2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ according to this comment it seems the answer is "No". $\endgroup$ – Carlo Beenakker Nov 3 '17 at 12:40
  • $\begingroup$ @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. $\endgroup$ – Wuchen Nov 3 '17 at 14:07
  • $\begingroup$ thanks for the clarification, I think it then follows rather simply from the Wirtinger differential, see answer below. $\endgroup$ – Carlo Beenakker Nov 5 '17 at 15:57
1
$\begingroup$

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:

$$df_n=\sum_m\left(\frac{\partial f_n}{\partial x_m}dx_m+\frac{\partial f_n}{\partial y_m}dy_m\right)$$ $$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dx_m+\frac{\partial f_n}{\partial \bar{z}_m}dx_m+i\frac{\partial f_n}{\partial z_m}dy_m-i\frac{\partial f_n}{\partial \bar{z}_m}dy_m\right)$$ $$\qquad=\sum_m\left(\frac{\partial f_n}{\partial z_m}dz_m+\frac{\partial f_n}{\partial \bar{z}_m}d\bar{z}_m\right).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.