Dependence of roots on parameters

Question

Let a function $f$ be holomorphic in a polydisk $U=U'\times U_n$,and suppose that for each fixed $z'\in U'$ it has a unique zero $z_n = \alpha(z')$ in the disk $U_n$. Then the function $\alpha(z')$ is holomorphic in $U'$.
On the proof of this proposition,it says firstly by Rouché's theorem for the family of functions $f(z',z_n)$ depending on the parameter $z'\in U'$ implies that $\alpha(z')$ is continuous...
I'm totally puzzled why $\alpha(z')$ is continuous.Can you tell me?
Thank you in advance!
Now I'm not satisfied with the above proposition,I want to prove this:
Let a function $f$ be holomorphic in a polydisk $D=D'\times D_n$,and suppose that for each fixed $z'\in D'$ it has in $D_n\subset\mathbb{C}^n$ exactly $m$ geometrically distinct zeros with multiplicities $k_1,\ldots,k_m$ respectively.Then these zeros $\alpha_1(z'),\ldots,\alpha_m(z')$ depend holomorphiclly on $z'$.
How to use the above proposition or ideas in the proof of the last proposition to prove this problem?
Thank you very much.

Answer 1

We don't need Rouché's theorem to see the continuity of $a(z')$.

The continuity - and even the holomorphy - of $a$ follows by the residue theorem, which yields the representation

$$a(z') = \frac{1}{2\pi i}\int_{\partial U_n} \zeta\frac{\frac{\partial f}{\partial z_n}(z',\zeta)}{f(z',\zeta)}\,d\zeta,\tag{1}$$

and standard results of integration theory (the Riemann integral suffices).

Since the integrand is continuous on $U'\times \partial U_n$, the continuity of $a$ is a classical result on parameter-dependent integrals.

Applying Morera's theorem for a coordinate of $z'$,

$$\begin{align} \int_\gamma a(z'_1,\dotsc, z'_{k-1}, \eta, z'_{k+1},\dotsc,z'_{n-1})\,d\eta &= \frac{1}{2\pi i}\int_\gamma \int_{\partial U_n} \zeta\frac{\frac{\partial f}{\partial z_n}(z'_1,\dotsc,\eta,\dotsc,z'_{n-1},\zeta)}{f(z'_1,\dotsc,\eta,\dotsc.z'_{n-1},\zeta)}\,d\zeta\,d\eta\\ &= \frac{1}{2\pi i}\int_{\partial U_n}\int_\gamma \zeta\frac{\frac{\partial f}{\partial z_n}(z'_1,\dotsc,\eta,\dotsc,z'_{n-1},\zeta)}{f(z'_1,\dotsc,\eta,\dotsc.z'_{n-1},\zeta)}\,d\eta\,d\zeta\\ &= \frac{1}{2\pi i}\int_{\partial U_n}0\,d\zeta\\ &= 0, \end{align}$$

where the change of the order of integration is justified by the continuity of the integrand on $\{(z'_1,\dotsc,z'_{k-1})\}\times\operatorname{Tr}\gamma \times \{(z'_{k+1},\dotsc,z'_{n-1})\}\times \partial U_n$ and compactness, shows that $a$ is separately holomorphic. A separately holomorphic continuous function is holomorphic, so the holomorphy of $a$ is established.

---

We can indeed use Rouché's theorem to see that $Z_f = \{z\in U : f(z) = 0\}$ has no limit points in $U'\times \partial U_n$, and the continuity of $a$, but that argument is less direct:

Fix $z'\in U'$. Since $U_n$ is open and $a(z') \in U_n$, $r := \inf \{\lvert z_n-a(z')\rvert : z_n \in \partial U_n\} > 0$. For every $0 < \varepsilon < r$, $f$ has no zero on $\{z'\}\times \{z_n : \lvert z_n-a(z')\rvert = \varepsilon\}$.

Let $\delta(\varepsilon) = \min \{\lvert f(z',z_n) : \lvert z_n-a(z')\rvert = \varepsilon\}$. By continuity of $f$, there is a neighbourhood $W_{\delta(\varepsilon)}$ of $z'$ with

$$\lvert f(w,z_n) - f(z',z_n)\rvert < \delta(\varepsilon)\tag{2}$$

for all $w\in W_{\delta(\varepsilon)}$ and all $z_n \in \partial U_n$. By Rouché's theorem, $(2)$ ensures that $\zeta \mapsto f(w,\zeta)$ has a unique zero in the disk $D_{\varepsilon}(a(z'))$ for every $w \in W_{\delta(\varepsilon)}$, and that means $\lvert a(w) - a(z')\rvert < \varepsilon$ for $w \in W_{\delta(\varepsilon)}$. That shows the continuity of $a$, and hence $a(z') \to \partial U_n \Longrightarrow z' \to \partial U'$, whence $Z_f$ has no limit points in $U'\times\partial U$.

To see the holomorphy of $a$, one can also differentiate under the integral sign in $(1)$, since $\lvert f(w,z_n)\rvert$ is bounded away from $0$ in a neighbourhood of $\{z'\}\times \partial U_n$, that is unproblematic.

Answer 2

Step 1.Similarly as in Proposition 1,$Z_f$has no limit points on $D'\times \partial D_n$.Let $\zeta'\in D'$,let $\alpha_1(\zeta'),\ldots,$$\alpha_m(\zeta')$be the zeros of $f(\zeta',z_n)$ in $D_n$,and let $U_{nj}:=\{|z_n-\alpha_j(\zeta')|\le r_{nj}\}$ are pairwise nonintersecting disks in $D_n$.Since $Z_f$ is closed in $D$,there exists a polydisk $U'\ni\zeta'$ such that $U'\times\partial U_{nj}\subset D\setminus Z_f$ for all $j=1,\ldots,m.$Since the number of distinct zeros of $f(Z',_n)$ in $D_n$ is $m$ by assumption,for fixed $z'\in U'$,in each $U_{nj}$,there is exactly one zero $\alpha_j(z')$ of $f(z',z_n)$ with multiplicity $k_j$.By Proposition 1 (in the polydisk $U'\times U_{nj}$), $\alpha_j(z')$ is a holomorphic function in $U'$.
Step 2.This argument can be repeated in a neighborhood of any point $\zeta'\in D'$.By construction, the analytic elements obtained coincide, after a suitable renumbering,on the intersections of the corresponding polydisks in $D'$,and can therefore be analytically continued along any path in $D'$.Since $D'$ is simply connected, the monodromy Theorem that these elements make up the $m$ holomorphic functions $a_j(z')$.
Step 3.In each polydisk $U^j=U'\times U_{nj}$,we represent $f$ as $f(z)=(z_n-a_j(z'))^{k_j}\varphi_j(z)$ by Weierstrass preparation theorem,where $\varphi_j$ is holomorphic and zero free in $U^j$.Define $\varphi(z)=f(z)/\Pi_1^m(z_n-\alpha_j(z'))^{k_j}$.Then $\varphi$ is defined and holomorphic outside $Z_f$.In $U^j/Z_j,\varphi(z)=f(z)/\Pi_1^m(z_n-\alpha_j(z'))^{k_j}=\varphi_j(z)/\Pi_{i\ne j}(z_n-\alpha_i(z'))^{k_i}.$By the Riemann extension theorem,it can be defined holomorphically in all of $U_j$.Since the $U^j$ cover $Z_f \cap \{z'\in U'\}$,$\varphi$ is holomorphic and zero free in $U'\times D_n.$By covering $D'$ with such polydisk $U'$.We obtain that $\varphi$ is holomorphic and zero free in $D$.By the uniqueness of Weierstrass preparation theorem,we have $\alpha_1(\zeta'),\ldots,$$\alpha_m(\zeta')$ are holomorphic.