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Let $U \subset \mathbf C$ be an open subset of the complex plane and suppose we have a differential operator of order 1, $L: \mathcal C^{\infty}(U) \to C^{\infty}(U)$ such that $Lu = 0$ if and only if $u$ is holomorphic in $U$. Is it true that $L$ must be the Cauchy-Riemann operator $\frac{\partial}{\partial \bar z}$?

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I guess up to a non-zero multiplicative constant. Is the differential operator defined as $a(z)\frac{\partial}{\partial z}+b(z)\frac{\partial}{\partial \bar z}$ where $a$ and $b$ are $C^{\infty}$ functions? – Davide Giraudo May 14 '12 at 20:55
Assuming that $\partial/\partial \overline{z}$ is defined as $1/2(\partial_x + i \partial_y) = 1/2(\partial_x u - \partial _y v) + i/2(\partial_x v + \partial_y u)$, acting on $f = u + iv$, then, by the C.-R. equations, $f$ is holomorphic if and only if $\partial_{\overline{z}} f = 0$ (since $f$ is holomorphic if and only if it satisfies the C.-R. equations). – William May 15 '12 at 1:39
@William I know that $f$ is holomorphic iff $\partial _{\bar z} f = 0$. What I want to know is whether $\partial _{\bar z}$ is the only differential operator of first order that has the holomorphic functions as its kernel. – Lucas Kaufmann May 17 '12 at 13:23
@DavideGiraudo, could you expand a bit on this? – Lucas Kaufmann May 17 '12 at 13:24
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@Lucas: If by first order differential you mean something of the form $au_x + bu_y + cv_x + cv_y$, $a,b,c,d\in C^\infty(\mathbb{C})$, and $u, v\in C^\infty(\mathbb{C})$ such that $f = u + iv$, then the answer is of course no, since $u_x - v_y$ is a counterexample. If you meant in the sense of what Davide suggested, then the answer is yes up to multiplication by a smooth function: take $f(z) = z$ and apply $a(z)\partial_z + b(z)\partial_{\overline{z}}$ to it; you'll get $a(z) = 0$, hence the original operator is $b(z)\partial_{\overline{z}}$. If you meant something else, please clarify. – William May 18 '12 at 2:52
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