Hyperbolic geodesic cap ($Q(I)$)

For a proper arc $I$ of the boundary of the unit disc, $Q(I)$ is the side of the hyperbolic geodesic between its endpoints adjacent to $I$. For short arcs this is a curved boundary cap. For arcs longer than a semicircle it contains the center. With endpoints $e^{\pm i\alpha }$ and $0<\alpha <\pi /2$, the geodesic is the circle with center $\sec \alpha$ and radius $\tan \alpha$, orthogonal to the unit circle. It is a diameter at $\alpha =\pi /2$.

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.