Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 9 / 5 / a / Solution

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.