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?


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
  • $\begingroup$ 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 at 9:38

There's also the Looman–Menchoff theorem.

  • 5
    $\begingroup$ The original answer was not clickable. There was no indication that there even was a URL. I just made it clickable. $\endgroup$
    – Aryabhata
    Dec 30, 2010 at 17:12
  • $\begingroup$ Where can I find proof of this theorem? $\endgroup$
    – Abhay
    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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.