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This supremum function takes a supremum over a cone approaching the boundary point $x$; the positive number $a$ is its aperture. For an analytic upper-half-plane $H^1$ function, it belongs to $L^1$. To prove this without first assuming boundary absolute continuity, shift the function upwards, apply subharmonic Poisson comparison to its square root, and use the strong $L^2$ bound for the Hardy-Littlewood maximal function. Fatou lemma removes the shift.

We prove sufficiency by a non-tangential maximal function and the layer cake representation. Write $K=C(\mu )$. First $\mu (\mathbb{D})\le CK$: three hyperbolic geodesic caps corresponding to arcs of normalized length $2/3$, with equally spaced centers, cover the unit disc. This handles interior points and large arcs, not only a thin boundary strip.

Fix aperture two and set $Nf(\xi )={\sup }_{z\in {\Gamma }_2(\xi )}|f(z)|$. The crucial H1 non-tangential maximal inequality on the disk isHere is its endpoint proof. For $0<R<1$ let $f_R(z)=f(Rz)$, which is holomorphic across the closed unit disc. Since modulus powers of holomorphic functions are subharmonic, Poisson comparison givesThe estimate that a Poisson maximal function is dominated by the Hardy-Littlewood maximal function yields $(N{f}_R{)}^{1/2}\le CM{g}_R$. The strong $L^2$ Hardy-Littlewood maximal inequality now givesFor each interior point $f_R(z)\to f(z)$, hence $Nf\le {\mathrm{liminf}}_{R\uparrow 1}N{f}_R$. The Fatou lemma proves the maximal inequality. This uses a genuine $L^2$ maximal bound; the Hardy-Littlewood maximal function is not strongly bounded on $L^1$.

For $t>0$, the set $O_t=\{\xi :Nf(\xi )>t\}$ is open, since each interior point witnessing this strict inequality has an open boundary shadow. Write its disjoint arc components as $I_j$. If $|f(z)|>t$ and $\delta =1-|z|$, the arc centered at $z/|z|$ with angular half-width $\delta$ lies in $O_t$, because $|z-\xi |<2\delta$ there. Thus it lies in one component $I_j$, and $z$ has angular projection in $I_j$ with $\delta \le \pi m(I_j)$. The geodesic-cap and Carleson-box comparison puts all such points in $Q(\kappa I_j)$ for a fixed enlargement factor $\kappa$ whenever $I_j$ is small. For large $I_j$, use $\mu (\mathbb{D})\le CK\le C^{\prime }Km(I_j)$. If $z=0$ or $O_t$ is the entire circle, that same total-mass estimate applies. Summing, without needing the enlarged caps to be disjoint, givesThe layer cake representation and the proved H1 non-tangential maximal inequality on the disk finish the argument:Thus the two conditions are equivalent, with their least admissible constants comparable: $C(\mu )\le C_{1}N(\mu )$ and $N(\mu )\le C_{2}C(\mu )$. This is the Carleson embedding theorem on the disk, proved at the analytic $H^1$ endpoint.

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

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

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

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

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

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

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

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

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

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

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

Write $dm=dt/(2\pi )$ in angular coordinates on the unit circle. The centered Hardy-Littlewood maximal function iswhere $I_r(\omega )$ is the centered circular arc of angular radius $r$, with $r=\pi$ giving the whole circle. The Hardy-Littlewood maximal inequality in this one-dimensional setting is the weak-type estimateThe precise constant is not important for differentiation, but we prove this one explicitly.

Let $E=\{Mf>\alpha \}$ and choose a compact subset $F\subset E$. For each point of $F$, choose a centered arc whose average of $|f|$ exceeds $\alpha$. A finite subfamily covers $F$. Order these arcs by decreasing length and retain an arc precisely when it is disjoint from the previously retained arcs. Every rejected arc meets a retained arc of at least its length and lies in the latter's triple, interpreted on the circle and truncated to the whole circle if necessary. If the retained disjoint arcs are $I_1,\dots ,I_N$, thenThis is the interval version of the Wiener covering lemma. The superlevel set is measurable: the average at each fixed radius is continuous in the center, by translation continuity in $L^1$, so $E$ is open. Inner regularity of Lebesgue measure now passes the compact-set bound to $E$, proving the Hardy-Littlewood maximal inequality. The same finite-interval argument gives the line version $|\{Mf>\alpha \}|\le 3\Vert f{\Vert }_1/\alpha$ for integrable functions on the real line.

To prove differentiation, first recall that continuous functions are dense in $L^1(m)$ on the circle. Approximate an integrable function by simple functions; for each of their measurable sets, inner and outer regularity give a compact subset and an open superset with arbitrarily small measure difference. A continuous cutoff between these sets approximates the indicator in $L^1$. Combining finitely many cutoffs proves the claimed density.

For a representative of $f$ finite almost everywhere, define its local absolute oscillation byIf $h$ is continuous and $u=f-h$, the triangle inequality and continuity of $h$ giveThus the Hardy-Littlewood maximal inequality and the elementary integral bound for a superlevel set yield, for every $\epsilon >0$,Choose continuous $h$ with $\Vert f-h{\Vert }_1$ arbitrarily small. It follows that $\Delta f=0$ outside a null set, by using positive rational $\epsilon$. In particular,for almost every $\omega$. This proves the requested Lebesgue differentiation theorem, including the stronger absolute-oscillation formulation, rather than assuming pointwise convergence of the approximating continuous functions.

For failure at a specified point, use coordinates $-\pi <t\le \pi$ and put $a_n=2^{-2^n}$. Let $f$ be one on the annuli $a_{2j+1}<|t|\le a_{2j}$ for $j\ge 0$, and zero elsewhere, including at zero. This bounded measurable function is integrable. Since $a_{n+1}/a_n\to 0$, the average $A_n$ over the centered interval $[-a_n,a_n]$ satisfiesIndeed, the outermost annulus occupies all but the fraction $a_{n+1}/a_n$ of that interval, and has value one or zero according to parity. The even-radius subsequence tends to one and the odd-radius subsequence to zero, so no centered-average limit exists at $\omega =1$. The normalization of $m$ cancels from these averages. This is how oscillating annuli obstruct differentiation at a point without contradicting almost-everywhere differentiation.

One potential form of the Hardy–Littlewood–Sobolev inequality is: if $0<\alpha <n$, $1<p<n/\alpha$, and $1/r=1/p-\alpha /n$, thenA normalization constant for the Riesz potential only changes $C$. Equivalently, for $0<\lambda <n$, $p,s>1$ and $1/p+1/s+\lambda /n=2$,The exponent relation is necessary under dilation. We prove the potential estimate and then derive the bilinear version.

First establish the needed maximal estimate, so it is not an unproved step in the argument. Define the centered Hardy-Littlewood maximal function by $Mf(x)={\sup }_{t>0}|B_t{|}^{-1}{\int }_{B_t(x)}|f|$. For integrable $f$, cover a compact subset of $\{Mf>a\}$ by finitely many balls whose averages exceed $a$. Order these balls by decreasing radius and greedily retain a ball if it is disjoint from those already retained. Every discarded ball intersects a retained ball of at least its radius, so lies in its triple. HenceThe passage from compact subsets uses inner regularity of Lebesgue measure; the superlevel set is open because fixed-radius local averages are continuous. For $f\in L^p$, split $f=f{1}_{\{|f|>a/2\}}+f{1}_{\{|f|\le a/2\}}$. The second summand's maximal function is at most $a/2$, and the first is integrable. ThereforeThe layer cake representation and Tonelli theorem giveThus the Strong Lp bound for the Hardy-Littlewood maximal function is proved for every $p>1$.

For $I_{\alpha }|f|(x)$, split at radius $R$. Dyadic shells with radii $2^{-j}R$ giveFor the outer part, Hölder's inequality givesThe last integral is finite exactly because $p<n/\alpha$. Balance the two bounds by $R^{n/p}=\Vert f{\Vert }_p/Mf(x)$ to obtain the Hedberg inequalityThe zero cases follow directly or by a limit; $Mf$ is finite almost everywhere by the maximal estimate. Since $r(1-\theta )=p$, integration and the proved maximal bound yield $\Vert I_{\alpha }f{\Vert }_r\le C\Vert f{\Vert }_p$. Taking $\alpha =n-\lambda$ and pairing $I_{\alpha }|f|$ with $|g|\in L^{r^{\prime }}$ proves the bilinear form by Hölder's inequality, because $s=r^{\prime }$. All integrals are justified first for nonnegative truncations and then by monotone convergence and the resulting finite bound. The strong estimate does not include $p=1$ or $p=n/\alpha$.

The relation $1/q=1/p-\alpha /n>0$ implies $\alpha p<n$. Fix $r>0$ and split the Riesz potential into the regions $|x-y|<r$ and $|x-y|\ge r$.

For the near part, decompose the ball into the dyadic annuliThe definition of the Hardy-Littlewood maximal function then givesFor the far part, Hölder's inequality giveswhere convergence follows from $\alpha p<n$.

If $Mf(x)>0$, chooseThe two bounds then have the same size. If $Mf(x)=0$, the result is immediate. Consequently the Hedberg inequality isHere the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.

Use the centered Hardy-Littlewood maximal functionIf $x>0$, choosing $0<r<x$ makes the whole interval lie in the positive half-line, so $M\theta (x)=1$. If $x=0$, every centered interval has exactly half its length in the positive half-line, so $M\theta (0)=1/2$.

If $x<0$, the average is zero for $r\le -x$. For $r>-x$, it isand these values tend to $1/2$ as $r\to \infty$. HenceFor the uncentered maximal-function convention, intervals extending arbitrarily far to the right make the supremum equal to $1$ at every $x$.

The centered Hardy-Littlewood maximal function isThe Hardy-Littlewood maximal inequality gives, for example,This valid constant follows from the disjoint-interval covering argument: a compact subset of the superlevel set has a finite cover by intervals of average above $t$; select disjoint intervals whose triples cover it, bound their total length by $\Vert f{\Vert }_1/t$, and use inner regularity. An optimal constant is not required.

For integrable boundary data $g$ on the unit circle, this follows from the Poisson kernel bound $C_A\delta /({\delta }^2+d^2)$ in a fixed Stolz region. The arc of radius $\delta$ and successive dyadic annuli contribute at most $C_A{2}^{-k}Mg(\xi )$, which sum to the displayed estimate. This links harmonic extension to the Hardy-Littlewood maximal function and its strong $L^p$ bounds for $p>1$.

If $k(x)=\varphi (|x|)\ge 0$ is decreasing in radius and has integral one, then $|k_t\ast f(x)|\le Mf(x)$, where $k_t(x)=t^{-d}k(x/t)$ and $M$ is the centered Hardy-Littlewood maximal function of $|f|dx$. The layer cake representation expresses $k_t$ as a positive mixture of ball indicators; each corresponding integral is bounded by its ball volume times $Mf(x)$. Integrating these volumes gives exactly one. This applies to the Poisson kernel for the upper half-space and supplies a weak type (1,1) domination for its boundary convergence.

For $1<p\le \infty$, the Hardy-Littlewood maximal function is bounded on $L^p({\mathbb{R}}^n)$. A direct proof for finite $p$ applies the weak $(1,1)$ estimate to $f{1}_{\{|f|>s/2\}}$, integrates the resulting distribution-function bound against $p{s}^{p-1}$, and uses Tonelli theorem.

Let $S$ be a sublinear operator with positive constants and $\mu \{Sf>a\}\le C_1\Vert f{\Vert }_1/a$ and $\Vert Sf{\Vert }_{\infty }\le C_{\infty }\Vert f{\Vert }_{\infty }$. Split $f$ into its part above $a/(2C_{\infty })$ and its bounded remainder. The latter contributes at most $a/2$, so the weak bound controls the superlevel measure by $2C_1{a}^{-1}{\int }_{|f|>a/(2C_{\infty })}|f|$. Integrating this estimate with layer cake representation proves the displayed bound for $1<p<\infty$ and $q=p/(p-1)$. In particular it gives the strong $L^p$ estimate for the Hardy-Littlewood maximal function.