# The solution of Cauchy-Riemann equation

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
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

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
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

With respect to the new bounty:

Here's a way to see this. Suppose $u$ is a compactly supported solution to $\bar{\partial}u=f$. Then we have, for large enough $R>0$

$$0=\int_{\partial D(0,R)} u(z)dz = 2i\int_{D(0,R)} \bar{\partial}u(z)\: d\bar{z}\wedge dz=2i\int_{D(0,R)} f(z)\:d\bar{z}\wedge dz$$ Taking $R\to\infty$, this implies that if a compactly supported solution to this equation exists, then $$\int_{\mathbb{C}}f(z)\:d\bar{z}\wedge dz=0$$Then, consider the standard $\mathscr{C}^{\infty}_0(\mathbb{C})$ bump function $\varphi$ such that $\varphi\geq0, \text{supp}(\varphi)\subset \mathbb{D}$, and $$\int_{\mathbb{C}} \varphi(z)\: d\bar{z}\wedge dz=1$$ (the construction of which is in most graduate real analysis texts). Then there exists no compactly supported $u$ such that $\bar{\partial}u=\varphi$.

Edit: Here's the explanation of the comment. Suppose $\int_{\mathbb{C}}z^nf(z)dA=0$ for all $n\geq0$. Let $D$ be a large enough ball centered at $0$ such that $\text{supp}(f)\subset D$. Consider the solution

$$u(z)=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z-\omega}dA(\omega)$$If $|z|>|\omega|$ (i.e. $z\not\in \bar{D}$). Then \begin{align*}u(z)&=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z-\omega}dA(\omega)=\frac{1}{\pi}\int_{D}\frac{f(\omega)}{z(1-\frac{\omega}{z})}dA(\omega)=\frac{1}{\pi}\int_{D}f(\omega)\sum_{n=0}^{\infty}\omega^nz^{-n-1}dA(\omega)\\&=\frac{1}{\pi}\sum_{n=0}^{\infty}\left(\int_{D}f(\omega)\omega^ndA(\omega)\right)z^{-n-1}=\frac{1}{\pi}\sum_{n=0}^{\infty}\left(\int_{\mathbb{C}}f(\omega)\omega^ndA(\omega)\right)z^{-n-1}\\&=0\end{align*}So $u=0$ outside of $D$, so $\text{supp}(u)\subset D$ and thus $u$ is compactly supported.

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I think I'm missing something. How do I see that $\int z^n f(z) dA = 0$ implies compact support? – Alfred Yerger Apr 28 at 4:34
I think that's what was said in the comment I responded to. That there was u with compact support if and only if that held. In the bounty I specified I wanted to see that part if the answer expanded upon. – Alfred Yerger Apr 28 at 4:39
Sorry, missed that. Edited. – Moya Apr 28 at 4:49
This is exactly what I was missing. Thanks so much. I can't award the bounty until tomorrow, but I upvoted your solution, and I will award it in the near future. – Alfred Yerger Apr 28 at 6:21

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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If u is compactly-supported, why $\lim_{|\zeta| \to \infty}u(\zeta) = \infty$ ? – student May 23 '14 at 20:29