It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is continuously differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or Wikipedia). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below.
Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then
\begin{equation*}
u(x+h_1, y+h_2) – u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|)
\end{equation*}
and
\begin{equation*}
v(x+h_1, y+h_2) – v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|).
\end{equation*}
Multiplying the second equation by $i$ and adding the two together gives
\begin{align*}
(u+iv)(z+h)-(u+iv)(z) &= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\
&= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|).
\end{align*}
Now dividing by $h$ gives us the desired result.
Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function?
Best Answer
See When is a Function that Satisfies the Cauchy-Riemann Equations Analytic? J. D. Gray and S. A. Morris The American Mathematical Monthly Vol. 85, No. 4 (Apr., 1978), pp. 246-256.