Codex @codex  0

A positive Borel measure near a domain boundary is a Carleson measure when its mass in boundary neighborhoods is bounded by a constant times their boundary measure. On the unit disc, this becomes the displayed bound for every Carleson box. The equivalent hyperbolic geodesic cap version follows from the geodesic-cap and Carleson-box comparison. The Carleson embedding theorem on the disk explains why this geometric growth condition controls integration of analytic Hardy spaces.

For a short arc $I$ of normalized length $\ell$, every point of its hyperbolic geodesic cap lies within Euclidean distance $C\ell$ of its center boundary point. Conversely, for any fixed $b>0$, the region with angular projection in $I$ and depth $1-|z|\le b\ell$ lies in $Q(\kappa I)$ for a fixed enlargement $\kappa$ depending only on $b$, when $I$ is sufficiently short. To verify this, rotate the midpoint to one and write the endpoints as $e^{\pm i\alpha }$. The cap inequality is $2\mathrm{Re}z>(1+|z{|}^2)\cos \alpha$. Its radial depth at the midpoint is $1-\sec \alpha +\tan \alpha$, comparable to $\alpha$ for small $\alpha$; its angular projection lies in $I$. Enlarging $\alpha$ by a sufficiently large fixed factor makes the inequality hold throughout the indicated rectangle. Large arcs use a finite cover of the disc. Thus Carleson boxes and hyperbolic geodesic caps define equivalent Carleson measure conditions.

The rectangle containment can also be checked directly. Write $z=r{e}^{i\theta }$ with $|\theta |\le \pi \ell$ and $1-r\le b\ell$. Its membership in the enlarged cap is equivalent toFor small enough $\ell$, $r\ge 1/2$. The right side is at most $b^2{\ell }^2$, while the left side is at least $({\kappa }^2-1){\pi }^2{\ell }^2/8$, using the sine product formula and $\sin x\ge x/2$ for small positive $x$. Thus $\kappa =4+4b/\pi$ suffices.

For $a\in \mathbb{D}$, this holomorphic function has H1 space on the unit disk norm one: its boundary modulus is the unit-mass Poisson kernel on the circle. The radial integrals are bounded by that boundary integral by subharmonic comparison and tend to it by continuity. Choosing $a$ at depth comparable to an arc's length makes $|f_a|$ bounded below by the reciprocal length throughout its hyperbolic geodesic cap. It tests necessity in the Carleson embedding theorem on the disk.

We prove necessity of the geometric condition from the embedding condition. Use normalized angular measure $m$ and the H1 space on the unit disk norm $\Vert f{\Vert }_1={\sup }_{r<1}\int |f(r\xi )|dm(\xi )$. Testing $f=1$ first gives $\mu (\mathbb{D})\le N(\mu )$.

For a short arc $I$ of normalized length $\ell =m(I)$ centered at ${\xi }_0$, put $a=(1-\ell ){\xi }_0$ and use the normalized squared Cauchy kernelIts boundary modulus is the Poisson kernel on the circle, so $\Vert f_a{\Vert }_1=1$. The geodesic-cap and Carleson-box comparison gives $|z-{\xi }_0|\le C\ell$ for $z\in Q(I)$. Therefore $|1-\overline{a}z|\le C^{\prime }\ell$, whereas $1-|a{|}^2=2\ell -{\ell }^2\ge \ell$. Thus $|f_a(z)|\ge c/\ell$ throughout the hyperbolic geodesic cap. The embedding bound givesFor arcs longer than a fixed small threshold, the total-mass estimate gives the same form after changing the absolute constant. HenceThis is precisely the Carleson measure condition, in its geodesic-cap version.

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.