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Can you give me an example that there is a $f \in C_0^{\infty}(\mathbb C)$, such that the equation $\bar \partial u=f$ has no $C_0^{\infty}(\mathbb C)$ solution?

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Could you explain the notation please? I'm not familiar with it. Is $C^{\infty}_0(\mathbb{C})$ the space of $C^{\infty}$ function germs $(\mathbb{C},0) \to \mathbb{C}$? What about $\overline{\partial}u$ is that some derivative of the real part of $f$? – Fly by Night Sep 7 '12 at 21:34
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Maybe $C_0^\infty(\mathbb{C})$ means the $C^\infty$ function on $\mathbb{C}$ with compact support? – Paul Sep 7 '12 at 21:42
@Paul But all such functions are constant, unless the OP means smooth as a function on $\mathbb R^2$. – Alex Becker♦ Sep 7 '12 at 22:01
Do you mean "Cauchy-Riemann equations"? – Sasha Sep 7 '12 at 22:02
To clarify, $\bar \partial u$ means $\frac{\partial }{{\partial \bar z}}u = \frac{1}{2}(\frac{\partial }{{\partial x}} + i\frac{\partial }{{\partial y}})u$ and $C_0^{\infty}(\mathbb C)$ means a smooth function on $\mathbb R^2$ with compact set. – Hezudao Sep 8 '12 at 3:15

2 Answers

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If $f\in C^\infty_0(\mathbb{C})$, then $\displaystyle u:\zeta\mapsto \frac{1}{2i\pi}\int_{\mathbb{C}}\frac{f(z)}{z-\zeta}dz\wedge d\bar{z}$ is also in $C^\infty_0(\mathbb{C})$ and is a solution to $\bar{\partial}u=f$

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$u$ may not be compactly supported. – Hezudao Sep 8 '12 at 3:16
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Indeed, I wrote my answer too quickly. Such a solution exists iff $\displaystyle\int_\mathbb{C} z^n f(z)dz\wedge d\bar{z}=0$ for all $n\geq 0$. – girianshiido Sep 8 '12 at 8:17
up vote 0 down vote accepted

We take a $f\in C^\infty_0(\mathbb{C})$ such that supp$f$ is contained in the unit ball and $f \geq 0$ and $f=1$ on some smaller ball.

If we have a compactly-supported solution for the equation, it has to be $\displaystyle u:\zeta\mapsto \frac{1}{2i\pi}\int_{\mathbb{C}}\frac{f(z)}{z-\zeta}dz\wedge d\bar{z}$ since $\mathop {\lim }\limits_{\left| \zeta \right| \to \infty } u(\zeta ) = \infty $.

But if we examine the integral using polar coordinate more carefully, we will find Re$u(x,0)>0$ when $x$ is sufficiently large.

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