Lemma 6.11.2D7BCCD73BD13

Proof. We bound the norm of $H$ from below by a weighted norm of $H\circ \gamma $ on a disc, and then use that $H\circ \gamma =F\circ \gamma $ at the points $t$ with $t^ q=a$.

Choose $\rho >0$ with $\rho ^ q<R$ such that $\gamma $ is defined on a neighborhood of the closed disc $|t|\leq \rho $, $F\circ \gamma $ has no zeros for $0<|t|\leq \rho $, and, since $\gamma $ is continuous with $\gamma (0)=0$, $(\gamma _2(t),\ldots ,\gamma _ n(t))$ lies in a fixed polydisc $P'\Subset P$ centered at $0$ for $|t|\leq \rho $. Let $\delta >0$ be half the distance from $P'$ to the complement of $P$. For $0<|z_1|<\rho ^ q$, each of the $q$ points $\gamma (t)$ with $t^ q=z_1$ has the form $(z_1,w)$ with $w\in P'$, and (6.8.7) for the holomorphic function $H(z_1,\cdot )$ on the ball $B(w,\delta )\subset P$ gives $|H(z_1,w)|^2\leq C_\delta \int _ P|H(z_1,w')|^2\, \mathrm dV(w')$, where $C_\delta ^{-1}$ is the volume of a ball of radius $\delta $ in $\mathbb C^{n-1}$. Summing over the $q$ points, we obtain $\sum _{t^ q=z_1}|H(\gamma (t))|^2\leq qC_\delta \int _ P|H(z_1,w')|^2\, \mathrm dV(w')$.

Now we integrate over $|z_1|<\rho ^ q$. The map $t\mapsto t^ q$ maps each of the $q$ half-open sectors $\{ 0<|t|<\rho ,\ 2\pi k/q\leq \arg t<2\pi (k+1)/q\} $ bijectively onto the punctured disc $0<|z_1|<\rho ^ q$, and on the interior of the sector it is a diffeomorphism onto the complement of a ray, with real Jacobian $|qt^{q-1}|^2$. By the change of variables formula, applied on each sector, $\int _{|t|<\rho }\Phi (t)q^2|t|^{2q-2}\, \mathrm dV_2(t)=\int _{|z_1|<\rho ^ q}\sum _{t^ q=z_1}\Phi (t)\, \mathrm dV_2(z_1)$ for every nonnegative measurable $\Phi $. Since $\Delta _{\rho ^ q}\times P\subset \Omega $, it follows that

with $c_0=1/(qC_\delta )$. If $n=1$, the same inequality holds with $c_0=1/q$, without the submean step.

Write $F\circ \gamma (t)=t^ mv(t)$ with an integer $m\geq 0$ and $v$ without zeros on $|t|\leq \rho $, and put $h=(H/F)\circ \gamma $. This function is holomorphic for $0<|t|<\rho $, because $F\circ \gamma $ has no zeros there, and by hypothesis it extends holomorphically across $0$ with $h(0)=0$; thus $h(t)=\sum _{\ell \geq 1}h_\ell t^\ell $ for $|t|<\rho $. Then $H\circ \gamma =t^ mvh$, and $|v|$ is bounded below by a positive constant on the compact disc $|t|\leq \rho $, being continuous and without zeros there. On each circle $|t|=s<\rho $ the power series of $h$ converges uniformly, and the monomials $t^\ell $ are orthogonal there, so $\int _0^{2\pi }|h(s\mathrm e^{\mathrm i\theta })|^2\, \mathrm d\theta =2\pi \sum _\ell |h_\ell |^2s^{2\ell }$. Integrating in polar coordinates and using the monotone convergence theorem, we obtain

with $c_1=\pi q^2c_0\min _{|t|\leq \rho }|v(t)|^2$. Now let $0<|a|<\rho ^ q$. For each of the $q$ roots $b$ of $b^ q=a$, the point $\gamma (b)$ lies on the slice $\{ z_1=a\} $, where $H=F$, and $F(\gamma (b))\neq 0$; hence $h(b)=1$. The average $q^{-1}\sum _{b^ q=a}b^\ell $ equals $a^{\ell /q}$ if $q$ divides $\ell $ and $0$ otherwise, so averaging $h$ over these roots gives $\sum _{j\geq 1}h_{qj}a^ j=1$. By the Cauchy–Schwarz inequality,

If $|a|\leq \rho ^ q/2$, the last sum equals $\rho ^{-2(m+q)}\sum _{j\geq 1}(qj+m+q)x^ j$ with $x=|a|^2\rho ^{-2q}\leq 1/4$, and this series is at most a constant times $x$. Hence (6.11.3) holds for $0<|a|<r=\rho ^ q/2$, with $c$ independent of $a$ and $H$. $\square$