Integral Representations

By Sébastien Boisgérault, Mines ParisTech, under CC BY-NC-SA 4.0

September 30, 2019

Contents

Exercises

Functions of Several Complex Variables

Question

Let \(n \geq 2\), let \(\Omega\) be an open subset of \(\mathbb{C}^n\) and let \(f: \Omega \mapsto \mathbb{C}\) a continuous function. Show that \(f\) is complex-differentiable in \(\Omega\) if and only if for any \((z_1, \dots, z_n) \in \Omega\), the partial function \[ f_{k,z}: w \mapsto f(z_1, \dots, z_{k-1}, w, z_{k+1}, \dots, z_n) \] is holomorphic.

Answer

We may define the embedding functions \(e_{k,z}: \mathbb{C} \to \mathbb{C}^n\) by \[ e_{k,z}(w) = (z_1, \dots, z_{k-1}, w, z_{k+1}, \dots, z_n). \] It is plain that the \(e_{k,z}\) are continuous. A function \(f_{k,z}\) is defined on the preimage of the open set \(\Omega\) by \(e_{k,z}\) which is therefore an open set.

Assume that \(f\) is complex-differentiable; it is continuous. Additionally, \(f_{k,z} = f \circ e_{k,z}\); as the function \(e_{k,z}\) is complex-linear, it is complex-differentiable and \(f_{k,z}\) is complex-differentiable (or holomorphic) as the composition of complex-differentiable functions.

Conversely, if \(f\) if continuous and every partial function \(f_{k,z}\) is complex-differentiable, the function \(f\) itself is complex-differentiable as every partial derivative \(z \in \Omega \mapsto (\partial f/\partial z_k) (z)\) is continuous – not merely as a function of its \(k\)-th variable which is plain, but as a function of all its variables.

Let \(z=(z_1,\dots, z_n) \in \Omega\), let \(c \in \mathbb{C}\) and \(r>0\) such that \[ \forall \, w \in \mathbb{C}, \; |w - c| \leq r \; \to \; (z_1,\dots,z_{k-1},w,z_k,\dots, z_n) \in \Omega. \] Cauchy’s formula, applied to the partial function \(f_{k,z}\) for the path \(\gamma =c + r[\circlearrowleft]\), provides \[ f(z_1,\dots, z_n) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z_1, \dots, z_{k-1}, w, z_{k+1}, \dots, z_n)}{w-z_k} dw \] The integrand is continuous with respect to the pair \((z_1,w_1)\) and complex-differentiable with respect to \(z_1\), thus we may compute the partial derivative of \(f\) with respect to \(z_k\) satisfies by differentiation under the integral sign: \[ \frac{\partial f}{\partial z_k} (z_1,\dots, z_n) = \frac{1}{i2\pi} \int_{\gamma} \frac{f(z_1, \dots, z_{k-1}, w, z_{k+1}, \dots, z_n)}{(w-z_k)^2} dw. \] As the function \(f\) is continuous, the partial derivative is also continous.