Proof. We need to check that the ratio

is bounded for $(z,P)\in{\cal M}$.

It suffices to prove the Lemma assuming that $U_{1}$ is the parabolic domain $\{(x,y)\in{{\bf R}}^{2}|y>x^{2}\}$ and $P$ is the upper half-plane. The vector field $v_{(z,P)}$ is given for $z=(x,y)$ by the formulas

The denominator is equal to $g(n_{P},v_{(z,P)})=\langle dy,v_{(z,P)}\rangle\cdot\sqrt{g(\partial/\partial y,\partial/\partial y)}=\sqrt{g(\partial/\partial y,\partial/\partial y)}=\exp(O(1))$ near $(0,0)$.

The numerator is equal to

where $f_{1}(x,y)=(\nabla_{\partial/\partial x}\,g)(n_{P},n_{P})$ and $f_{2}(x,y)=(\nabla_{\partial/\partial y}\,g)(n_{P},n_{P})$ are two $C^{\infty}$-functions.

By assumption of the Lemma we have $f_{1}(0,0)=0$. Therefore $|f_{1}(x,y)|\leq const\,\max\{|x|,|w|\}$ where $w=x^{2}-y$ is a convenient local coordinate near the point $(0,0)$. Notice also that $f_{2}(x,y)=O(1)$.

Now we can estimate first summand of the numerator assuming that $|x|$ and $|w|$ are sufficiently small. As we have seen, it is bounded by

There are three cases which we need to consider.

a) If $0<w<x^{2}$ then $I={x\over{x^{2}}}O(|x|)=O(1)$.

b) if $x^{2}\leq w<x$ then $I={x\over w}O(|x|)=O(1)$.

c) If $x\leq w\leq 1$ the $I={x\over w}O(|w|)=O(1)$.

We see that the numerator is bounded. This concludes the proof of Lemma. $\blacksquare$