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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.