Minimum and maximum of $|f(z)|$, analytic but doesn't satisfy Cauchy-Riemann

Question

Maximum modulus principle

Let $D\subset \mathbb{C}$ be a limited domain (connected open set) and $f(z)$ a complex holomorphic function in $D$, i.e. $f:\bar{D} \to \mathbb{C}$ such that

$$ z = x + i y \mapsto u(x, y) + i v(x, y) $$

• $u(x, \ y)$ and $v(x, \ y)$ are analytic functions of class $C^{\infty}$
• $u(x, \ y)$ and $v(x, \ y)$ satisfy the Cauchy-Riemann conditions \eqref{1}

$$\dfrac{\partial u}{\partial x} = \dfrac{\partial v}{\partial y} \ \ \ \ \ \text{and} \ \ \ \ \ \ \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x} \tag{1}\label{1}$$

The Maximum modulus principle says, if $f(z)$ is not constant, then the minimum and maximum of $|f(z)|$ are on $\partial D$.

Question:

By taking out the Cauchy-Riemann condition from $f(z)$, is there anything can be said about its minimum and maximum?

For example:

• The minimum and maximum are on $\partial D$, if and only if $u$ and $v$ satisfy Cauchy-Riemann.
• If $u$ and $v$ don't satisfy Cauchy-Riemann, it may have infinites min/max on $\partial D$, or finite values in $D$, not both
• From \eqref{1}, one can show \eqref{2}. But let's say $u$ and $v$ satisfy \eqref{2} but not \eqref{1}, then there are a finite number of minimum and maximum

$$\dfrac{\partial^2 u}{\partial x^2} +\dfrac{\partial^2 u}{\partial y^2} =0 \ \ \ \ \ \text{and} \ \ \ \ \ \ \dfrac{\partial^2 v}{\partial x^2} +\dfrac{\partial^2 v}{\partial y^2} = 0 \tag{2}\label{2}$$

Note 1: $f(z)$ is continuos on $\bar{D}$

Note 2: $\bar{D}$ is the closed set which is the union of the openset $D$ and its boundary $\partial D$

Note 3: It's possible to have $u$ and $v$ such \eqref{2} happens, but not \eqref{1}. Example: get $f_1(z) = z^2$ and $f_2(z) = z^3$, it's not hard to see that cauchy relation is not valid in $f_3(z) = u_1(z) + i \cdot v_2(z)$, even though $\nabla^2 u_1 = \nabla^2 v_2 = 0$

$$f_1(z) = \left(x^2-y^2\right) + i \cdot 2xy$$ $$f_2(z) = \left(x^3-3xy^2\right) + i \cdot \left(3x^2y-y^3\right)$$ $$\dfrac{\partial u_1}{\partial x} = 2x \ne 3\left(x^2-y^2\right) = \dfrac{\partial v_2}{\partial y}$$

Note 4: By $u(x, \ y)$ being analitic, I mean it can be expressed in Taylor's in two variables:

$$u(x, \ y) = \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} a_{ij} \cdot \left(x-x_0\right)^{i} \cdot \left(y-y_0\right)^{j}$$

Answer 1

If $u$ and $v$ are harmonic functions i.e. $(2)$ is assumed, then both functions satisfy the intermediate value property. That means for $\overline{B(z,r)}\subset D,$ $z=x+iy,$ $h=u$ or $h=v$ there holds $$h(x,y)={1\over 2\pi}\int\limits_0^{2\pi}h(x+r\cos t,y+r\sin t)\,dt$$ Hence $$f(z)=u(x,y)+iv(x,y)\\ ={1\over 2\pi}\int\limits_0^{2\pi}[u(x+r\cos t,y+r\sin t)+iv(x+r\cos t,y+r\sin t)]\,dt\\ ={1\over 2\pi}\int\limits_0^{2\pi}f(z+re^{it})\,dt$$ Hence $$|f(z)|\le {1\over 2\pi}|f(z+re^{it})|\,dt$$ which implies $$|f(z)|\le \max_{|w-z|=r}|f(w)|$$ Assume the maximum of $|f|$ is attained at a point $z_0\in D.$ Then $|f|$ is constant for $z$ in a ball centered at $z_0.$ By the connectedness of $D$ the function $|f|$ is constant, i.e. $u^2+v^2$ is constant. Thus $$0={1\over 2}\Delta(u^2+v^2)=\left ({\partial u\over \partial x}\right )^2+\left ({\partial u\over \partial y}\right )^2+\left ({\partial v\over \partial x}\right )^2+\left ({\partial v\over \partial y}\right )^2$$ Therefore $u$ and $v$ are constant.

Summarizing C-R equations $(1)$ are not necessary for the maximum modulus principle. The condition $(2)$ suffices.