请验证：函数$f:\mathbb{C}\rightarrow \mathbb{C}$ $f(x+iy)=\sqrt{|xy|}$在原点满足柯西-黎曼条件却不可导

Alright, let's break this down step by step to verify the claim about this function. This is a great example that highlights a critical nuance in complex analysis: satisfying the Cauchy-Riemann (C-R) conditions at a point doesn't guarantee the function is differentiable there.

First, Let's Clarify Key Definitions #

Before diving in, let's recap the basics so we're all aligned:

• Cauchy-Riemann Conditions: For a function $f(x+iy) = u(x,y) + iv(x,y)$ to satisfy C-R at $(a,b)$, the partial derivatives of $u$ and $v$ must exist at that point, and:
  • $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$
  • $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$
• Complex Differentiability: A function is differentiable at $z_0$ if the limit $\lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}$ exists and is finite, no matter what path $z$ takes to approach $z_0$.

For our function $f(x+iy) = \sqrt{|xy|}$, note that it maps every complex number to a real number. That means we can write it as $u(x,y) = \sqrt{|xy|}$ and $v(x,y) = 0$ for all $(x,y) \in \mathbb{R}^2$.

Step 1: Verify C-R Conditions at $(0,0)$ #

We need to compute the partial derivatives of $u$ and $v$ at the origin using the formal definition of partial derivatives (since we can't just differentiate the expression directly at $(0,0)$):

Partial derivatives of $u$ at $(0,0)$ #

• $\frac{\partial u}{\partial x}(0,0) = \lim_{h \to 0} \frac{u(h, 0) - u(0,0)}{h} = \lim_{h \to 0} \frac{\sqrt{|h \cdot 0|} - 0}{h} = \lim_{h \to 0} \frac{0}{h} = 0$
• $\frac{\partial u}{\partial y}(0,0) = \lim_{k \to 0} \frac{u(0, k) - u(0,0)}{k} = \lim_{k \to 0} \frac{\sqrt{|0 \cdot k|} - 0}{k} = 0$

Partial derivatives of $v$ at $(0,0)$ #

Since $v(x,y) = 0$ everywhere, all its partial derivatives are 0:

• $\frac{\partial v}{\partial x}(0,0) = 0$
• $\frac{\partial v}{\partial y}(0,0) = 0$

Check the C-R equations #

Now plug these into the C-R conditions:

• $\frac{\partial u}{\partial x} = 0 = \frac{\partial v}{\partial y}$ ✔️
• $\frac{\partial u}{\partial y} = 0 = -\frac{\partial v}{\partial x}$ (since $-\frac{\partial v}{\partial x} = 0$) ✔️

So the Cauchy-Riemann conditions are indeed satisfied at $(0,0)$.

Step 2: Prove $f$ is NOT Differentiable at $(0,0)$ #

To show non-differentiability, we just need to find two different paths approaching the origin where the limit $\lim_{z \to 0} \frac{f(z)}{z}$ gives different results. Here are two simple paths to test:

Path 1: Along the line $y = x$ #

Let $z = x + ix = x(1+i)$, where $x \to 0$. Then:
$$\frac{f(z)}{z} = \frac{\sqrt{|x \cdot x|}}{x(1+i)} = \frac{|x|}{x(1+i)}$$

• If $x \to 0^+$ (approaching from the positive real side), this simplifies to $\frac{x}{x(1+i)} = \frac{1}{1+i} = \frac{1 - i}{2}$
• If $x \to 0^-$ (approaching from the negative real side), this becomes $\frac{-x}{x(1+i)} = \frac{-1}{1+i} = \frac{-1 + i}{2}$

Already, we see the limit depends on the direction we approach along this line. But let's compare it to another path to drive the point home.

Path 2: Along the real axis ($y = 0$) #

Let $z = x$ (so $y=0$), $x \to 0$. Then:
$$\frac{f(z)}{z} = \frac{\sqrt{|x \cdot 0|}}{x} = \frac{0}{x} = 0$$

Since the limit is $0$ along the real axis, but $\frac{1-i}{2}$ when approaching from the positive $y=x$ direction, the overall limit $\lim_{z \to 0} \frac{f(z)}{z}$ does not exist. Therefore, $f$ is not differentiable at $(0,0)$.

Key Takeaway #

This example reinforces a crucial lesson: the Cauchy-Riemann conditions are a necessary condition for complex differentiability, but not sufficient. For a function to be differentiable at a point, it needs to satisfy C-R and have the limit defining the derivative exist (regardless of path), or equivalently, be differentiable as a real-valued function of two variables at that point.

内容的提问来源于stack exchange，提问作者Dan Balan