

For $f \in L^{1} (T, m)$ the Poisson integral $P [f] = P [f m]$ and the radial functions $(P_{r} * f) (ζ) = P [f] (r ζ) = \int_{T} P_{r} (ζ - η) f (η) d m (η)$ are defined by integration against the kernel; for real continuous data on $\partial D$ this agrees with the published continuous-data Poisson integral (The Poisson integral of a finite complex boundary measure).

[L2]

The torus $T$ carries the probability measure $m$ , which is invariant under translations $ζ \mapsto ζ - η$ ; the product space $T \times T$ is sigma-finite and Tonelli's theorem applies to nonnegative product-measurable functions (The one-dimensional torus and its normalized Haar integral, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[L3]

For $0 \leq r < 1$ the kernel satisfies $P_{r} (θ) = (1 - r^{2}) / (1 - 2 r \cos θ + r^{2}) > 0$ , $\int_{T} P_{r} d m = 1$ , and $P (z, η) \leq (1 + ∣ z ∣) / (1 - ∣ z ∣)$ for all $z \in D$ , $η \in T$ (The Poisson kernel is positive, has total mass one, and concentrates at a boundary point, The Poisson kernel on the unit disc).

[L4]

A nonnegative measurable density $w$ with $\int w d m = 1$ defines a probability measure $w d m$ on $T$ , and $\int g d (w d m) = \int g w d m$ for nonnegative measurable $g$ ; on this probability space Jensen's inequality applies to real $g \in L^{1}$ and a convex $φ$ with $φ \circ g \in L^{1}$ (The measure with density $f$ relative to $μ$ , Integrating against a density agrees with integrating the product, Jensen's integral inequality for a probability measure).

[L5]

On the probability space $T$ , $L^{p}$ Hölder with the constant function $1$ gives $\int_{T} ∣ f ∣ d m \leq ∥ f ∥_{p}$ for every $1 \leq p \leq \infty$ ; uniform convergence implies $L^{p}$ convergence for finite $p$ , and $∣ \int g d m ∣ \leq \int ∣ g ∣ d m$ (Complex Holder, Minkowski, and the quotient norm).

[L6]

For $1 \leq p < \infty$ the continuous complex functions on $T$ are dense in $L^{p} (T, m; C)$ (Continuous functions are dense in $L^{p}$ of finite tori and of bounded intervals).

[L7]

The Poisson integral of a continuous real boundary datum is harmonic on $D$ , and a locally uniform limit of harmonic functions is harmonic (Poisson integrals are harmonic on the unit disc, Locally uniform limits of harmonic functions are harmonic).

[L8]

The Poisson integral of continuous real boundary data converges to that data uniformly on $T$ as $r ↑ 1$ , equivalently the continuous-data Poisson integral extends continuously to the closed disc (The Poisson kernel is a boundary approximate identity, The Poisson integral gives the unique continuous harmonic extension on the closed unit disc).

Proof

technique · direct

1.1givenL1L5algebra

By [L5] (Hölder against the constant function $1$ , whose conjugate norm is $m (T)^{1 - 1 / p} = 1$ for finite $p$ and $∥ 1 ∥_{\infty} = 1$ for $p = \infty$ ), every $f \in L^{p} (T, m; C)$ satisfies $\int_{T} ∣ f ∣ d m \leq ∥ f ∥_{p} < \infty$ ; hence $f \in L^{1} (T, m)$ and the Poisson integral $P [f]$ of [L1] is defined.

1.2L1L2L3L5algebra

For every $0 \leq r < 1$ and every $ζ \in T$ , translation invariance of $m$ and [L3] give $\int_{T} P_{r} (ζ - η) d m (η) = \int_{T} P_{r} d m = 1$ ; since the kernel is positive, the elementary estimate $∣ \int g d m ∣ \leq \int ∣ g ∣ d m$ of [L5] applied to $g = P_{r} (ζ - \cdot) f$ gives $∣ (P_{r} * f) (ζ) ∣ \leq \int_{T} P_{r} (ζ - η) ∣ f (η) ∣ d m (η) .$

1.3L1L7algebra

If $g \in C (T, C)$ , write $g = u + i v$ with real continuous $u, v$ . By the agreement clause of [L1] the integrals $P [u]$ and $P [v]$ coincide with the published Poisson integrals of the real continuous boundary data $u \circ φ^{- 1}$ and $v \circ φ^{- 1}$ ; [L7] makes both real harmonic, and linearity of the integral gives $P [g] = P [u] + i P [v]$ , so $P [g]$ is complex harmonic.

2.1step 1.2L3

For $p = \infty$ : step 1.2 and the total mass from [L3] give $∣ (P_{r} * f) (ζ) ∣ \leq ∥ f ∥_{\infty} \int_{T} P_{r} (ζ - η) d m (η) = ∥ f ∥_{\infty}$ for every $ζ$ , so $∥ P_{r} * f ∥_{\infty} \leq ∥ f ∥_{\infty}$ and $\sup_{0 \leq r < 1} ∥ (P [f])_{r} ∥_{\infty} \leq ∥ f ∥_{\infty}$ .

2.2step 1.2L2L3algebra

For $p = 1$ : integrating the display of step 1.2 over $ζ$ and applying Tonelli's theorem [L2] to the nonnegative product-measurable integrand $(ζ, η) \mapsto P_{r} (ζ - η) ∣ f (η) ∣$ , then translation invariance of $m$ , gives $∥ P_{r} * f ∥_{1} \leq \int_{T} ⁣ \int_{T} P_{r} (ζ - η) ∣ f (η) ∣ d m (η) d m (ζ) = \int_{T} ∣ f (η) ∣ (\int_{T} P_{r} (ζ - η) d m (ζ)) d m (η) = ∥ f ∥_{1} .$

2.3step 1.2L2L3L4L5algebra

For $1 < p < \infty$ : fix $ζ$ and put $w_{ζ} (η) : = P_{r} (ζ - η)$ , a nonnegative measurable density with $\int w_{ζ} d m = 1$ by step 1.2, so [L4] makes $ν_{ζ} : = w_{ζ} d m$ a probability measure on $T$ with $\int ∣ f ∣ d ν_{ζ} = \int P_{r} (ζ - η) ∣ f (η) ∣ d m (η) \leq \sup_{η} P_{r} ∥ f ∥_{1} < \infty$ , so $∣ f ∣ \in L^{1} (ν_{ζ})$ and $∣ f ∣^{p} \in L^{1} (ν_{ζ})$ because $\int ∣ f ∣^{p} d ν_{ζ} \leq \sup_{η} P_{r} ∥ f ∥_{p}^{p}$ . Jensen's inequality [L4] applied to the convex function $t \mapsto t^{p}$ and $∣ f ∣$ gives, using step 1.2, $∣ (P_{r} * f) (ζ) ∣^{p} \leq (\int_{T} ∣ f ∣ d ν_{ζ})^{p} \leq \int_{T} ∣ f ∣^{p} d ν_{ζ} = \int_{T} P_{r} (ζ - η) ∣ f (η) ∣^{p} d m (η);$ integrating this display over $ζ$ and applying the same Tonelli and translation-invariance argument yields $∥ P_{r} * f ∥_{p}^{p} \leq ∥ f ∥_{p}^{p}$ , hence $∥ P_{r} * f ∥_{p} \leq ∥ f ∥_{p}$ .

2.4step 1.1step 1.3L1L3L6L7algebra

$P [f]$ is complex harmonic. Indeed $f \in L^{1}$ by step 1.1; choose continuous $g_{n}$ with $∥ f - g_{n} ∥_{1} \to 0$ [L6] at $p = 1$ . For $z \in D$ the bound of [L3] and step 1.1 give $∣ P [f] (z) - P [g_{n}] (z) ∣ = ∣ P [f - g_{n}] (z) ∣ \leq \frac{1 + ∣ z ∣}{1 - ∣ z ∣} ∥ f - g_{n} ∥_{1},$ and the factor $(1 + ∣ z ∣) / (1 - ∣ z ∣)$ is bounded on every compact subset of $D$ ; hence $P [g_{n}] \to P [f]$ locally uniformly on $D$ . Each $P [g_{n}]$ is complex harmonic by step 1.3, so the locally uniform limit $P [f]$ is complex harmonic by [L7].

2.5step 1.2L1L2L5L8algebra

Convergence for continuous data: let $g \in C (T, C)$ . Applying [L8] to the real and imaginary parts and using the identification of [L1] gives $∣ P [g] (r e^{i α}) - g (e^{i α}) ∣ \to 0$ uniformly in $α$ as $r ↑ 1$ ; consequently, for every $1 \leq p < \infty$ , $∥ P_{r} * g - g ∥_{p} \leq ∥ P_{r} * g - g ∥_{\infty} \to 0$ because $m$ is a probability measure [L2].

3.1step 2.1step 2.2step 2.3step 2.4

Combining steps 2.1, 2.2 and 2.3, for every $1 \leq p \leq \infty$ and every $0 \leq r < 1$ the contraction $∥ P_{r} * f ∥_{p} \leq ∥ f ∥_{p}$ holds. With the harmonicity proved in step 2.4 this yields $\sup_{0 \leq r < 1} ∥ (P [f])_{r} ∥_{p} \leq ∥ f ∥_{p} < \infty$ , so $P [f] \in h^{p} (D)$ with $∥ P [f] ∥_{h^{p}} \leq ∥ f ∥_{p}$ .

4.1step 2.5step 3.1L5L6algebra∎

Let $p < \infty$ and $ε > 0$ . By [L6] choose continuous $g$ with $∥ f - g ∥_{p} < ε / 3$ . For every $0 \leq r < 1$ , step 3.1 and the triangle inequality for the $L^{p}$ norm give $∥ P_{r} * f - f ∥_{p} \leq ∥ P_{r} * (f - g) ∥_{p} + ∥ P_{r} * g - g ∥_{p} + ∥ g - f ∥_{p} \leq 2 ∥ f - g ∥_{p} + ∥ P_{r} * g - g ∥_{p},$ and step 2.5 makes the last term smaller than $ε / 3$ for all $r$ sufficiently close to $1$ ; hence $∥ P_{r} * f - f ∥_{p} < 2 ε / 3 + ε / 3 = ε$ for those $r$ , so $∥ P_{r} * f - f ∥_{p} \to 0$ as $r ↑ 1$ . This proves the finite- $p$ norm limit, while at $p = \infty$ no norm-convergence claim is made; the contraction and the $h^{p}$ membership are step 3.1, completing the proof.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

h1 is isometric to finite regular complex boundary measures

Statement

Assume the Axiom of Choice. Every $u \in h^{1} (D)$ has a unique finite regular complex Borel measure $μ$ on $T$ with $u = P [μ]$ . Conversely every finite regular complex Borel measure $μ$ on $T$ gives an $h^{1}$ function $P [μ]$ , and $∥ P [μ] ∥_{h^{1}} = ∣ μ ∣ (T),$ with $(P [μ])_{r} m$ converging weak-star to $μ$ against $C (T)$ as $r ↑ 1$ . A general $h^{1}$ function need not have an $L^{1}$ density: its boundary measure need not be of the form $f m$ with $f \in L^{1} (T, m)$ .

Facts & Assumptions

Given: The Axiom of Choice, hence Countable Choice (The Axiom of Choice, AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )), a function $u \in h^{1} (D)$ with bound $M : = ∥ u ∥_{h^{1}}$ , and finite regular complex Borel measures $μ, ν$ on $T$ where they occur.

[L1]

$P [μ] (z) = \int_{T} P (z, ζ) d μ (ζ)$ for finite complex Borel measures and $P [f] = P [f m]$ for $f \in L^{1}$ ; the radial traces are $(P [μ])_{r} (ζ) = P [μ] (r ζ)$ (The Poisson integral of a finite complex boundary measure).

[L2]

The kernel is positive with $P (z, η) = (1 - ∣ z ∣^{2}) / ∣ η - z ∣^{2}$ , $\int_{T} P (z, η) d m (η) = 1$ , and $P_{r} * g \to g$ uniformly on $T$ for every continuous $g$ as $r ↑ 1$ ; $m$ is a probability measure on the compact metric space $T$ (The Poisson kernel is positive, has total mass one, and concentrates at a boundary point, The Poisson kernel is a boundary approximate identity, The one-dimensional torus and its normalized Haar integral).

[L3]

For $f \in L^{1} (μ)$ one has $∣ \int f d μ ∣ \leq \int ∣ f ∣ d ∣ μ ∣$ , and $∣ μ ∣$ is a finite measure (Integrals against signed or complex measures are bounded by total variation, The total variation of a signed or complex measure is a positive measure).

[L4]

Fubini's theorem applies to functions integrable for a product of sigma-finite measures, and Tonelli's theorem applies to nonnegative product-measurable functions (Fubini's theorem for L^1 functions on a sigma-finite product, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product).

[L5]

For $0 \leq r < 1$ the density measure $u_{r} m$ is a finite complex measure with $∣ u_{r} m ∣ (T) = \int_{T} ∣ u_{r} ∣ d m = ∥ u_{r} ∥_{1}$ ; the class $h^{1} (D)$ consists of the complex harmonic functions with $\sup_{0 \leq r < 1} ∥ u_{r} ∥_{1} < \infty$ and $∥ u ∥_{h^{1}}$ denotes that supremum (A complex L^1 density defines a complex measure whose total variation is |h| dmu, Harmonic Hardy classes on the unit disc).

[L6]

$C (T, R)$ has a countable dense family $(f_{j})_{j \geq 1}$ , namely rational polynomials in finitely many distance functions to an enumerated dense subset; passing to the double family $f_{j} + i f_{k}$ over the countable set $N \times N$ exhibits a countable dense family in $C (T, C)$ , so that space is separable (A countable dense family of continuous functions on a compact metric space, Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L7]

If $X$ is a separable real or complex normed space, then under the ultrafilter lemma every sequence in the dual unit ball has a weak-star convergent subsequence, and the limit functional is again an element of $X^{*}$ ; weak-star convergence is evaluation convergence on every element of $X$ (A separable predual has weak-star sequentially compact dual ball, Weak star convergence, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter).

[L8]

Assume Dependent Choice. Every bounded complex linear functional on $C_{0} (X; C)$ of an LCH space $X$ is represented uniquely by a finite regular complex Borel measure, and conversely every such measure defines a functional of norm $∣ μ ∣ (X)$ ; in particular for compact $T$ this gives $∣ μ ∣ (T) = \sup {∣ \int g d μ ∣ : g \in C (T), ∥ g ∥_{\infty} \leq 1}$ and uniqueness of the representing measure (The bounded complex dual of C_0(X) is regular complex measures).

[L9]

A harmonic function on an open set containing the closed disc of radius $r$ is recovered on it by the Poisson formula with the kernel $(r^{2} - ∣ z ∣^{2}) / ∣ r η - z ∣^{2}$ ; under the identification of $T$ with the unit circle this is the formula $w (z) = \int_{T} \frac{r^{2} - ∣ z ∣^{2}}{∣ r η - z ∣^{2}} w (r η) d m (η)$ for $∣ z ∣ < r$ (A harmonic function is recovered from its values on any containing circle by the Poisson formula, The one-dimensional torus and its normalized Haar integral).

[L10]

The Dirac measure $δ_{ζ_{0}}$ at a point of $T$ is a probability measure and a finite regular Borel measure, with $δ_{ζ_{0}} ({ζ_{0}}) = 1$ and $m ({ζ_{0}}) = 0$ ; for every $f \in L^{1} (T, m)$ the density measure satisfies $(f m) ({ζ_{0}}) = \int_{{ζ_{0}}} f d m = 0$ (The Dirac set function at a point, A Dirac set function is a probability measure, Locally finite Borel measures on second-countable LCH spaces are regular, The one-dimensional torus and its normalized Haar integral).

[L11]

Complex polynomials are holomorphic; their real and imaginary parts, being $C^{2}$ , are harmonic. Under Countable Choice, locally uniform limits of real harmonic functions are harmonic (Complex polynomials are entire with the power-rule derivative, and rational functions are holomorphic wherever their denominator is nonzero, The $C^{2}$ real and imaginary parts of a holomorphic function satisfy Laplace's equation and form a harmonic-conjugate pair, Locally uniform limits of harmonic functions are harmonic).

Proof

technique · direct

1.1givenL1L2L3L4L5L11algebra

Converse direction, norm bound. Let $μ$ be a finite regular complex Borel measure and put $v : = P [μ]$ . First establish harmonicity. Identifying $η \in T$ with its unit-circle point, the geometric series gives $P (z, η) = 1 + \sum_{k \geq 1} (z^{k} {\overline{η}}^{k} + {\overline{z}}^{k} η^{k}) .$ Put $a_{k} : = \int {\overline{η}}^{k} d μ (η)$ and $b_{k} : = \int η^{k} d μ (η)$ . These integrals exist by [L3]. The functions $v_{N} (z) : = μ (T) + \sum_{k = 1}^{N} (a_{k} z^{k} + b_{k} {\overline{z}}^{k})$ are complex harmonic by [L11] and linearity of the Laplacian. For $∣ z ∣ \leq ρ < 1$ , the kernel remainder and [L3] give $∣ v (z) - v_{N} (z) ∣ \leq \frac{2 ∣ μ ∣ (T) ρ^{N + 1}}{1 - ρ} .$ Thus $v_{N} \to v$ locally uniformly; [L11], applied to real and imaginary parts under the given Countable Choice, proves that $v$ is complex harmonic. For $0 \leq r < 1$ and $ζ \in T$ , [L1] and [L3] give $∣ v_{r} (ζ) ∣ \leq \int_{T} P (r ζ, η) d ∣ μ ∣ (η)$ ; integrating over $ζ$ and applying Tonelli's theorem [L4] to the nonnegative integrand, together with translation invariance of $m$ and the unit mass of the kernel from [L2], yields $\int_{T} ∣ v_{r} ∣ d m \leq \int_{T} (\int_{T} P (r ζ, η) d m (ζ)) d ∣ μ ∣ (η) = ∣ μ ∣ (T) < + \infty .$ Hence $\sup_{0 \leq r < 1} ∥ v_{r} ∥_{1} \leq ∣ μ ∣ (T)$ and $v \in h^{1} (D)$ with $∥ v ∥_{h^{1}} \leq ∣ μ ∣ (T)$ by [L5].

1.2givenL1L2L3L4L7algebra

Converse direction, weak-star convergence. Let $g \in C (T, C)$ . The function $(ζ, η) \mapsto g (ζ) P (r ζ, η)$ is bounded by $∥ g ∥_{\infty} \sup_{ζ} P (r ζ, η)$ and is integrable for the product of the probability measure $m$ and the finite measure $∣ μ ∣$ ; Fubini's theorem [L4] therefore gives $\int_{T} g v_{r} d m = \int_{T} (\int_{T} g (ζ) P (r ζ, η) d m (ζ)) d μ (η) .$ The inner integral equals $(P_{r} * g) (η)$ by translation invariance of $m$ and the symmetry $P_{r} (- θ) = P_{r} (θ)$ , and $P_{r} * g \to g$ uniformly by [L2]; hence $∣ \int g v_{r} d m - \int g d μ ∣ = ∣ \int (P_{r} * g - g) d μ ∣ \leq ∥ P_{r} * g - g ∥_{\infty} ∣ μ ∣ (T) \to 0$ . Thus $v_{r} m \overset{*}{⇀} μ$ against $C (T)$ as $r ↑ 1$ .

1.3givenL5L6L7L8algebra

Extraction of a boundary measure. Let $u \in h^{1} (D)$ with $M = ∥ u ∥_{h^{1}} < \infty$ and put $r_{j} : = 1 - 1 / (j + 1) ↑ 1$ . Each $u_{r_{j}} m$ is a finite complex measure with $∣ u_{r_{j}} m ∣ (T) = ∥ u_{r_{j}} ∥_{1} \leq M$ by [L5], so $(u_{r_{j}} m)$ is a bounded sequence in the dual of the separable space $C (T, C)$ by [L6]. The ultrafilter lemma gives a subsequence, relabelled $(u_{r_{j}} m)$ , and a finite regular complex Borel measure $μ$ with $u_{r_{j}} m \overset{*}{⇀} μ$ by [L7] and [L8]. For every $g \in C (T)$ with $∥ g ∥_{\infty} \leq 1$ , $∣ \int g d μ ∣ = \lim_{j} ∣ \int g u_{r_{j}} d m ∣ \leq {lim inf}_{j} ∥ u_{r_{j}} ∥_{1} \leq M,$ and taking the supremum over such $g$ gives $∣ μ ∣ (T) \leq M$ by the norm formula of [L8].

1.4givenL5L10algebra

The Dirac measure is not a density. Fix $ζ_{0} \in T$ . By [L10], $δ_{ζ_{0}}$ is a finite regular complex Borel measure with $δ_{ζ_{0}} ({ζ_{0}}) = 1$ , while $(f m) ({ζ_{0}}) = \int_{{ζ_{0}}} f d m = 0$ for every $f \in L^{1} (T, m)$ because $m ({ζ_{0}}) = 0$ . Hence there is no $f \in L^{1} (T, m)$ with $f m = δ_{ζ_{0}}$ : the boundary measure $δ_{ζ_{0}}$ admits no $L^{1}$ density.

2.1step 1.1step 1.2L5L8algebra

Converse direction, reverse norm inequality. For a finite regular complex Borel measure $μ$ and $v = P [μ]$ , the norm formula of [L8] gives $∣ μ ∣ (T) = \sup {∣ \int g d μ ∣ : ∥ g ∥_{\infty} \leq 1}$ ; for each such $g$ , step 1.2 yields $∣ \int g d μ ∣ = \lim_{r ↑ 1} ∣ \int g v_{r} d m ∣ \leq \sup_{r < 1} ∥ v_{r} ∥_{1} = ∥ v ∥_{h^{1}}$ . Taking the supremum over $g$ and combining with step 1.1 gives $∥ P [μ] ∥_{h^{1}} = ∣ μ ∣ (T)$ .

2.2step 1.3L1L2L5L9algebra

Identification of the Poisson integral. Let $u$ and $μ$ be as in step 1.3 and fix $z \in D$ . For all large $j$ one has $r_{j} > ∣ z ∣$ , and [L9] applied to the harmonic function $u$ on the disc of radius $r_{j}$ gives, in the torus normalization, $u (z) = \int_{T} \frac{r_{j}^{2} - ∣ z ∣^{2}}{∣ r_{j} η - z ∣^{2}} u_{r_{j}} (η) d m (η) = \int_{T} P (z, η) u_{r_{j}} (η) d m (η) + \int_{T} (P^{(j)} (z, η) - P (z, η)) u_{r_{j}} (η) d m (η),$ where $P^{(j)} (z, η) : = (r_{j}^{2} - ∣ z ∣^{2}) / ∣ r_{j} η - z ∣^{2} \to P (z, η)$ uniformly in $η$ as $j \to \infty$ because $∣ z ∣ < 1$ stays a positive distance from the boundary circle; the second term is bounded by $∥ P^{(j)} (z, \cdot) - P (z, \cdot) ∥_{\infty} ∥ u_{r_{j}} ∥_{1} \to 0$ . The first term tends to $\int_{T} P (z, η) d μ (η) = P [μ] (z)$ by step 1.3, since $η \mapsto P (z, η)$ is continuous. Hence $u (z) = P [μ] (z)$ for every $z \in D$ , that is, $u = P [μ]$ .

2.3step 1.2L8

Uniqueness of the boundary measure. If $P [μ] = P [ν]$ , then for every $g \in C (T)$ step 1.2 applied to $μ$ and to $ν$ gives $\int g d μ = \lim_{r} \int g (P [μ])_{r} d m = \lim_{r} \int g (P [ν])_{r} d m = \int g d ν$ . The uniqueness clause of the representation theorem [L8] then gives $μ = ν$ . In particular the measure produced by the weak-star subsequence in step 1.3 is the unique representing measure of $u$ , independently of the subsequence.

3.1step 1.3step 2.1step 2.2L5algebra

Norm equality on $h^{1}$ . Let $u \in h^{1} (D)$ with representing measure $μ$ as in steps 1.3 and 2.2. Step 1.3 gives $∣ μ ∣ (T) \leq ∥ u ∥_{h^{1}}$ , and step 2.2 gives $u = P [μ]$ , so step 2.1 yields $∥ u ∥_{h^{1}} = ∥ P [μ] ∥_{h^{1}} = ∣ μ ∣ (T)$ . Therefore the representation is an isometry, and every $h^{1}$ function has the same norm as its boundary measure.

3.2step 1.2step 2.2

Full-net weak-star convergence. Since $u = P [μ]$ by step 2.2, step 1.2 applied to the measure $μ$ shows that $u_{r} m = (P [μ])_{r} m \overset{*}{⇀} μ$ against $C (T)$ along the whole net $r ↑ 1$ , not merely along the subsequence selected in step 1.3.

4.1step 1.1step 1.3step 1.4step 2.1step 2.2step 2.3step 3.1step 3.2L7L8∎

Assembly. (i) If $u \in h^{1} (D)$ , steps 1.3 and 2.2 produce a finite regular complex Borel measure $μ$ with $u = P [μ]$ , step 2.3 shows it is unique, step 3.1 gives $∥ u ∥_{h^{1}} = ∣ μ ∣ (T)$ , and step 3.2 gives $u_{r} m \overset{*}{⇀} μ$ . Conversely, if $μ$ is a finite regular complex Borel measure, step 1.1 puts $P [μ]$ in $h^{1} (D)$ and step 2.1 gives $∥ P [μ] ∥_{h^{1}} = ∣ μ ∣ (T)$ , while step 1.2 gives the weak-star convergence of the radial measures; this proves both directions of the asserted isometric correspondence. (ii) For the final clause, the measure $δ_{ζ_{0}}$ of step 1.4 is finite and regular, so $P [δ_{ζ_{0}}]$ is an $h^{1}$ function whose boundary measure is not of the form $f m$ ; hence a general $h^{1}$ function need not have an $L^{1}$ density. (iii) The Axiom of Choice is used exactly as recorded: it gives the ultrafilter lemma used in the separable-predual sequential compactness theorem and Dependent Choice for the Riesz representation theorem, both cited in [L7] and [L8] and carried in the dependency list of this item.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

h^p is the Poisson image of Lp for 1<p<=infinity

Statement

Assume the Axiom of Choice. Let $1 < p \leq \infty$ and let $u$ be complex harmonic on $D$ with $u \in h^{p} (D)$ . Then there is a unique $f \in L^{p} (T, m; C)$ with $u = P [f]$ , and $∥ u ∥_{h^{p}} = ∥ f ∥_{p} .$ Moreover, if $p < \infty$ then $∥ (P [f])_{r} - f ∥_{p} \to 0$ as $r ↑ 1$ , while for $p = \infty$ the radial functions converge weak-star to $f$ in $L^{\infty} (T, m) = L^{1} (T, m)^{*}$ , that is, $\int_{T} g u_{r} d m ⟶ \int_{T} g f d m (g \in L^{1} (T, m))$ as $r ↑ 1$ . No $L^{\infty}$ norm-convergence of the radial functions is asserted.

Facts & Assumptions

Given: the Axiom of Choice, an exponent $1 < p \leq \infty$ with conjugate exponent $q$ , $r_{j} : = 1 - \frac{1}{j + 1}$ for $j \geq 1$ , and a complex harmonic $u \in h^{p} (D)$ with $M : = ∥ u ∥_{h^{p}}$ .

[L1]

$h^{p} (D)$ consists of the complex harmonic functions with $∥ u ∥_{h^{p}} = \sup_{0 \leq r < 1} ∥ u_{r} ∥_{p} < \infty$ , where $u_{r} (ζ) = u (r ζ)$ ; for $f \in L^{p} (T, m; C)$ the Poisson integral $P [f] = P [f m]$ is defined and complex harmonic, $(P [f])_{r} = P_{r} * f$ satisfies $∥ P_{r} * f ∥_{p} \leq ∥ f ∥_{p}$ , so $P [f] \in h^{p} (D)$ with $∥ P [f] ∥_{h^{p}} \leq ∥ f ∥_{p}$ , and for $p < \infty$ one has $∥ P_{r} * f - f ∥_{p} \to 0$ as $r ↑ 1$ (Harmonic Hardy classes on the unit disc, The Poisson integral of a finite complex boundary measure, Poisson extension is an Lp contraction and converges in finite Lp).

[L2]

If $w$ is harmonic on an open set containing the closed disc of radius $R < 1$ about $0$ , then for $∣ z ∣ < R$ one has $w (z) = \int_{T} \frac{R^{2} - ∣ z ∣^{2}}{∣ R η - z ∣^{2}} w (R η) d m (η)$ , the integral being taken over the normalized torus measure (A harmonic function is recovered from its values on any containing circle by the Poisson formula, The one-dimensional torus and its normalized Haar integral).

[L3]

Every $v \in h^{1} (D)$ has a unique finite regular complex Borel measure $μ$ on $T$ with $v = P [μ]$ , $∥ v ∥_{h^{1}} = ∣ μ ∣ (T)$ , and $\int_{T} g d μ = \lim_{r ↑ 1} \int_{T} g v_{r} d m$ for every continuous $g$ (h1 is isometric to finite regular complex boundary measures).

[L4]

Under ${A C}_{ω}$ the space $L^{p} (T, m; C)$ is reflexive for $1 < p < \infty$ ; under the ultrafilter lemma, DC and HB every norm-bounded sequence in a reflexive space has a weakly convergent subsequence; weak convergence $x_{j} ⇀ x$ means $Λ (x_{j}) \to Λ (x)$ for every bounded linear functional, and for $L^{p}$ the functionals $h \mapsto \int h g d m$ with $g \in L^{q}$ are bounded with $∥ h \mapsto \int h g d m ∥ \leq ∥ g ∥_{q}$ (Reflexivity of Lp for one less p less infinity, Reflexivity is equivalent to weak subsequential compactness of bounded sequences, The functional $Λ_{g}$ has norm $∥ g ∥_{q}$ ; for $q = \infty$ assume $μ$ is semifinite).

[L5]

For a measurable $f$ with $f s$ integrable for every complex finite simple $s$ of finite-measure support, $∥ f ∥_{p} = \sup {∣ \int_{T} f s d m ∣ : ∥ s ∥_{p^{'}} \leq 1}$ , where $p^{'}$ is conjugate to $p$ and $1 \leq p \leq \infty$ ; and Hölder gives $∣ \int h g d m ∣ \leq ∥ h ∥_{p} ∥ g ∥_{p^{'}}$ for conjugate exponents (Complex Lq norm recovery from finite simple dual tests, Complex Holder, Minkowski, and the quotient norm).

[L6]

The measure space $(T, B (T), m)$ is sigma-finite; every bounded real linear functional on the real space $L^{1} (T, m; R)$ is integration against a unique real $g \in L^{\infty} (T, m)$ with equal norms; the complex continuous functions are dense in $L^{1} (T, m; C)$ ; a bounded linear map from a dense normed subspace into a Banach space extends uniquely to the whole space with the same norm (On a sigma-finite measure space, every bounded linear functional on $L^{p}$ is integration against a unique $L^{q}$ function, Continuous functions are dense in $L^{p}$ of finite tori and of bounded intervals, A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm, The class $L^{1} (μ)$ of integrable functions, Complex Lp classes and Euclidean test-function conventions).

[L7]

Integration against every continuous function determines a finite regular complex Borel measure on $T$ uniquely; the density measure $f m$ of $f \in L^{1}$ is a finite Borel measure with $∣ f m ∣ (E) = \int_{E} ∣ f ∣ d m$ , and every finite Borel measure on the second-countable space $T$ is regular (The bounded complex dual of C_0(X) is regular complex measures, Locally finite Borel measures on second-countable LCH spaces are regular, The Poisson integral of a finite complex boundary measure).

[L8]

Fubini's theorem applies to integrable functions on the product of the sigma-finite spaces $(T, m)$ and $(T, m)$ , and $m$ is a translation invariant probability measure (Fubini's theorem for L^1 functions on a sigma-finite product, The one-dimensional torus and its normalized Haar integral).

[L9]

The Axiom of Choice implies DC and ${A C}_{ω}$ ; it implies the ultrafilter lemma; and it implies the dominated-extension principle HB (AC implies DC implies countable choice, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, Hahn-Banach dominated extension theorem for real vector spaces, The real dominated-extension principle as an additional hypothesis over ZF).

Proof

technique · direct

1.1givenL1L5

Setup. Let $u \in h^{p} (D)$ with $M = ∥ u ∥_{h^{p}}$ and let $r_{j} = 1 - \frac{1}{j + 1} ↑ 1$ . By [L1] the function $u$ is complex harmonic, hence continuous, on $D$ , and $u_{r} (ζ) = u (r ζ)$ is measurable with $∥ u_{r} ∥_{p} \leq M$ for every $0 \leq r < 1$ ; if $p = \infty$ this reads $∣ u_{r} ∣ \leq M$ everywhere, and then the probability measure $m$ gives $∥ u_{r} ∥_{1} \leq ∥ u_{r} ∥_{\infty} \leq M$ by [L5].

1.2givenL4L9

Choice bookkeeping. By [L9] the Axiom of Choice supplies ${A C}_{ω}$ , the ultrafilter lemma, DC and HB, so the reflexivity and weak-subsequence hypotheses recorded in [L4] are met.

2.1step 1.1step 1.2L4

The finite exponent case: a weak limit. Assume $1 < p < \infty$ . The sequence $(u_{r_{j}})$ is norm-bounded by $M$ in the reflexive space $L^{p} (T, m; C)$ by steps 1.1 and 1.2, so [L4] provides a subsequence, relabelled $(u_{r_{j}})$ , and an element $f \in L^{p} (T, m; C)$ with $u_{r_{j}} ⇀ f$ ; explicitly $\int_{T} u_{r_{j}} g d m \to \int_{T} f g d m$ for every $g \in L^{q} (T, m; C)$ .

2.2step 1.1L1L3L5

The exponent infinity: a boundary measure. Assume $p = \infty$ . By step 1.1, $∥ u_{r} ∥_{1} \leq M$ for every $r$ , so $u \in h^{1} (D)$ with $∥ u ∥_{h^{1}} \leq M$ ; [L3] therefore gives a unique finite regular complex Borel measure $μ$ on $T$ with $u = P [μ]$ , $∥ u ∥_{h^{1}} = ∣ μ ∣ (T)$ and $\int_{T} g d μ = \lim_{r ↑ 1} \int_{T} g u_{r} d m$ for every continuous $g$ .

3.1step 2.1L2L5algebra

The finite exponent case: identification of $u$ . Assume $1 < p < \infty$ and let $f$ be the weak limit of step 2.1. Fix $z \in D$ . For every $j$ with $r_{j} > ∣ z ∣$ the function $u$ is harmonic on an open set containing the closed disc of radius $r_{j}$ , so [L2] gives $u (z) = \int_{T} P^{(j)} (z, η) u_{r_{j}} (η) d m (η), P^{(j)} (z, η) : = \frac{r_{j}^{2} - ∣ z ∣^{2}}{∣ r_{j} η - z ∣^{2}},$ and writing $P (z, η) = (1 - ∣ z ∣^{2}) / ∣ η - z ∣^{2}$ this becomes $u (z) = \int_{T} P (z, η) u_{r_{j}} (η) d m (η) + \int_{T} (P^{(j)} (z, η) - P (z, η)) u_{r_{j}} (η) d m (η) .$ The function $P (z, \cdot)$ is continuous on $T$ , hence belongs to $L^{q}$ , so step 2.1 gives $\int_{T} P (z, \cdot) u_{r_{j}} d m \to \int_{T} P (z, \cdot) f d m = P [f] (z)$ ; the second integral is bounded in modulus by $∥ P^{(j)} (z, \cdot) - P (z, \cdot) ∥_{q} ∥ u_{r_{j}} ∥_{p}$ by [L5], and this tends to $0$ because $r_{j} ↑ 1$ , the point $z$ stays at positive distance from $T$ , and $∥ u_{r_{j}} ∥_{p} \leq M$ . Hence $u (z) = P [f] (z)$ for every $z \in D$ , that is $u = P [f]$ .

3.2step 2.2L5L6algebra

The exponent infinity: extension of the boundary functional. Assume $p = \infty$ and let $μ$ be the measure of step 2.2. For continuous $g$ and $0 \leq r < 1$ , [L5] gives $∣ \int_{T} g u_{r} d m ∣ \leq ∥ g ∥_{1} ∥ u_{r} ∥_{\infty} \leq M ∥ g ∥_{1}$ ; letting $r ↑ 1$ along the convergence of step 2.2 yields $∣ \int_{T} g d μ ∣ \leq M ∥ g ∥_{1}$ . Hence $W (g) : = \int_{T} g d μ$ is a complex linear functional on the dense subspace $C (T, C)$ of $L^{1} (T, m; C)$ satisfying $∣ W (g) ∣ \leq M ∥ g ∥_{1}$ , and the bound in particular shows that $W$ vanishes on continuous functions that are $m$ -almost everywhere zero, so $W$ is well defined on the corresponding subspace of the quotient; [L6] therefore extends it uniquely to a bounded complex linear functional $\tilde{W}$ on $L^{1} (T, m; C)$ with $∥ \tilde{W} ∥ \leq M$ .

4.1step 3.2L1L6L7algebra

The exponent infinity: the density. Keep $p = \infty$ , $\tilde{W}$ as in step 3.2. The functional $h \mapsto Re \tilde{W} (h)$ is real linear and bounded on the real space $L^{1} (T, m; R)$ with norm at most $∥ \tilde{W} ∥ \leq M$ , so [L6] (applied with $p = 1$ on the sigma-finite space $(T, m)$ ) provides $f_{1} \in L^{\infty} (T, m; R)$ with $Re \tilde{W} (h) = \int_{T} h f_{1} d m$ for all real $h \in L^{1}$ and $∥ f_{1} ∥_{\infty} \leq M$ ; applying the same theorem to $h \mapsto Im \tilde{W} (h)$ gives $f_{2} \in L^{\infty} (T, m; R)$ with $∥ f_{2} ∥_{\infty} \leq M$ . Put $f : = f_{1} + i f_{2} \in L^{\infty} (T, m; C)$ : by complex linearity of $\tilde{W}$ and of the integral, $\tilde{W} (h) = \int_{T} h f d m$ for every $h \in L^{1} (T, m; C)$ , and in particular $\int_{T} g d μ = \int_{T} g f d m$ for every continuous $g$ . Both $μ$ and the density measure $f m$ are finite Borel measures on the second-countable space $T$ , hence regular by [L7], and they agree on all continuous functions, so the uniqueness clause of [L7] gives $μ = f m$ ; consequently $u = P [μ] = P [f m] = P [f]$ by [L1].

4.2step 2.1step 3.1L1L5algebra

The finite exponent case: norm equality. Assume $1 < p < \infty$ , let $f$ be the weak limit of step 2.1 and keep the identification $u = P [f]$ of step 3.1. For every complex finite simple $s$ of finite-measure support with $∥ s ∥_{q} \leq 1$ , step 2.1 gives $\int_{T} f s d m = \lim_{j} \int_{T} u_{r_{j}} s d m$ , and [L5] bounds $∣ \int_{T} u_{r_{j}} s d m ∣ \leq ∥ u_{r_{j}} ∥_{p} ∥ s ∥_{q} \leq M$ ; the norm identity of [L5] therefore gives $∥ f ∥_{p} \leq M$ . Since $u = P [f]$ , the contraction in [L1] gives $∥ u ∥_{h^{p}} = ∥ P [f] ∥_{h^{p}} \leq ∥ f ∥_{p}$ , and hence $∥ f ∥_{p} = M = ∥ u ∥_{h^{p}}$ .

5.1step 3.1step 4.2L1

The finite exponent case: uniqueness and norm convergence. Assume $1 < p < \infty$ and let $g \in L^{p} (T, m; C)$ satisfy $P [g] = u = P [f]$ . Then $P [f - g] = 0$ , and the convergence clause of [L1] gives $∥ (P [f - g])_{r} - (f - g) ∥_{p} \to 0$ , so $f - g = 0$ almost everywhere and $f$ is the unique representing function. The same convergence clause applied to $f$ yields $∥ u_{r} - f ∥_{p} = ∥ (P [f])_{r} - f ∥_{p} \to 0$ as $r ↑ 1$ .

5.2step 3.2step 4.1L5algebra

The exponent infinity: the sharp norm bound. Keep $p = \infty$ and $f = f_{1} + i f_{2}$ from step 4.1. For every complex finite simple $s$ of finite-measure support with $∥ s ∥_{1} \leq 1$ one has $s \in L^{1} (T, m; C)$ , so step 4.1 gives $\int_{T} f s d m = \tilde{W} (s)$ and hence $∣ \int_{T} f s d m ∣ \leq ∥ \tilde{W} ∥ ∥ s ∥_{1} \leq M$ . The norm identity of [L5] at $p = \infty$ , whose hypothesis $\int ∣ f s ∣ d m < \infty$ holds because $f \in L^{\infty}$ and $s$ is bounded with finite-measure support, gives $∥ f ∥_{\infty} \leq M$ .

6.1step 4.1step 5.2L1algebra

The exponent infinity: norm equality and uniqueness. Assume $p = \infty$ . By step 4.1, $u = P [f]$ with $f \in L^{\infty} (T, m; C)$ , so the contraction of [L1] at $p = \infty$ gives $∥ u ∥_{h^{\infty}} = ∥ P [f] ∥_{h^{\infty}} \leq ∥ f ∥_{\infty} \leq M = ∥ u ∥_{h^{\infty}}$ , and therefore $∥ f ∥_{\infty} = ∥ u ∥_{h^{\infty}}$ . If also $P [g] = u$ with $g \in L^{\infty}$ , then $f - g \in L^{1}$ and $P [f - g] = 0$ , so the finite exponent convergence clause of [L1] at $p = 1$ gives $∥ (P [f - g])_{r} - (f - g) ∥_{1} \to 0$ , whence $f = g$ almost everywhere.

7.1step 4.1step 6.1L1L5L8algebra

The exponent infinity: weak-star convergence of the radial functions. Keep $p = \infty$ and $f$ as in step 4.1. For $g \in L^{1} (T, m; C)$ and $0 \leq r < 1$ one has $u_{r} = P_{r} * f$ by [L1], and the product integrand $(ζ, η) \mapsto g (ζ) P_{r} (ζ - η) f (η)$ is integrable for the product of the probability measure $m$ with itself, because $∣ f ∣ \leq ∥ f ∥_{\infty}$ and $\int_{T} P_{r} (ζ - η) d m (ζ) = 1$ for every $η$ ; Fubini [L8] and the translation invariance of $m$ therefore give $\int_{T} g u_{r} d m = \int_{T} (\int_{T} g (ζ) P_{r} (ζ - η) d m (ζ)) f (η) d m (η) = \int_{T} (P_{r} * g) (η) f (η) d m (η) .$ Consequently [L5] bounds $∣ \int_{T} g u_{r} d m - \int_{T} g f d m ∣ \leq ∥ P_{r} * g - g ∥_{1} ∥ f ∥_{\infty}$ , which tends to $0$ as $r ↑ 1$ by the $L^{1}$ convergence clause of [L1]; hence $u_{r} m \overset{*}{⇀} f$ in $σ (L^{\infty} (T, m), L^{1} (T, m))$ .

8.1step 1.2step 2.1step 2.2step 3.1step 4.1step 4.2step 5.1step 5.2step 6.1step 7.1L1L3L4∎

Assembly. If $1 < p < \infty$ , steps 2.1, 3.1, 4.2 and 5.1 produce a unique $f \in L^{p} (T, m; C)$ with $u = P [f]$ , the norm identity $∥ u ∥_{h^{p}} = ∥ f ∥_{p}$ and the $L^{p}$ convergence $∥ u_{r} - f ∥_{p} \to 0$ . If $p = \infty$ , steps 2.2, 3.2, 4.1, 5.2, 6.1 and 7.1 produce a unique $f \in L^{\infty} (T, m; C)$ with $u = P [f]$ , the norm identity $∥ u ∥_{h^{\infty}} = ∥ f ∥_{\infty}$ and the weak-star convergence of the radial functions against $L^{1}$ ; no norm convergence is claimed at $p = \infty$ , in accordance with the fact that [L1] asserts norm convergence only for finite exponents. The Axiom of Choice was used exactly through step 1.2: ${A C}_{ω}$ for reflexivity of $L^{p}$ and the ultrafilter lemma, DC and HB for the weak-subsequence criterion of [L4], while the case $p = \infty$ additionally rests on the $h^{1}$ representation theorem [L3], itself licensed by AC. This proves all the assertions of the Statement.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

The circle maximal function and nontangential approach regions

Definition

Assume countable choice. Identify the torus $T = R / Z$ with the Euclidean unit circle through $φ ([t]) = e^{2 π i t}$ , and write $m$ for the normalized Haar measure of $T$ (The one-dimensional torus and its normalized Haar integral), a probability measure on the Borel sets of $T$ . For $ζ, η \in T$ put $d (ζ, η) : = \min {∣ s - t - k ∣ : k \in Z}, ζ = [t], η = [s],$ the circular distance; it is well defined because replacing $t$ or $s$ by an integer translate does not change the set of numbers $∣ s - t - k ∣$ .

Centered arcs. For $ζ \in T$ and $0 < h \leq \frac{1}{2}$ put $I_{h} (ζ) : = {η \in T : d (ζ, η) < h} (0 < h < \frac{1}{2}), I_{1 / 2} (ζ) : = T .$ Thus $I_{h} (ζ)$ is the open circular arc of radius $h$ centered at $ζ$ , the point $ζ$ itself included, and the half-circle case is deliberately the whole circle so that the antipode is not lost. The normalization is the one used throughout this pair: for $0 < h \leq \frac{1}{2}$ , $m (I_{h} (ζ)) = 2 h .$ For $0 < h < \frac{1}{2}$ one has $I_{h} ([0]) = q ((- h, h))$ for the quotient map $q : R \to T$ , and $q^{- 1} q ((- h, h)) = ⋃_{k \in Z} (- h + k, h + k)$ meets $[0, 1)$ in $[0, h) \cup (1 - h, 1)$ , a set of measure $2 h$ ; translation invariance of $m$ (proved with the measure $m$ in The one-dimensional torus and its normalized Haar integral) moves this identity to every center. The case $h = \frac{1}{2}$ reads $m (T) = 1 = 2 \cdot \frac{1}{2}$ by the convention $I_{1 / 2} (ζ) = T$ .

Circle maximal function. Let $μ$ be a finite regular complex Borel measure on $T$ (Regular complex Borel measures), with total variation $∣ μ ∣$ (The total variation |nu|(E) from countable measurable partitions, The total variation of a signed or complex measure is a positive measure). Define $M_{T} μ (ζ) : = \sup_{0 < h \leq 1 / 2} \frac{∣ μ ∣ (I_{h} (ζ))}{m (I_{h} (ζ))}, ζ \in T .$ Each quotient is finite because $∣ μ ∣ (T) < + \infty$ , and each is nonnegative. Their supremum is therefore a well-defined extended nonnegative real: $M_{T} μ : T \to [0, + \infty]$ . It may equal $+ \infty$ , for example at an atom of $∣ μ ∣$ . For $f \in L^{1} (T, m)$ (The class $L^{1} (μ)$ of integrable functions) define likewise $M_{T} f (ζ) : = \sup_{0 < h \leq 1 / 2} \frac{1}{m (I_{h} (ζ))} \int_{I_{h} (ζ)} ∣ f ∣ d m .$ The two definitions agree when $μ = f m$ is the density measure of $f$ , that is when $μ (E) = \int_{E} f d m$ : then $∣ μ ∣ (E) = \int_{E} ∣ f ∣ d m$ (A complex L^1 density defines a complex measure whose total variation is |h| dmu), so $∣ μ ∣ (I_{h} (ζ)) = \int_{I_{h} (ζ)} ∣ f ∣ d m$ and $M_{T} μ = M_{T} f$ . The assignment $f \mapsto M_{T} f$ is unchanged if $f$ is replaced by an almost-everywhere equal function, because the integrals over the arc agree.

Nontangential regions. For $A > 1$ and $ζ \in T$ put $Γ_{A} (ζ) : = {z \in D : ∣ z - ζ ∣ < A (1 - ∣ z ∣)} \subseteq D,$ and for $v : D \to C$ let $N_{A} v (ζ) : = \sup_{z \in Γ_{A} (ζ)} ∣ v (z) ∣ \in [0, + \infty] .$ Here $∣ z - ζ ∣$ is the Euclidean modulus after identifying $T$ with the unit circle. The sets are nested: $Γ_{A} (ζ) \subseteq Γ_{B} (ζ)$ for $1 < A \leq B$ , and $0 \in Γ_{A} (ζ)$ for every $A > 1$ because $∣ 0 - ζ ∣ = 1 < A$ . A point $z \in D ∖ {ζ}$ lies in $Γ_{A} (ζ)$ as soon as $A > ∣ z - ζ ∣ / (1 - ∣ z ∣)$ , and $∣ z - ζ ∣ \geq 1 - ∣ z ∣$ for every $z \in D$ , so every such $z$ lies in some $Γ_{A} (ζ)$ and $⋃_{A > 1} Γ_{A} (ζ) = D ∖ {ζ}$ . A complex-valued $v$ on $D$ has nontangential limit $L$ at $ζ$ if for every $A > 1$ and every $ε > 0$ there is $δ > 0$ with $∣ v (z) - L ∣ < ε$ whenever $z \in Γ_{A} (ζ)$ and $∣ z - ζ ∣ < δ$ . Because the regions increase with $A$ , it suffices to verify this for every integer $A \geq 2$ : an arbitrary $A > 1$ satisfies $Γ_{A} (ζ) \subseteq Γ_{m} (ζ)$ for every integer $m \geq A$ .

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The circle maximal function is weak type one one for finite measures

Statement

Assume countable choice. For every finite regular complex Borel measure $μ$ on $T$ and every real $λ > 0$ , the superlevel set ${M_{T} μ > λ}$ is Borel measurable and $m ({M_{T} μ > λ}) \leq \frac{3 ∣ μ ∣ (T)}{λ} .$ In particular, for $f \in L^{1} (T, m)$ one has $m ({M_{T} f > λ}) \leq 3 ∥ f ∥_{1} / λ$ with $∥ f ∥_{1} = \int_{T} ∣ f ∣ d m$ .

Facts & Assumptions

Given: Countable choice, a finite regular complex Borel measure $μ$ on $T$ , and a real number $λ > 0$ .

[L1]

For $ζ \in T$ and $0 < h \leq \frac{1}{2}$ the set $I_{h} (ζ)$ is the centered open arc of radius $h$ (the whole circle when $h = \frac{1}{2}$ ) and $m (I_{h} (ζ)) = 2 h$ ; moreover $M_{T} μ (ζ) = \sup_{0 < h \leq 1 / 2} \frac{∣ μ ∣ (I_{h} (ζ))}{m (I_{h} (ζ))}, M_{T} f (ζ) = \sup_{0 < h \leq 1 / 2} \frac{1}{m (I_{h} (ζ))} \int_{I_{h} (ζ)} ∣ f ∣ d m$ (The circle maximal function and nontangential approach regions).

[L2]

For a complex measure the total variation $∣ μ ∣$ is a measure, so monotonicity gives $∣ μ ∣ (E) \leq ∣ μ ∣ (T) < + \infty$ for every Borel $E$ (The total variation |nu|(E) from countable measurable partitions, The total variation of a signed or complex measure is a positive measure).

[L3]

Fatou's lemma: for nonnegative measurable functions on a measure space, $\int {lim inf}_{n} f_{n} d ν \leq {lim inf}_{n} \int f_{n} d ν$ (Fatou's lemma).

[L4]

The normalized Haar measure $m$ is a probability measure on the compact second-countable Hausdorff space $T$ , and every Borel set $E$ satisfies $m (E) = \sup {m (K) : K \subseteq E compact}$ (The one-dimensional torus and its normalized Haar integral, Locally finite Borel measures on second-countable LCH spaces are regular).

[L5]

For $f \in L^{1} (T, m)$ the density measure $f m$ is a complex measure with $∣ f m ∣ (E) = \int_{E} ∣ f ∣ d m$ and $∥ f ∥_{1} = \int_{T} ∣ f ∣ d m$ , and $M_{T} (f m) = M_{T} f$ (A complex L^1 density defines a complex measure whose total variation is |h| dmu, The circle maximal function and nontangential approach regions, Complex Holder, Minkowski, and the quotient norm).

[L6]

A compact metric subspace admits a finite subcover from every family of ambient open sets covering it (A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Proof

technique · direct

1.1givenL1L2L3algebra

First let $0 < h < \frac{1}{2}$ . If $ζ_{n} \to ζ$ in $T$ , then for every $η \in I_{h} (ζ)$ the triangle inequality for the circular distance gives $d (ζ_{n}, η) \leq d (ζ_{n}, ζ) + d (ζ, η) < h$ for all large $n$ , so the indicators satisfy $1_{I_{h} (ζ)} \leq {lim inf}_{n} 1_{I_{h} (ζ_{n})}$ pointwise; applying [L3] to the finite measure $∣ μ ∣$ of [L2] gives $∣ μ ∣ (I_{h} (ζ)) \leq {lim inf}_{n} ∣ μ ∣ (I_{h} (ζ_{n}))$ , that is, the map $ζ \mapsto ∣ μ ∣ (I_{h} (ζ))$ is lower semicontinuous. For $h = \frac{1}{2}$ , [L1] gives $I_{h} (ζ) = T$ for every centre, so the mass function is constant and hence lower semicontinuous. Since [L1] makes $m (I_{h} (ζ)) = 2 h$ independent of $ζ$ for every $0 < h \leq \frac{1}{2}$ , the set $A_{h} : = {ζ \in T : ∣ μ ∣ (I_{h} (ζ)) > λ m (I_{h} (ζ))}$ is the superlevel set of a lower semicontinuous function and is therefore open.

2.1step 1.1L1

Because $M_{T} μ$ is the supremum of the quotients over $0 < h \leq \frac{1}{2}$ , the identity ${M_{T} μ > λ} = ⋃_{0 < h \leq 1 / 2} A_{h}$ holds; it is a union of open sets, so ${M_{T} μ > λ}$ is open and in particular Borel measurable.

3.1step 2.1L6givenconstruct

Let $K \subseteq {M_{T} μ > λ}$ be compact. If $K = \emptyset$ , then $m (K) = 0 \leq 3 ∣ μ ∣ (T) / λ$ ; assume henceforth that $K \neq \emptyset$ . The family of all open arcs $I_{h} (ζ)$ with $h \in (0, \frac{1}{2}]$ and $∣ μ ∣ (I_{h} (ζ)) > λ m (I_{h} (ζ))$ is a family of open subsets of $T$ , described by a formula and hence requiring no selection, that covers $K$ by step 2.1; [L6] provides a finite subcover $I_{1}, \dots, I_{N}$ of $K$ by such arcs, each satisfying $∣ μ ∣ (I_{j}) > λ m (I_{j})$ .

4.1step 3.1L1algebra

Relabel the finite list so that the radii satisfy $h_{1} \geq h_{2} \geq \dots \geq h_{N}$ , and pass through it once, keeping an arc exactly when it is disjoint from every previously kept arc. The kept arcs are pairwise disjoint and each still satisfies $∣ μ ∣ (I_{i}) > λ m (I_{i})$ . If $h_{1} = \frac{1}{2}$ , the first kept arc is $T$ and contains every arc of the subcover; set ${\hat{I}}_{1} = T$ , whose measure is at most $3 m (I_{1})$ . Otherwise all radii are strictly less than $\frac{1}{2}$ . If $I_{j}$ with center $c_{j}$ is rejected, it meets a kept arc $I_{i}$ with center $c_{i}$ and $i < j$ , so $h_{i} \geq h_{j}$ and $d (c_{i}, c_{j}) \leq h_{i} + h_{j} \leq 2 h_{i}$ ; every $η \in I_{j}$ therefore satisfies $d (η, c_{i}) \leq d (η, c_{j}) + d (c_{j}, c_{i}) < h_{j} + 2 h_{i} \leq 3 h_{i}$ . Writing ${\hat{I}}_{i} : = {η : d (η, c_{i}) < 3 h_{i}}$ , every arc of the subcover lies in ${\hat{I}}_{i}$ for some kept arc $I_{i}$ , and $m ({\hat{I}}_{i}) \leq 3 m (I_{i})$ : if $3 h_{i} < \frac{1}{2}$ this reads $6 h_{i} = 3 \cdot 2 h_{i}$ , while if $3 h_{i} \geq \frac{1}{2}$ then $m ({\hat{I}}_{i}) = 1 \leq 6 h_{i} = 3 m (I_{i})$ ; at equality the antipode is excluded but has measure zero.

5.1step 4.1L2algebra

The kept arcs are pairwise disjoint, so their $m$ -measures add and their $∣ μ ∣$ -values add; by step 4.1 and [L2], $λ m (⋃_{i} I_{i}) = λ \sum_{i} m (I_{i}) < \sum_{i} ∣ μ ∣ (I_{i}) = ∣ μ ∣ (⋃_{i} I_{i}) \leq ∣ μ ∣ (T) .$

6.1step 4.1step 5.1algebra

The arcs $I_{1}, \dots, I_{N}$ cover $K$ and each lies in some ${\hat{I}}_{i}$ of a kept arc, so step 4.1 and step 5.1 give $m (K) \leq m (⋃_{i} {\hat{I}}_{i}) \leq \sum_{i} m ({\hat{I}}_{i}) \leq 3 \sum_{i} m (I_{i}) < \frac{3 ∣ μ ∣ (T)}{λ} .$

7.1step 2.1step 6.1L2L4

By [L4] the measure of the Borel set ${M_{T} μ > λ}$ is the supremum of $m (K)$ over compact $K \subseteq {M_{T} μ > λ}$ ; step 2.1 supplies the measurability and step 6.1 bounds every such $m (K)$ by $3 ∣ μ ∣ (T) / λ$ , so $m ({M_{T} μ > λ}) \leq 3 ∣ μ ∣ (T) / λ$ .

8.1step 7.1L5∎

Let $f \in L^{1} (T, m)$ and apply step 7.1 to the finite complex measure $f m$ : [L5] gives $∣ f m ∣ (T) = \int_{T} ∣ f ∣ d m = ∥ f ∥_{1}$ and $M_{T} (f m) = M_{T} f$ , so $m ({M_{T} f > λ}) \leq 3 ∥ f ∥_{1} / λ$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Poisson nontangential maximal function is controlled by circle maximal averages

Statement

Assume countable choice. For every finite regular complex Borel measure $μ$ on $T$ , every $A > 1$ and every $ζ \in T$ , $N_{A} (P [μ]) (ζ) \leq (A + 1)^{2} M_{T} μ (ζ) .$ In particular $N_{A} (P [f]) (ζ) \leq (A + 1)^{2} M_{T} f (ζ)$ for every $f \in L^{1} (T, m)$ , where $P [f] = P [f m]$ and $M_{T} f = M_{T} (f m)$ .

Facts & Assumptions

Given: Countable choice, a finite regular complex Borel measure $μ$ on $T$ , a real $A > 1$ , and a point $ζ \in T$ .

[L1]

The sets $Γ_{A} (ζ) = {z \in D : ∣ z - ζ ∣ < A (1 - ∣ z ∣)}$ and $N_{A} v (ζ) = \sup_{z \in Γ_{A} (ζ)} ∣ v (z) ∣$ are the nontangential regions and maximal functions, and $M_{T} μ (ζ) = \sup_{0 < h \leq 1 / 2} ∣ μ ∣ (I_{h} (ζ)) / m (I_{h} (ζ))$ takes values in $[0, + \infty]$ ; moreover $M_{T} f = M_{T} (f m)$ and $P [f] = P [f m]$ (The circle maximal function and nontangential approach regions, The Poisson integral of a finite complex boundary measure).

[L2]

For $z \in D$ and $η \in T$ the kernel is $P (z, η) = (1 - ∣ z ∣^{2}) / ∣ η - z ∣^{2} > 0$ ; for $z = r ζ$ and $η = ζ e^{2 π i ε}$ with $∣ ε ∣ \leq \frac{1}{2}$ one has $∣ η - r ζ ∣ = ∣ e^{2 π i ε} - r ∣$ , so $∣ η - r ζ ∣^{2} = 1 - 2 r \cos (2 π ε) + r^{2}$ (The Poisson kernel on the unit disc).

[L3]

Cosine is strictly decreasing on $[0, π]$ , and sine and cosine are continuous (indeed $1$ -Lipschitz) (Signs, monotonicity intervals, and ranges of sine and cosine, Sine and cosine are $1$ -Lipschitz on $R$ ).

[L4]

For every $f \in L^{1} (μ)$ the bound $∣ \int f d μ ∣ \leq \int ∣ f ∣ d ∣ μ ∣$ holds, where $∣ μ ∣$ is a measure with $∣ μ ∣ (E) \leq ∣ μ ∣ (T) < + \infty$ (Integrals against signed or complex measures are bounded by total variation, The total variation of a signed or complex measure is a positive measure).

[L5]

The normalized Haar measure $m$ is a probability measure with $m (I_{h} (ζ)) = 2 h$ for $0 < h \leq \frac{1}{2}$ , and $\int_{T} P (z, η) d m (η) = 1$ for every $z \in D$ ; equality in the second display of [L1] holds for every arc radius (The one-dimensional torus and its normalized Haar integral, The Poisson kernel is positive, has total mass one, and concentrates at a boundary point, The Poisson integral of a finite complex boundary measure).

Proof

technique · direct

1.1givenL1L2algebra

Let $z \in Γ_{A} (ζ)$ , $r = ∣ z ∣$ , and $η \in T$ . Put $u : = Re (\overline{ζ} z) \leq r$ . Since $∣ r ζ - z ∣^{2} = 2 r^{2} - 2 r u$ and $∣ ζ - z ∣^{2} = 1 + r^{2} - 2 u$ , the identity $2 r^{2} - 2 r u \leq 1 + r^{2} - 2 u$ is equivalent to $(1 - r) (1 + r - 2 u) \geq 0$ , which holds because $1 - r > 0$ and $2 u \leq 2 r \leq 1 + r$ ; hence $∣ r ζ - z ∣ \leq ∣ ζ - z ∣$ . Therefore $∣ r ζ - η ∣ \leq ∣ r ζ - z ∣ + ∣ z - η ∣ \leq ∣ ζ - z ∣ + ∣ z - η ∣ \leq A (1 - r) + ∣ z - η ∣ \leq (A + 1) ∣ z - η ∣,$ where the cone condition gives $∣ ζ - z ∣ < A (1 - r)$ and the reverse triangle inequality gives $∣ z - η ∣ \geq 1 - ∣ z ∣ = 1 - r$ . The case $z = 0$ is included: then $r = 0$ , $u = 0$ and $∣ r ζ - η ∣ = 1 \leq A + 1 = (A + 1) ∣ z - η ∣$ .

1.2givenL2L3L5choosealgebra

Fix $z = r ζ$ with $0 < r < 1$ and put $g (δ) : = (1 - r^{2}) / (1 - 2 r \cos (2 π δ) + r^{2})$ for $δ \in [0, \frac{1}{2}]$ . By [L2], $P (z, η) = g (d (ζ, η))$ for every $η \in T$ , and by [L3] the function $g$ is continuous and strictly decreasing on $[0, \frac{1}{2}]$ . Given $ε > 0$ , choose $0 = δ_{0} < δ_{1} < \dots < δ_{m} = \frac{1}{2}$ with $g (δ_{j - 1}) - g (δ_{j}) \leq ε$ for all $j$ (continuity on a compact interval); put $c_{0} : = g (\frac{1}{2})$ and $c_{j} : = g (δ_{j - 1}) - g (δ_{j}) \geq 0$ . Then for every $δ \in [0, \frac{1}{2}]$ one has $g (δ) \leq ϕ (δ) : = c_{0} + \sum_{j = 1}^{m} c_{j} 1_{[0, δ_{j})} (δ) \leq g (δ) + ε,$ because on $[δ_{k - 1}, δ_{k})$ the function $ϕ$ equals $g (δ_{k - 1})$ while $g (δ_{k}) \leq g (δ) \leq g (δ_{k - 1})$ and $g (δ_{k - 1}) - g (δ_{k}) = c_{k} \leq ε$ ; on the single point $δ = \frac{1}{2}$ both sides equal $g (\frac{1}{2})$ . Consequently $P (z, η) \leq c_{0} + \sum_{j} c_{j} 1_{{d (ζ, η) < δ_{j}}} (η)$ for every $η$ , while integrating the two-sided bound against $m$ and using $m (T) = 1$ together with $\int_{T} g (d (ζ, η)) d m (η) = \int_{T} P (z, η) d m (η) = 1$ gives $c_{0} + \sum_{j = 1}^{m} c_{j} m ({d (ζ, \cdot) < δ_{j}}) \leq 1 + ε .$

1.3givenL1L4L5algebra

By definition of $M_{T} μ$ as a supremum over $h \in (0, \frac{1}{2}]$ , every arc satisfies $∣ μ ∣ (I_{h} (ζ)) \leq M_{T} μ (ζ) m (I_{h} (ζ)) = 2 h M_{T} μ (ζ)$ ; in particular $∣ μ ∣ (T) = ∣ μ ∣ (I_{1 / 2} (ζ)) \leq M_{T} μ (ζ)$ . Moreover each set ${d (ζ, \cdot) < δ}$ with $δ \in (0, \frac{1}{2}]$ satisfies $m ({d < δ}) = 2 δ$ and $∣ μ ∣ ({d < δ}) \leq M_{T} μ (ζ) m ({d < δ})$ : for $δ < \frac{1}{2}$ the set is the arc $I_{δ} (ζ)$ , and for $δ = \frac{1}{2}$ it is $T$ minus the antipode, whose $m$ -measure is $1$ and whose $∣ μ ∣$ -measure is at most $∣ μ ∣ (T) \leq M_{T} μ (ζ)$ .

2.1step 1.2step 1.3L1L4algebra

Put $M : = M_{T} μ (ζ)$ . If $M = + \infty$ , the desired bound is automatic. Assume $M < + \infty$ . For $0 < r < 1$ the pointwise bound of step 1.2 and the estimates of step 1.3 give $∣ P [μ] (r ζ) ∣ \leq \int_{T} P (r ζ, η) d ∣ μ ∣ (η) \leq c_{0} ∣ μ ∣ (T) + \sum_{j = 1}^{m} c_{j} ∣ μ ∣ ({d (ζ, \cdot) < δ_{j}}) \leq M (c_{0} + \sum_{j = 1}^{m} c_{j} m ({d (ζ, \cdot) < δ_{j}})) \leq (1 + ε) M,$ where [L4] supplies the first inequality. Since $ε > 0$ was arbitrary, $\int_{T} P (r ζ, η) d ∣ μ ∣ (η) \leq M$ and hence $∣ P [μ] (r ζ) ∣ \leq M$ . For $r = 0$ , $P (0, η) = 1$ and step 1.3 give $\int_{T} P (0, η) d ∣ μ ∣ (η) = ∣ μ ∣ (T) \leq M$ , while [L4] gives $∣ P [μ] (0) ∣ = ∣ μ (T) ∣ \leq ∣ μ ∣ (T)$ .

2.2step 1.1L2algebra

For $z \in Γ_{A} (ζ)$ with $∣ z ∣ = r$ and every $η \in T$ , step 1.1 gives $∣ η - r ζ ∣ \leq (A + 1) ∣ η - z ∣$ , hence $P (z, η) = \frac{1 - r^{2}}{∣ η - z ∣^{2}} \leq (A + 1)^{2} \frac{1 - r^{2}}{∣ η - r ζ ∣^{2}} = (A + 1)^{2} P (r ζ, η) .$

3.1step 2.1step 2.2L1L4algebra∎

Combining steps 2.1 and 2.2, for every $z \in Γ_{A} (ζ)$ with $∣ z ∣ = r$ , $∣ P [μ] (z) ∣ \leq \int_{T} P (z, η) d ∣ μ ∣ (η) \leq (A + 1)^{2} \int_{T} P (r ζ, η) d ∣ μ ∣ (η) \leq (A + 1)^{2} M_{T} μ (ζ),$ the first inequality by [L4]. Taking the supremum over $z \in Γ_{A} (ζ)$ gives $N_{A} (P [μ]) (ζ) \leq (A + 1)^{2} M_{T} μ (ζ)$ . For $f \in L^{1} (T, m)$ the identities $P [f] = P [f m]$ and $M_{T} f = M_{T} (f m)$ of [L1] give $N_{A} (P [f]) (ζ) \leq (A + 1)^{2} M_{T} f (ζ)$ , completing the proof.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Fatou limits for Poisson extensions of L1 boundary data

Statement

Assume countable choice. If $f \in L^{1} (T, m; C)$ , then for $m$ -almost every $ζ \in T$ one has $P [f] (z) \to f (ζ)$ as $z \to ζ$ within every fixed nontangential region $Γ_{A} (ζ)$ , $A > 1$ . The assertion uses an almost-everywhere representative of $f$ and makes no claim about arbitrary tangential paths.

Facts & Assumptions

Given: Countable choice, a function $f \in L^{1} (T, m; C)$ , and the nontangential regions $Γ_{A} (ζ) = {z \in D : ∣ z - ζ ∣ < A (1 - ∣ z ∣)}$ of [L1].

[L1]

The region $Γ_{A} (ζ)$ , the nontangential maximal function $N_{A} v (ζ) = \sup_{z \in Γ_{A} (ζ)} ∣ v (z) ∣$ , the circle maximal function $M_{T} f$ and the definition $P [f] = P [f m]$ are as in the two definitions cited; the regions increase with the aperture, so verifying a nontangential limit for every integer aperture $m \geq 2$ verifies it for every $A > 1$ (The circle maximal function and nontangential approach regions, The Poisson integral of a finite complex boundary measure).

[L2]

Weak type: $m ({M_{T} u > λ}) \leq 3 ∥ u ∥_{1} / λ$ for every $u \in L^{1} (T, m)$ and $λ > 0$ (The circle maximal function is weak type one one for finite measures).

[L3]

Nontangential maximal bound: $N_{A} (P [ν]) (ζ) \leq (A + 1)^{2} M_{T} ν (ζ)$ for every finite complex Borel measure $ν$ , every $A > 1$ and every $ζ$ (Poisson nontangential maximal function is controlled by circle maximal averages).

[L4]

For $f \in L^{1}$ the Poisson integral $P [f]$ is complex harmonic, hence continuous on $D$ ; for continuous $g$ the radial functions satisfy $∥ P_{r} * g - g ∥_{\infty} \to 0$ (Poisson extension is an Lp contraction and converges in finite Lp, The Poisson kernel is a boundary approximate identity).

[L5]

The kernel satisfies $P (z, η) = (1 - ∣ z ∣^{2}) / ∣ η - z ∣^{2} > 0$ , $\int_{T} P (z, η) d m (η) = 1$ , and $\sup_{δ \leq ∣ θ ∣ \leq π} P_{r} (θ) \to 0$ as $r ↑ 1$ for every $δ \in (0, π]$ (The Poisson kernel is positive, has total mass one, and concentrates at a boundary point, The Poisson kernel on the unit disc, The one-dimensional torus and its normalized Haar integral).

[L6]

Continuous complex functions on $T$ are dense in $L^{1} (T, m; C)$ (Continuous functions are dense in $L^{p}$ of finite tori and of bounded intervals).

[L7]

The set $Q^{2} \cap D$ is countable and dense in $D$ ; every nonempty open subset of $D$ therefore contains a point of it ( $Q^{n}$ is a countable dense subset of $R^{n}$ , and rational open boxes form a countable basis).

[L8]

Chebyshev's inequality: $m ({∣ u ∣ > t}) \leq ∥ u ∥_{1} / t$ for $u \in L^{1}$ and $t > 0$ (Chebyshev-Markov inequality for the integral).

[L9]

A countable union of measurable $m$ -null sets is $m$ -null (Finite and countable subadditivity of measures).

Proof

technique · direct

1.1givenL1L5algebra

Continuous data converge in every cone. Let $g \in C (T, C)$ , $ζ \in T$ and $A > 1$ , and let $ε > 0$ ; choose $δ > 0$ with $∣ g (η) - g (ζ) ∣ < ε$ for $d (ζ, η) < δ$ . For $z \in Γ_{A} (ζ)$ with $r = ∣ z ∣$ and $u : = Re (\overline{ζ} z) \leq r$ one has $∣ r ζ - z ∣^{2} = 2 r (r - u) \leq 1 + r^{2} - 2 u = ∣ ζ - z ∣^{2}$ because $(1 - r) (1 + r - 2 u) \geq 0$ , so for every $η \in T$ the triangle inequality gives $∣ η - r ζ ∣ \leq ∣ η - z ∣ + ∣ ζ - z ∣ \leq (1 + A) ∣ η - z ∣$ , using $∣ η - z ∣ \geq 1 - r$ and the cone condition. Hence, by [L5] and the unit mass of the kernel, $∣ P [g] (z) - g (ζ) ∣ \leq \int_{T} P (z, η) ∣ g (η) - g (ζ) ∣ d m (η) \leq ε + 2 ∥ g ∥_{\infty} (A + 1)^{2} \sup_{2 π δ \leq ∣ θ ∣ \leq π} P_{r} (θ),$ and the last supremum tends to $0$ as $r ↑ 1$ ; since $z \to ζ$ inside $Γ_{A} (ζ)$ forces $r \to 1$ , this is less than $2 ε$ for $z$ close enough to $ζ$ . Thus $P [g] (z) \to g (ζ)$ along $Γ_{A} (ζ)$ , and in particular the cone limsup $L_{m} (g) (ζ) : = {lim sup}_{z \to ζ, z \in Γ_{m} (ζ)} ∣ P [g] (z) - g (ζ) ∣$ is $0$ for every integer $m \geq 2$ .

1.2givenL1L4L7algebra

The cone limsup is Borel measurable. Fix an integer $m \geq 2$ and let $D_{0} : = Q^{2} \cap D$ , countable and dense in $D$ by [L7]. For $j \geq 1$ put $S_{j} (ζ) : = \sup {∣ P [f] (z) - f (ζ) ∣ : z \in D_{0}, z \in Γ_{m} (ζ), ∣ z - ζ ∣ < 1 / j}$ . For fixed $z \in D_{0}$ the summand is the product of the constant $∣ P [f] (z) - f (ζ) ∣$ restricted to the Borel set ${ζ \in T : ∣ z - ζ ∣ < m (1 - ∣ z ∣), ∣ z - ζ ∣ < 1 / j}$ ; a countable supremum of Borel measurable functions is Borel measurable, so every $S_{j}$ is Borel measurable and so is $L_{m} : = \inf_{j} S_{j}$ . Moreover, since $P [f]$ is continuous on $D$ by [L4] and $D_{0}$ is dense, the supremum over the points of $D_{0}$ in the open set $U_{j} (ζ) : = Γ_{m} (ζ) \cap {∣ z - ζ ∣ < 1 / j}$ equals the supremum over all of $U_{j} (ζ)$ : every point of $U_{j} (ζ)$ is a limit of points of $D_{0} \cap U_{j} (ζ)$ . Therefore $L_{m} (ζ) = \inf_{j} S_{j} (ζ) = {lim sup}_{z \to ζ, z \in Γ_{m} (ζ)} ∣ P [f] (z) - f (ζ) ∣$ is exactly the cone limsup, and it is Borel measurable.

2.1step 1.1L1L3algebra

Pointwise error bound. Let $g \in C (T, C)$ and let $m \geq 2$ be an integer. By step 1.1, $L_{m} (g) (ζ) = 0$ for every $ζ$ , and limsup subadditivity gives $L_{m} (f) (ζ) \leq L_{m} (f - g) (ζ) + L_{m} (g) (ζ) \leq N_{m} (P [f - g]) (ζ) + ∣ f - g ∣ (ζ) \leq (m + 1)^{2} M_{T} (f - g) (ζ) + ∣ f - g ∣ (ζ),$ where the middle inequality uses ${lim sup}_{z} ∣ P [f - g] (z) - (f - g) (ζ) ∣ \leq {lim sup}_{z} ∣ P [f - g] (z) ∣ + ∣ f - g ∣ (ζ)$ and the last one is [L3] with aperture $m$ and measure $ν = (f - g) m$ , together with $P [f - g] = P [(f - g) m]$ and $M_{T} (f - g) = M_{T} ((f - g) m)$ from [L1].

3.1step 1.2step 2.1L2L6L8algebra

Small measure of the bad sets. Fix an integer $m \geq 2$ and $t > 0$ . If a point $ζ$ satisfies $(m + 1)^{2} M_{T} (f - g) (ζ) \leq t$ and $∣ f - g ∣ (ζ) \leq t$ , then step 2.1 gives $L_{m} (f) (ζ) \leq 2 t$ ; hence ${L_{m} > 2 t} \subseteq {(m + 1)^{2} M_{T} (f - g) > t} \cup {∣ f - g ∣ > t} .$ By [L2] and [L8], applied to the $L^{1}$ function $f - g$ , the first set has measure at most $3 (m + 1)^{2} ∥ f - g ∥_{1} / t$ and the second at most $∥ f - g ∥_{1} / t$ , so $m ({L_{m} > 2 t}) \leq (3 (m + 1)^{2} + 1) ∥ f - g ∥_{1} / t$ for every continuous $g$ . Given $ε > 0$ , [L6] supplies a continuous $g$ with $∥ f - g ∥_{1} < ε$ ; hence $m ({L_{m} > 2 t}) \leq (3 (m + 1)^{2} + 1) ε / t$ for every $ε > 0$ , and consequently $m ({L_{m} > 2 t}) = 0$ .

4.1step 1.2step 3.1L9

The exceptional set is null. For each integer $m \geq 2$ , the set ${L_{m} > 0} = ⋃_{k \geq 1} {L_{m} > 1 / (2 k)}$ is a countable union of Borel sets of $m$ -measure zero by steps 1.2 and 3.1, hence is $m$ -null by [L9]; the union $B : = ⋃_{m \geq 2} {L_{m} > 0}$ over the countably many integers $m \geq 2$ is then $m$ -null as well.

5.1step 1.1step 4.1L1∎

Conclusion. Let $ζ \notin B$ , so that the complement of $B$ has full measure. Then $L_{m} (ζ) = 0$ for every integer $m \geq 2$ : for every $ε > 0$ there is $δ > 0$ with $∣ P [f] (z) - f (ζ) ∣ < ε$ for all $z \in Γ_{m} (ζ)$ with $∣ z - ζ ∣ < δ$ . Given $A > 1$ , choose an integer $m \geq A$ ; since $Γ_{A} (ζ) \subseteq Γ_{m} (ζ)$ by [L1], the same $δ$ witnesses $P [f] (z) \to f (ζ)$ as $z \to ζ$ within $Γ_{A} (ζ)$ . Thus the nontangential limit exists and equals $f (ζ)$ for every $ζ$ outside the null set $B$ . If $f^{'} = f$ almost everywhere is another representative, then $L_{m}^{'} \leq L_{m} + ∣ f - f^{'} ∣$ for the corresponding cone limsups, so the bad set for $f^{'}$ is contained in $B \cup {∣ f - f^{'} ∣ > 0}$ , and the latter is a countable union of null sets by [L8] and [L9]; hence the assertion is independent of the representative.

CorollaryStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Bounded harmonic functions have L-infinity Fatou boundary data

Statement

Assume the Axiom of Choice. Let $u : D \to C$ be complex harmonic and bounded, and put $M : = \sup_{z \in D} ∣ u (z) ∣ < + \infty$ . Then there is a unique $f \in L^{\infty} (T, m; C)$ with $u = P [f]$ , and $∥ u ∥_{h^{\infty}} = \sup_{z \in D} ∣ u (z) ∣ = ∥ f ∥_{\infty} .$ Moreover, for $m$ -almost every $ζ \in T$ one has $u (z) \to f (ζ)$ as $z \to ζ$ within every fixed nontangential region $Γ_{A} (ζ)$ , $A > 1$ .

Facts & Assumptions

Given: The Axiom of Choice; a complex harmonic function $u : D \to C$ with $M : = \sup_{z \in D} ∣ u (z) ∣ < + \infty$ ; the notation $u_{r} (ζ) = u (r ζ)$ and $P [f] = P [f m]$ .

[L1]

Under countable choice $h^{\infty} (D)$ consists of the complex harmonic $u$ on $D$ with $\sup_{0 \leq r < 1} \sup_{ζ \in T} ∣ u (r ζ) ∣ < + \infty$ , and $∥ u ∥_{h^{\infty}} = \sup_{0 \leq r < 1} \sup_{ζ \in T} ∣ u (r ζ) ∣ = \sup_{z \in D} ∣ u (z) ∣;$ every element of $h^{\infty} (D)$ is continuous on $D$ (Harmonic Hardy classes on the unit disc).

[L2]

Under the Axiom of Choice, for every $1 < p \leq \infty$ and every $u \in h^{p} (D)$ there is a unique $f \in L^{p} (T, m; C)$ with $u = P [f]$ , and $∥ u ∥_{h^{p}} = ∥ f ∥_{p}$ (h^p is the Poisson image of Lp for 1<p<=infinity).

[L3]

Under countable choice, if $f \in L^{1} (T, m; C)$ , then for $m$ -almost every $ζ \in T$ one has $P [f] (z) \to f (ζ)$ as $z \to ζ$ within every fixed nontangential region $Γ_{A} (ζ)$ , $A > 1$ ; the assertion uses an almost-everywhere representative of $f$ (Fatou limits for Poisson extensions of L1 boundary data).

[L4]

The Axiom of Choice implies dependent choice, which implies countable choice (AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice).

[L5]

The normalized Haar measure $m$ on $T$ is a probability measure: $m (T) = 1$ and $\int_{T} 1 d m = 1$ , so the class of the constant function $1$ has $∥ 1 ∥_{L^{1} (T, m)} = 1$ (The one-dimensional torus and its normalized Haar integral).

[L6]

For conjugate exponents $p, p^{'} \in [1, \infty]$ and $f \in L^{p} (T, m; C)$ , $g \in L^{p^{'}} (T, m; C)$ one has $\int_{T} ∣ f g ∣ d m \leq ∥ f ∥_{p} ∥ g ∥_{p^{'}}$ ; the $L^{p}$ norms are the quotient norms of the spaces $L^{p} (T, m; C)$ (Complex Holder, Minkowski, and the quotient norm, Complex Lp classes and Euclidean test-function conventions).

Proof

technique · direct

1.1givenL1L2L3L4

Class membership and choice bookkeeping. The hypothesis says that $u$ is complex harmonic with $M = \sup_{z \in D} ∣ u (z) ∣ < + \infty$ , so [L1] gives $u \in h^{\infty} (D)$ and $∥ u ∥_{h^{\infty}} = M$ . By [L4] the Axiom of Choice supplies countable choice, so the choice hypotheses of [L2] (the Axiom of Choice) and of [L3] (countable choice) are met.

2.1step 1.1L2

The boundary data. Apply [L2] with $p = \infty$ , which is allowed because $1 < \infty \leq \infty$ : there is a unique $f \in L^{\infty} (T, m; C)$ with $u = P [f]$ , and $∥ u ∥_{h^{\infty}} = ∥ f ∥_{\infty}$ . Together with step 1.1 this gives $∥ f ∥_{\infty} = M = \sup_{z \in D} ∣ u (z) ∣$ , so $f$ is the promised boundary datum and the norm identity holds.

3.1step 2.1L5L6

The boundary datum is integrable. Apply the Hölder inequality of [L6] with the conjugate pair $(p, p^{'}) = (\infty, 1)$ , to $f \in L^{\infty} (T, m; C)$ and to the constant function $1 \in L^{1} (T, m; C)$ : one has $∥ f ∥_{1} = \int_{T} ∣ f \cdot 1 ∣ d m \leq ∥ f ∥_{\infty} ∥ 1 ∥_{1} = ∥ f ∥_{\infty}$ , where the last equality is [L5]. Since $∥ f ∥_{\infty} = M < + \infty$ by step 2.1, this shows $f \in L^{1} (T, m; C)$ .

4.1step 1.1step 2.1step 3.1L3

Nontangential convergence almost everywhere. By step 3.1 the function $f$ lies in $L^{1} (T, m; C)$ and by step 1.1 countable choice is available, so [L3] applies: there is a set $N \subseteq T$ with $m (N) = 0$ such that for every $ζ \in T ∖ N$ and every $A > 1$ one has $P [f] (z) \to f (ζ)$ as $z \to ζ$ within $Γ_{A} (ζ)$ . Since $u = P [f]$ by step 2.1, the same convergence holds with $u$ in place of $P [f]$ ; the exceptional set does not depend on $A$ , and the almost-everywhere representative used is the class $f$ of step 2.1.

5.1step 1.1step 2.1step 3.1step 4.1

Assembly. Steps 2.1, 3.1 and 4.1 produce a unique $f \in L^{\infty} (T, m; C)$ with $u = P [f]$ and $∥ u ∥_{h^{\infty}} = ∥ f ∥_{\infty}$ , and show that $u (z) \to f (ζ)$ as $z \to ζ$ within every fixed nontangential region $Γ_{A} (ζ)$ , $A > 1$ , for $m$ -almost every $ζ \in T$ ; by step 1.1 the norm $∥ u ∥_{h^{\infty}}$ equals $\sup_{z \in D} ∣ u (z) ∣$ , so all clauses of the Statement hold. The Axiom of Choice is used exactly in step 1.1: it supplies the hypothesis of the representation theorem [L2] and, through dependent and countable choice, the hypothesis of the Fatou theorem [L3]. ∎

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Positive harmonic boundary measures and compact normalized families

Statement

Assume the Axiom of Choice.

(a) If $u : D \to R$ is nonnegative and harmonic, then there is a unique finite nonnegative regular Borel measure $μ$ on $T$ with $u = P [μ]$ , and necessarily $μ (T) = u (0)$ .

(b) Conversely, for every finite nonnegative regular Borel measure $μ$ on $T$ , the function $P [μ]$ is nonnegative and harmonic and satisfies $P [μ] (0) = μ (T)$ .

(c) Every sequence $(u_{n})_{n \geq 1}$ of nonnegative harmonic functions on $D$ with $u_{n} (0) = 1$ for all $n$ has a subsequence converging locally uniformly on $D$ to a nonnegative harmonic function $u$ with $u (0) = 1$ .

Facts & Assumptions

Given: The Axiom of Choice; a nonnegative real harmonic function $u$ on $D$ where it occurs; a finite nonnegative regular Borel measure $ν$ on $T$ where it occurs; and a sequence $(u_{n})$ of nonnegative harmonic functions on $D$ with $u_{n} (0) = 1$ where it occurs.

[L1]

Under the Axiom of Choice every $u \in h^{1} (D)$ has a unique finite regular complex Borel measure $μ$ on $T$ with $u = P [μ]$ and $∥ u ∥_{h^{1}} = ∣ μ ∣ (T)$ ; conversely every finite regular complex Borel measure $μ$ on $T$ gives an $h^{1}$ function $P [μ]$ with $∥ P [μ] ∥_{h^{1}} = ∣ μ ∣ (T)$ ; and $u_{r} m$ converges weak-star to $μ$ against $C (T)$ as $r ↑ 1$ (h1 is isometric to finite regular complex boundary measures).

[L2]

A complex-valued function on $D$ is harmonic exactly when its real and imaginary parts are real harmonic; $h^{1} (D)$ consists of the complex harmonic functions with $\sup_{0 \leq r < 1} ∥ u_{r} ∥_{L^{1} (T, m)} < + \infty$ , and $∥ u ∥_{h^{1}}$ denotes that supremum. The zero function is harmonic (Harmonic Hardy classes on the unit disc, Plane harmonic functions).

[L3]

A real harmonic function satisfies the circle mean-value property $u (a) = \frac{1}{2 π} \int_{0}^{2 π} u (a + r e^{i θ}) d θ$ for every closed disc $\overline{D (a, r)}$ contained in its domain (Plane harmonic functions satisfy the mean-value property, The circle and disc mean-value properties).

[L4]

The normalized Haar integral on $T = R / Z$ satisfies $\int_{T} F d m = \int_{[0, 1)} F \circ q d λ_{1}$ ; for continuous $G$ on $T$ the change of variables $t \mapsto 2 π t$ gives $\int_{T} G d m = \frac{1}{2 π} \int_{0}^{2 π} G (e^{i θ}) d θ$ (The one-dimensional torus and its normalized Haar integral, A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions).

[L5]

For a finite regular complex Borel measure $μ$ on $T$ one has $P [μ] (z) = \int_{T} P (z, ζ) d μ (ζ)$ ; the kernel is $P (z, ζ) = (1 - ∣ z ∣^{2}) / ∣ φ (ζ) - z ∣^{2} > 0$ with $P (0, ζ) = 1$ and $ζ \mapsto P (z, ζ)$ continuous on $T$ ; for $f \in L^{1} (T, m)$ the density measure satisfies $P [f] = P [f m]$ and $\int_{T} g d (u_{r} m) = \int_{T} g u_{r} d m$ for bounded measurable $g$ (The Poisson integral of a finite complex boundary measure, The Poisson kernel on the unit disc).

[L6]

Assume Dependent Choice. Every bounded complex linear functional on $C (T, C) = C_{0} (T, C)$ is integration against a unique finite regular complex Borel measure $μ$ , and $∥ \cdot ∥ = ∣ μ ∣ (T)$ (The bounded complex dual of C_0(X) is regular complex measures).

[L7]

Assume Dependent Choice. Every bounded positive real-linear functional $L$ on $C_{0} (X; R)$ for LCH $X$ satisfies $L (f) = \int f d ρ$ for a unique finite regular Borel measure $ρ \geq 0$ (Positive C_0(X) functionals have finite regular representing measures).

[L8]

The integral against a signed or complex measure is defined as the limit of simple integrals along $L^{1} (∣ ν ∣)$ -approximating complex simple functions and is independent of the chosen approximating sequence; a finite measure is a finite signed measure and a finite complex measure; a finite regular Borel measure $ρ \geq 0$ is a finite regular complex Borel measure with $∣ ρ ∣ = ρ$ (Integration against a signed or complex measure, and the class L^1(nu) = L^1(|nu|), The simple integral against a signed or complex measure, The total variation |nu|(E) from countable measurable partitions, A signed measure is countably additive and takes at most one infinite value, Measures on sigma-algebras, A complex measure is a finite-valued countably additive set function, Regular Borel measure on an LCH space, Regular complex Borel measures).

[L9]

On a finite measure space a bounded Borel function $h \geq 0$ lies in $L^{1}$ with $\int ∣ h ∣ d ρ \leq M ρ (X)$ where $M = \sup h$ ; nonnegative measurable functions admit increasing nonnegative simple approximations; integrals of integrands bounded by an $L^{1}$ majorant may be passed to the limit (The class $L^{1} (μ)$ of integrable functions, The nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, The nonnegative integral agrees with the simple integral on simple functions, Every nonnegative measurable function is the increasing limit of simple measurable functions, Dominated convergence).

[L10]

For $f \in L^{1} (ν)$ one has $∣ \int f d ν ∣ \leq \int ∣ f ∣ d ∣ ν ∣ \leq ∥ f ∥_{\infty} ∣ ν ∣ (X)$ (Integrals against signed or complex measures are bounded by total variation).

[L11]

The Axiom of Choice implies Dependent Choice, which implies countable choice (AC implies DC implies countable choice, The Axiom of Choice, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L12]

$T$ is compact Hausdorff and $φ : T \to S^{1}$ , $φ ([t]) = (\cos 2 π t, \sin 2 π t)$ , is a homeomorphism onto the Euclidean unit circle $S^{1}$ , a closed and bounded hence compact subset of $R^{2}$ (Finite tori are compact Hausdorff spaces separated by characters, The one-dimensional torus and its normalized Haar integral, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L13]

Assume countable choice. Every nonempty compact metric space $K$ admits a sequence in $C (K, R)$ dense for the supremum norm; $N \times N$ is at most countable; a space is separable when it has an at most countable dense subset (A countable dense family of continuous functions on a compact metric space, Countable unions of at most countable sets, assuming ${A C}_{ω}$ , Separability: the existence of an at most countable dense subset).

[L14]

Under the ultrafilter lemma every sequence in the dual unit ball of a separable real or complex normed space has a weak-star convergent subsequence whose limit is an element of the dual; weak-star convergence is evaluation convergence on every element of the predual (A separable predual has weak-star sequentially compact dual ball, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter, Weak star convergence).

[L15]

A compact metric subspace admits a finite subcover from every family of ambient open sets covering it; this includes families indexed by points or natural numbers (Open cover, subcover, compact metric space, and compact subset of a metric space, A subset of a metric space is open in the subspace metric exactly when it is the trace of an open set of the ambient space, and it is compact as a metric space in its own right exactly when every family of open subsets of the ambient space covering it, indexed or not, has finitely many members covering it).

Proof

technique · direct

1.1givenL8L9algebra

Nonnegative integrands have nonnegative integrals against finite nonnegative measures. Let $ρ$ be a finite measure on $T$ and let $h : T \to R$ be bounded Borel with $h \geq 0$ , $M : = \sup_{T} h < \infty$ . Because $ρ$ is countably additive with values in $[0, \infty)$ it is a finite signed measure, and for every countable Borel partition of a Borel set $E$ one has $\sum_{j} ∣ ρ (E_{j}) ∣ = \sum_{j} ρ (E_{j}) = ρ (E)$ , so $∣ ρ ∣ (E) = ρ (E)$ and $h \in L^{1} (ρ) = L^{1} (∣ ρ ∣)$ since $\int_{T} ∣ h ∣ d ρ \leq M ρ (T) < \infty$ . Let $s_{n}$ be increasing nonnegative simple functions with $s_{n} ↑ h$ and $0 \leq s_{n} \leq h$ . Then $∣ h - s_{n} ∣ \leq M$ with the constant $M$ integrable for the finite measure $ρ$ , so $\int_{T} ∣ h - s_{n} ∣ d ρ \to 0$ , and $(s_{n})$ is an admissible approximating sequence in the definition of $\int_{T} h d ρ$ ; hence $\int_{T} h d ρ = \lim_{n} \int_{T} s_{n} d ρ$ . Writing $s_{n} = \sum_{j} c_{j} 1_{E_{j}}$ in canonical form, all $c_{j} > 0$ and all $ρ (E_{j}) \geq 0$ , so $\int_{T} s_{n} d ρ = \sum_{j} c_{j} ρ (E_{j}) \geq 0$ for every $n$ and therefore $\int_{T} h d ρ \geq 0$ .

1.2givenL2L3L4

The nonnegative harmonic $u$ lies in $h^{1} (D)$ with $∥ u ∥_{h^{1}} = u (0)$ . For $0 \leq r < 1$ the function $u_{r} (ζ) = u (r ζ)$ is continuous on $T$ and $u_{r} \geq 0$ . If $r = 0$ then $\int_{T} u_{0} d m = u (0)$ . If $r > 0$ , then $\int_{T} u_{r} d m = \frac{1}{2 π} \int_{0}^{2 π} u (r e^{i θ}) d θ = u (0)$ , the first equality by the torus identification combined with the change of variables $t \mapsto 2 π t$ and the second by the circle mean-value property applied to the harmonic $u$ on the disc $\overline{D (0, r)} \subseteq D$ . Hence $∥ u_{r} ∥_{L^{1} (T, m)} = \int_{T} ∣ u_{r} ∣ d m = \int_{T} u_{r} d m = u (0)$ for every radius, so $\sup_{0 \leq r < 1} ∥ u_{r} ∥_{1} = u (0) < \infty$ : $u$ is complex harmonic, its imaginary part being the harmonic zero function, and therefore $u \in h^{1} (D)$ with $∥ u ∥_{h^{1}} = u (0)$ .

1.3givenL5L15algebra

Kernel Lipschitz bound on compacta. Let $K \subseteq D$ be compact. For $K = \emptyset$ the estimate is vacuous; assume $K \neq \emptyset$ . The open discs $U_{n} = {z \in C : ∣ z ∣ < 1 - 1 / n}$ , $n \geq 2$ , cover $D$ and hence $K$ , so [L15] gives $N \geq 2$ with $K \subseteq U_{N}$ ; thus $∣ φ (ζ) - z ∣ \geq 1 - ∣ z ∣ > 1 / N$ for all $z \in K$ and $ζ \in T$ . Fix $z, z^{'} \in K$ and $w : = φ (ζ)$ for $ζ \in T$ , and set $a : = ∣ w - z ∣^{2}$ , $b : = ∣ w - z^{'} ∣^{2}$ . Then $P (z, ζ) - P (z^{'}, ζ) = [(1 - ∣ z ∣^{2}) b - (1 - ∣ z^{'} ∣^{2}) a] / (a b)$ and the numerator equals $(1 - ∣ z ∣^{2}) (b - a) + (∣ z^{'} ∣^{2} - ∣ z ∣^{2}) a$ , where $∣ b - a ∣ \leq 4 ∣ z - z^{'} ∣$ and $∣ (∣ z^{'} ∣^{2} - ∣ z ∣^{2}) a ∣ \leq 2 ∣ z - z^{'} ∣ \cdot 4$ ; hence the numerator has modulus at most $12 ∣ z - z^{'} ∣$ while $a b \geq N^{- 4}$ . Therefore $∣ P (z, ζ) - P (z^{'}, ζ) ∣ \leq 12 N^{4} ∣ z - z^{'} ∣$ for every $ζ \in T$ .

1.4givenL11L12L13

$C (T, C)$ is separable. By [L12] the torus $T$ is homeomorphic to the compact metric space $S^{1}$ , hence is itself a compact metric space; by [L13], and countable choice is available by [L11], there is a sequence $(f_{j})$ in $C (T, R)$ dense for the supremum norm. The family ${f_{j} + i f_{k} : j, k \geq 1}$ is at most countable by [L13] and is dense in $C (T, C)$ : for $g = v + i w$ and $ε > 0$ choose $j, k$ with $∥ v - f_{j} ∥_{\infty} < ε / 2$ and $∥ w - f_{k} ∥_{\infty} < ε / 2$ , so that $∥ g - (f_{j} + i f_{k}) ∥_{\infty} < ε$ . Hence $C (T, C)$ has an at most countable dense subset, that is, it is separable.

2.1step 1.2L1

Representation of $u$ . By [L1] applied to $u \in h^{1} (D)$ there is a unique finite regular complex Borel measure $μ$ on $T$ with $u = P [μ]$ , $∣ μ ∣ (T) = ∥ u ∥_{h^{1}} = u (0)$ , and $u_{r} m \overset{*}{⇀} μ$ against $C (T)$ as $r ↑ 1$ .

2.2step 1.1L1L5L8

Converse direction (b). Let $ρ$ be a finite nonnegative regular Borel measure on $T$ . It is a finite regular complex Borel measure, so by the converse clause of [L1] the function $P [ρ]$ lies in $h^{1} (D)$ and is harmonic. For $z \in D$ the function $P (z, \cdot)$ is continuous by [L5], hence bounded Borel, and positive, so step 1.1 with the finite measure $ρ$ and $h = P (z, \cdot)$ gives $P [ρ] (z) = \int_{T} P (z, ζ) d ρ (ζ) \geq 0$ . Moreover $P [ρ] (0) = \int_{T} P (0, ζ) d ρ (ζ) = \int_{T} 1 d ρ (ζ) = ρ (T)$ , because $P (0, ζ) = 1$ and the constant function $1$ is an admissible simple approximant in the definition of the integral.

3.1step 2.1step 1.1L5

Testing the boundary measure. For every $g \in C (T)$ with $g \geq 0$ one has $\int_{T} g d μ = \lim_{r ↑ 1} \int_{T} g d (u_{r} m) = \lim_{r ↑ 1} \int_{T} g u_{r} d m \geq 0$ : the first equality is the weak-star convergence recorded in step 2.1, the second uses the density-measure pairing $g d (u_{r} m) = g u_{r} d m$ of [L5], and for each $0 \leq r < 1$ the function $g u_{r}$ is a bounded nonnegative Borel function on the probability space $(T, m)$ , so step 1.1 gives $\int_{T} g u_{r} d m \geq 0$ ; limits of nonnegative numbers are nonnegative.

4.1step 3.1step 2.1L6L7L8L11

The boundary measure is nonnegative. Define $Λ (g) : = \int_{T} g d μ$ for real $g \in C (T)$ ; by step 3.1 these values are real, $Λ$ is real-linear and bounded, and $Λ (g) \geq 0$ whenever $g \geq 0$ . By [L7], with Dependent Choice available from [L11], there is a finite regular Borel measure $ρ$ on $T$ with $Λ (g) = \int_{T} g d ρ$ for every real $g \in C (T)$ . The complex-linear functionals $g \mapsto \int_{T} g d μ$ and $g \mapsto \int_{T} g d ρ$ agree on real-valued functions and hence, by complex linearity, on all of $C (T, C)$ ; the uniqueness clause of [L6] therefore gives $μ = ρ$ . Thus $μ$ is a nonnegative measure, and since $μ$ is nonnegative and $∣ μ ∣ (T) = u (0)$ by step 2.1, also $μ (T) = u (0)$ .

5.1step 2.1step 2.2step 4.1L1

Part (a). This proves (a): $u = P [μ]$ for the finite nonnegative regular Borel measure $μ$ of step 4.1 with $μ (T) = u (0)$ , and if $ρ$ is any further finite nonnegative regular Borel measure with $u = P [ρ]$ , then $ρ$ is in particular a finite regular complex Borel measure representing $u$ , so $ρ = μ$ by the uniqueness clause of [L1] recorded in step 2.1. The converse direction (b) is step 2.2.

6.1step 5.1

Normalized measures. For each $n$ the function $u_{n}$ is nonnegative harmonic with $u_{n} (0) = 1$ , so part (a) as proved in step 5.1 gives a unique finite nonnegative regular Borel measure $μ_{n}$ on $T$ with $u_{n} = P [μ_{n}]$ and $μ_{n} (T) = 1$ ; in particular $∣ μ_{n} ∣ (T) = 1$ for every $n$ , so $(μ_{n})$ is a sequence of probability measures.

7.1step 6.1step 1.4L6L11L14

Weak-star subsequence. By step 1.4 the space $C (T, C)$ is separable and the probability measures $μ_{n}$ lie in the closed unit ball of its dual, so [L14] -- the ultrafilter lemma being available from the Axiom of Choice by [L11] -- provides a subsequence $(μ_{n_{k}})$ and an element of the dual which [L6] identifies with a finite regular complex Borel measure $μ$ on $T$ such that $\int_{T} g d μ_{n_{k}} \to \int_{T} g d μ$ for every $g \in C (T, C)$ .

8.1step 7.1step 4.1step 1.1

The limit measure is a probability measure. For real $g \in C (T)$ with $g \geq 0$ step 7.1 gives $\int_{T} g d μ = \lim_{k} \int_{T} g d μ_{n_{k}} \geq 0$ , the inequality by step 1.1 applied to the finite measures $μ_{n_{k}}$ and the bounded nonnegative Borel function $g$ . The identification argument of step 4.1, with this positivity in place of step 3.1, now makes $μ$ a nonnegative measure, and testing the constant function $1$ gives $μ (T) = \lim_{k} μ_{n_{k}} (T) = 1$ , hence $∣ μ ∣ (T) = 1$ .

9.1step 8.1step 2.2step 1.1L1L5

The limit function. Put $u : = P [μ]$ . By the converse clause of [L1] the function $u$ is harmonic and lies in $h^{1} (D)$ ; by step 1.1 applied to the finite measure $μ$ and the bounded nonnegative Borel function $P (z, \cdot)$ one has $u (z) = \int_{T} P (z, ζ) d μ (ζ) \geq 0$ for every $z \in D$ ; and $u (0) = \int_{T} P (0, ζ) d μ (ζ) = \int_{T} 1 d μ = μ (T) = 1$ , exactly as in step 2.2.

10.1step 7.1step 9.1L5

Pointwise convergence. For each $z \in D$ the function $ζ \mapsto P (z, ζ)$ is continuous on $T$ by [L5], so the weak-star convergence of step 7.1 gives $u_{n_{k}} (z) = \int_{T} P (z, ζ) d μ_{n_{k}} (ζ) \to \int_{T} P (z, ζ) d μ (ζ) = u (z)$ .

11.1step 1.3step 6.1step 7.1step 8.1step 10.1L10L15

Local uniform convergence. Fix a compact $K \subseteq D$ and $ε > 0$ . Uniform convergence on $K = \emptyset$ is vacuous; assume $K \neq \emptyset$ . By step 1.3 choose $δ > 0$ with $∣ P (z, ζ) - P (z^{'}, ζ) ∣ < ε$ whenever $z, z^{'} \in K$ satisfy $∣ z - z^{'} ∣ < δ$ and $ζ \in T$ , and by [L15] applied to the ambient balls $B (z, δ)$ indexed by $z \in K$ choose $z_{1}, \dots, z_{m} \in K$ with $K \subseteq ⋃_{j = 1}^{m} B (z_{j}, δ)$ . For $z \in K$ pick $j$ with $∣ z - z_{j} ∣ < δ$ and split $u_{n_{k}} (z) - u (z) = \int_{T} (P (z, ζ) - P (z_{j}, ζ)) d μ_{n_{k}} (ζ) + \int_{T} P (z_{j}, ζ) d (μ_{n_{k}} - μ) (ζ) + \int_{T} (P (z_{j}, ζ) - P (z, ζ)) d μ (ζ) .$ By [L10] together with $∣ μ_{n_{k}} ∣ (T) = 1$ from step 6.1 and $∣ μ ∣ (T) = 1$ from step 8.1, the first and third terms have modulus at most $ε$ , while the middle term tends to $0$ as $k \to \infty$ by step 7.1 applied to the continuous function $P (z_{j}, \cdot)$ . Hence ${lim sup}_{k} \sup_{z \in K} ∣ u_{n_{k}} (z) - u (z) ∣ \leq 2 ε + \max_{j \leq m} \lim_{k} ∣ \int_{T} P (z_{j}, ζ) d (μ_{n_{k}} - μ) (ζ) ∣ \leq 2 ε$ , and since $ε > 0$ is arbitrary the subsequence converges to $u$ uniformly on $K$ ; as every compact subset of $D$ arises this way, the convergence is locally uniform on $D$ .

12.1step 2.2step 5.1step 11.1L11

Assembly. Steps 5.1, 2.2 and 6.1 through 11.1 prove the three assertions: (a) a nonnegative harmonic $u$ is $P [μ]$ for a unique finite nonnegative regular Borel measure $μ$ with $μ (T) = u (0)$ ; (b) conversely every finite nonnegative regular Borel measure gives a nonnegative harmonic $P [μ]$ with $P [μ] (0) = μ (T)$ ; (c) every sequence of nonnegative harmonic functions normalized by $u_{n} (0) = 1$ has a subsequence converging locally uniformly on $D$ to a nonnegative harmonic function with $u (0) = 1$ . The Axiom of Choice is used exactly as recorded: it supplies Dependent Choice and countable choice by [L11], for the Riesz representation [L6], the positive-functional lemma [L7] and the countable dense family [L13], and it supplies the ultrafilter lemma for [L14]; no other choice was made. ∎

5 · Examples, counterexamples and false statements

None yet.

Harmonic Hardy Classes and Fatou Boundary Limits

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The Poisson integral of a finite complex boundary measure

Definition

Harmonic Hardy classes on the unit disc

Definition

Poisson extension is an Lp contraction and converges in finite Lp

Statement

Facts & Assumptions

Proof

h1 is isometric to finite regular complex boundary measures

Statement

Facts & Assumptions

Proof

h^p is the Poisson image of Lp for 1<p<=infinity

Statement

Facts & Assumptions

Proof

The circle maximal function and nontangential approach regions

Definition

The circle maximal function is weak type one one for finite measures

Statement

Facts & Assumptions

Proof

Poisson nontangential maximal function is controlled by circle maximal averages

Statement

Facts & Assumptions

Proof

Fatou limits for Poisson extensions of L1 boundary data

Statement

Facts & Assumptions

Proof

Bounded harmonic functions have L-infinity Fatou boundary data

Statement

Facts & Assumptions

Proof

Positive harmonic boundary measures and compact normalized families

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources