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For an analytic H1 space on the unit disk function, its fixed-aperture non-tangential maximal function obeys the displayed strong $L^1$ bound. First replace $f$ by $f_R(z)=f(Rz)$, $R<1$. The fact that modulus powers of holomorphic functions are subharmonic gives $|f_R{|}^{1/2}\le P[|f(R\xi ){|}^{1/2}]$. Since a Poisson maximal function is dominated by the Hardy-Littlewood maximal function, the strong $L^2$ Hardy-Littlewood maximal inequality yields $\int N_A{f}_R\le C_A\int |f(R\xi )|\le C_A\Vert f{\Vert }_1$. Pointwise interior convergence and the Fatou lemma remove $R$. Analyticity permits this endpoint estimate; general $L^1$ boundary data have only a weak $L^1$ maximal bound.

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.

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.

The layer cake representation with $\phi (t)=t^p$ givesFor each $s>0$, splitSince $M{f}_1\le s/2$ and the maximal operator is sublinear,The weak $(1,1)$ Hardy-Littlewood maximal inequality therefore givesSubstitution into the distribution formula and another application of Tonelli theorem yieldThus the Strong Lp bound for the Hardy-Littlewood maximal function holds with

Approximate $f$ in $L^1$ by continuous $g$. The limsup local oscillation of $f$ is bounded by the maximal function of $f-g$ plus the vanishing oscillation of $g$. The weak $(1,1)$ Hardy-Littlewood maximal inequality makes the exceptional set have measure at most $C\Vert f-g{\Vert }_1/ϵ$; let the approximation error tend to zero.

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.