Show that \(f = u+iv\) with \(u, v\in C^1(\RR)\) satisfying the Cauchy-Riemann equations on \(\Omega\), then \(f\) is holomorphic on \(\Omega\) with \[ f'(z) = \dd{f}{x} = {1\over i} \dd{f}{y} = {1\over 2}\qty{u_x + iv_x} .\] Conversely, show that if \(f\) is holomorphic, then \(f\) satisfies the Cauchy-Riemann equations.
The last expression in the statement has a spurious factor: \(f'(z)=\dd{f}{x}=u_x+iv_x\). The factor \({1\over2}\) belongs to the Wirtinger derivative \({\partial f\over\partial z}={1\over2}\qty{\dd{f}{x}-i\dd{f}{y}}\), which equals \(f'(z)\) for holomorphic \(f\).
Solution
Direction 1: Holomorphic \(\implies\) CR.
Suppose \(f'(z_0)\) exists for some \(z_0 = x_0 + iy_0 \in \Omega\). Since the limit \[f'(z_0) = \lim_{h\to 0,\, h\in \CC} \frac{f(z_0 + h) - f(z_0)}{h}\] exists, we may approach along the real and imaginary axes.
Along \(h = t \in \RR\): \[f'(z_0) = \lim_{t\to 0} \frac{f(x_0 + t, y_0) - f(x_0, y_0)}{t} = f_x(x_0, y_0).\] Along \(h = it\), \(t \in \RR\): \[f'(z_0) = \lim_{t\to 0} \frac{f(x_0, y_0 + t) - f(x_0, y_0)}{it} = \frac{1}{i} f_y(x_0, y_0).\] Equating the two expressions for \(f'(z_0)\) and writing \(f = u + iv\): \[f_x = \frac{1}{i} f_y \implies i(u_x + iv_x) = u_y + iv_y \implies -v_x + iu_x = u_y + iv_y.\] Comparing real and imaginary parts: \(u_x = v_y\) and \(u_y = -v_x\).
Direction 2: CR \(\implies\) holomorphic.
Suppose \(u, v \in C^1(\Omega)\) satisfy \(u_x = v_y\) and \(u_y = -v_x\) on \(\Omega\). Write \(h = \Delta x + i\Delta y\). By the multivariable Taylor theorem (since \(u, v\) are \(C^1\)): \[u(x_0 + \Delta x, y_0 + \Delta y) - u(x_0, y_0) = u_x \Delta x + u_y \Delta y + o(|h|),\] \[v(x_0 + \Delta x, y_0 + \Delta y) - v(x_0, y_0) = v_x \Delta x + v_y \Delta y + o(|h|),\] where the partial derivatives are evaluated at \((x_0, y_0)\). Using the CR equations \(u_y = -v_x\) and \(v_y = u_x\): \[f(z_0 + h) - f(z_0) = (u_x \Delta x - v_x \Delta y) + i(v_x \Delta x + u_x \Delta y) + o(|h|).\] The first two terms factor as \((u_x + iv_x)(\Delta x + i\Delta y) = (u_x + iv_x) h\), so: \[\frac{f(z_0 + h) - f(z_0)}{h} = (u_x + iv_x) + \frac{o(|h|)}{h}.\] Since \(|o(|h|)/h| \to 0\) as \(h \to 0\), the limit exists and \(f'(z_0) = u_x + iv_x = u_x - iu_y\).