# Mean value theorem in terms of Wirtinger calculus?

The mean value theorem for vector-valued function in the real domain $f: \mathcal{R}^n \rightarrow \mathcal{R}^d$ can be expressed as $$f(x)-f(y)=\int_{0}^{1}\nabla f(x(\tau))d \tau \cdot (x -y),$$ where $x(\tau) = y + \tau (x - y)$.

I wonder if there exists similar results for vector-valued functions in the complex domain. Suppose $f:\mathcal{C}^n \rightarrow \mathcal{C}^d$, do we have $$f(z_1)-f(z_2)=\int_{0}^{1}\nabla f(z(\tau))d \tau \cdot \left(\begin{array}{c} z_{1}-z_{2}\\ \overline{z_{1}}-\overline{z_{2}} \end{array}\right),$$ where $z(\tau) = z_2 + \tau (z_1 - z_2)$, $\overline{z_1} - \overline{z_2}$ denotes the conjugate of $z_1 -z_2$ and $\nabla f(z(\tau))$ is the Wirtinger Jacobian.

• according to this comment it seems the answer is "No". – Carlo Beenakker Nov 3 '17 at 12:40
• @CarloBeenakker Thank you for the pointer. However, $f'$ there means the complex derivative for complex differentiable functions. What I want to find is the mean value theorem for Wirtinger derivative, not necessarily for complex differentiable functions. – Wuchen Nov 3 '17 at 14:07
• thanks for the clarification, I think it then follows rather simply from the Wirtinger differential, see answer below. – Carlo Beenakker Nov 5 '17 at 15:57

The desired relation can be written in components as $$f_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int_0^1 d\tau\,\sum_{m}\left(\frac{\partial f_n}{\partial z_m}(z_m^{(1)}-z_m^{(2)})+\frac{\partial f_n}{\partial \bar{z}_m}(\bar{z}_m^{(1)}-\bar{z}_m^{(2)})\right),\;\;[1]$$ with the prescribed $\tau$ dependence: $$\mathbf{z}(\tau)=\mathbf{z}^{(2)}+\tau(\mathbf{z}^{(1)}-\mathbf{z}^{(2)}),\;\; \mathbf{\bar z}(\tau)=\mathbf{\bar z}^{(2)}+\tau(\mathbf{\bar z}^{(1)}-\mathbf{\bar z}^{(2)}).\;\;[2]$$ We start from the equation $$f_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int_0^1 d\tau\,\frac{d}{d\tau}f_n(\mathbf{z},\mathbf{\bar z}).\;\;[3]$$ Equation [3] implies equation [1] if $$df_n=\sum_m\left(\frac{\partial f_n}{\partial z_m}dz_m+\frac{\partial f_n}{\partial \bar{z}_m}d\bar{z}_m\right).\;\;[4]$$ This is indeed a property of the Wirtinger differential.$^\ast$
$^\ast$ Proof
Denote $z=x+iy$, $\bar{z}=x-iy$ and insert the definitions $\partial/\partial x=\partial/\partial z+\partial/\partial \bar{z}$, $\partial/\partial y=i\partial/\partial z-i\partial/\partial \bar{z}$ of the Wirtinger derivatives:
$$df_n=\sum_m\left(\frac{\partial f_n}{\partial x_m}dx_m+\frac{\partial f_n}{\partial y_m}dy_m\right)$$ $$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dx_m+\frac{\partial f_n}{\partial \bar{z}_m}dx_m+i\frac{\partial f_n}{\partial z_m}dy_m-i\frac{\partial f_n}{\partial \bar{z}_m}dy_m\right)$$ $$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dz_m+\frac{\partial f_n}{\partial \bar{z}_m}d\bar{z}_m\right).$$