# Proof of implicit function theorem for a complex function not necessarily holomorphic

Let $$U,V\subset\mathbb{C}$$ be domains, and $$F(z,w):U\times V\to\mathbb{C}$$ be continuous, and holomorphic in $$z$$ for every fixed $$w\in V$$. Let $$(z_0,w_0)\in U\times V$$ be s.t $$F(z_0,w_0)=0$$ and $$\frac{\partial F}{\partial z}(z_0,w_0)=F_1(z_0,w_0)\neq0$$. Let $$r>0$$ be s.t for every $$z_0\neq z\in\overline{\mathbb{D}_r(z_0)}, F(z,w_0)\neq0$$. Prove that there exists $$\delta>0$$ s.t for every $$w\in\mathbb{D}_\delta(w_0)$$ there exists a unique $$z=g(w)\in\overline{\mathbb{D}_r(z_0)}$$ s.t $$F(z,w)=0$$.

Edit: I now see that I cannot use the implicit function theorem for real functions as $$F$$ is not necessarily continuiously differentiable as a real function. I tried using the argument theorem but this didnt get me too far. Any help would be appreciated.

• You will need some assumption on the regularity of $F$ with respect to the second variable. The real version of the implicit function theorem includes a uniqueness result, so that should not be an issue. Am I overlooking something here? Commented Jun 6, 2022 at 10:08
• @Thomas What do you mean regulariry of $F$ wrt to the second variable? Also, it seems youre correct and its given that $z$ is unique. Commented Jun 6, 2022 at 10:14
• To apply the (real) implicit function theorem, $F$ must be continuously differentiable as a function of it's real variables, in your case as a map from $\mathbb{R}^4$ to $\mathbb{R}^2$. Commented Jun 6, 2022 at 10:20
• @Thomas I see. I now realize I cant use the theorem here. Do you have any idea how I can go about it? Commented Jun 6, 2022 at 10:20
• @AndrewD.Hwang Sorry, I added that $F$ is continuous (and as was stated, holomorphic in the first variable). Commented Jun 6, 2022 at 12:58

## 1 Answer

For any $$z \in \mathbb C$$ and $$b \gt 0$$, let $$C_b(z)$$ be the circle centered on $$z$$ of radius $$b$$ and $$D_b(z)$$ the associated open disk.

Let's first prove that it exists $$r_1,r_2 \gt 0$$ such that $$F$$ doesn't vanish on $$C_{r_1}(z_0) \times D_{r_2}(w_0)$$

Let $$a = \frac{\partial F}{\partial z}(z_0,w_0)= F_1(z_0,w_0) \neq 0$$. As $$F(\cdot, w_0)$$ is supposed to be holomorphic, from $$F(z_0,w_0) = 0$$ and $$a \neq 0$$, it follows that it exists $$r_1 \gt 0$$ such that $$F(z,w_0) \neq 0$$ in $$\overline{D_{r_1}(z_0)} \setminus \{z_0\}$$. As $$C_{r_1}(z_0)$$ is compact, $$\lvert F \rvert$$ attains a positive minimum $$b$$ on $$C_{r_1}(z_0)$$. From the continuity of $$F$$ on $$U \times V$$ and again the compactness of $$C_{r_1}(z_0)$$, we can find $$r_2 \gt 0$$ such that $$\lvert F(z,w) \rvert \gt b/2 \gt 0$$ for $$(z,w) \in C_{r_1}(z_0) \times D_{r_2}(w_0)$$.

About the desired result

From the formula

$$\frac{\partial F}{\partial z}(z,w) = \frac{1}{2 i \pi} \int_{C_{\delta}(z_0)} \frac{F(\zeta,w)}{(\zeta -z)^2} \ d\zeta$$ which holds for $$\delta \gt 0$$ small enough, $$z \in D_\delta(z_0)$$ and the continuity of $$F$$, it follows that $$\frac{\partial F}{\partial z}(z,w)$$ is also continuous. Therefore the map $$\eta(w)=\frac{1}{2\pi i}\int_{C_{r_1}(z_0)}\frac{F_1(z,w)\,dz}{F(z,w)}$$ is well defined and continuous in the open disk $$D_{r_2}(w_0)$$. According to the argument principle, $$\eta(w)$$ is the number of zeros of the equation $$F(z,w)=0$$ in the open disk $$D_{r_1}(z_0)$$. Also $$\eta(w_0) = 1$$ as $$z_0$$ is the only root of $$F(z,w_0) = 0$$ in $$D_{r_1}(z_0)$$. We finally get the desired conclusion that for any $$w \in D_{r_2}(w_0)$$, the equation $$F(z,w) = 0$$ has a unique solution in the open disk $$D_{r_1}(z_0)$$.