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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{du}{x} h_1 + \frac{du}{dy}h_2 + o(|h|) \end{equation*} and \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{dv}{dx} h_1 + \frac{dv}{dy} 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{du}{dx} h_1 + i \frac{dv}{dx} h_1 + \frac{du}{dy} h_2 + i \frac{dv}{dy} h_2 + o(|h|)\\ &= \left( \frac{du}{dx} + i \frac{dv}{dx} \right) (h_1+h_2i) + 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?

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up vote 11 down vote accepted

See this.

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So that is why the question's title sounded vaguely familiar... :D –  J. M. Dec 30 '10 at 2:20
    
Fabulous! Thank you. –  Tony Dec 30 '10 at 20:40
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There's also the Looman–Menchoff theorem.

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The original answer was not clickable. There was no indication that there even was a URL. I just made it clickable. –  Aryabhata Dec 30 '10 at 17:12
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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.

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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**. –  t.b. Jun 20 '11 at 16:51
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