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?

  • $\begingroup$ The $C^1$ condition in this proof is used as follows: $C^1 \implies$ (Frechet) differentiable $\implies$ the "directional derivative" local linear approximation dot product formula $[\frac{\partial u}{\partial x}, \frac{\partial u}{\partial x}]\cdot [h_1,h_2]+o(|h|)$ (see e.g. Theorems 2.17 and 2.19 in Folland's Advanced Calculus) $\endgroup$
    – D.R.
    Feb 2, 2023 at 7:25

3 Answers 3


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.

  • 5
    $\begingroup$ So that is why the question's title sounded vaguely familiar... :D $\endgroup$ Dec 30, 2010 at 2:20
  • 1
    $\begingroup$ I wonder if there have been any advancements in the past 44 years since this paper (and 11 years since this answer). $\endgroup$
    – Mark S.
    Mar 23, 2022 at 9:38

There's also the Looman–Menchoff theorem.

  • $\begingroup$ Where can I find proof of this theorem? $\endgroup$
    – chesslad
    Dec 7, 2020 at 11:12
  • $\begingroup$ @Abhay, see the references in the Wikipedia page $\endgroup$
    – lhf
    Dec 7, 2020 at 11:18

Thinking of the Cauchy-Riemann operator as an elliptic partial differential operator, the basic elliptic regularity result implies that any distribution satisfying the C-R equation is a holomorphic function. For example, locally integrable suffices. This result was used in Gunning' "Riemann Surfaces", for example, in the discussion of Serre duality.

  • 1
    $\begingroup$ TeX only works inside $-signs here. In order to get italic text and bold text text you can enclose it in single and double asterisks: *italic text* and **bold text**. $\endgroup$
    – t.b.
    Jun 20, 2011 at 16:51
  • $\begingroup$ Just for completeness: $L^p_{\text{loc}}$ suffices for any $p>0$ as proven in link.springer.com/article/10.1007/BF02807221 $\endgroup$
    – Bananach
    Sep 27, 2016 at 10:37

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