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Let $u:\mathbb{C} \rightarrow \mathbb{R}$.

Prove that $u$ is an harmonic function iff $u(z) = f(z) + g(\bar{z})$ where $f,g$ are holomorphic functions.

I'd be happy for a hint


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The "if" part follows from Cauchy-Riemann equations immediately. To show the "only if" part, by considering the Cauchy-Riemann equations, please try to find the conjugate harmonic function $v$ of $u$, which satisfies that $h=u+iv$ is holomorphic.

Remark: There are several ways to construct $v$(or $h$ equivalently) once $u$ is given. One way is to define $v$ as a line integral: $v(z)=\int_0^z(u_ydx-u_xdy)$, where the value of the integral is independent of the choice of path, because $u$ is harmonic on $\mathbb{C}$. An alternative way is based on the following observation. If $h=u+iv$ is holomorphic, then $h'=u_x+iv_x=u_x-iu_y$. Now given $u$ harmonic on $\mathbb{C}$, $u_x-iu_y$ satisfies Cauchy-Riemann equations, so it must be holomorphic. Then there is a unique $h:\mathbb{C}\to\mathbb{C}$, holomorphic, such that $h'=u_x-iu_y$ and $h(0)=u(0)$. Note that $u$ is the real part of $h$, so $f$ and $g$ can be constructed in terms of $h$.

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thank you. I think I understand the if part. about the only if: I think that $v$ must be $\int u_{x} dy - \int u_{y} dx $ and so the imaginary part of any $g$ must be $\int - u_{x} dy + \int u_{y} dx $ but I'm not sure how to procceed. – catch22 Nov 12 '12 at 9:51
@Roy: Yes, you are almost done, but there is a little error. You should write $v$ as $\int(u_xdy-u_ydx)$, because neither $\int u_xdy$ nor $\int u_ydx$ is well defined(they are dependent on the path), but $u$ is harmonic implies that $\int(u_xdy-u_ydx)$ is independent of path, and hence well defined. Then $f=\frac{1}{2}(u+iv)$ is holomorphic and $g(z)=\bar{f}(\bar{z})$ is also holomorphic. Moreover, $u(z)=f(z)+g(\bar{z})$ – 23rd Nov 12 '12 at 10:35
thanks again, but i'm not sure what $\int(u_xdy-u_ydx)$ even means... I guess they want us to solve this question in another way. – catch22 Nov 12 '12 at 13:23
@Roy: Are you familiar with line integral and Green's theorem? If not, I will give an alternative answer. – 23rd Nov 12 '12 at 13:27
No, I'm not. This will be highly appreciated. – catch22 Nov 12 '12 at 14:08

There is another explanation of why this should be true.

First consider the corresponding "real-valued" problem. If you have a smooth function of two variables, $U:\mathbb{R}^2 \to \mathbb{R}$, such that $\frac{\partial^2}{\partial x \partial y} U(x,y) = 0$, it is easy to show that $U(x,y) = F(x) + G(y)$ for some $F,G:\mathbb{R} \to \mathbb{R}$ by constructing such functions explicitly.

Now, since $\Delta = 4 \frac{\partial^2}{\partial z \partial \overline z} $, and $z$ and $\overline z$ are algebraically independent, you may try to do the same with a harmonic function. The only technical thing you need is that any harmonic function is real analytic, hence it has an analytic continuation to a neighborhood of $\mathbb{R}^2 \subset \mathbb{C}^2$. Now you should just express this analytic function $U$ of $x$ and $y$ in terms of $z=x+iy$ and $z^\prime=x-iy$, essentially treating $z$ and $z^\prime$ as independent (of course, in $\mathbb{R}^2$ we have $z^\prime = \overline{z}$, but this no longer holds in $\mathbb{C}^2$). Observe that $\frac{\partial^2}{\partial z \partial z^\prime} U = 0$, and try to do the same thing that you would do in the real case, which should yield $U(z,z^\prime) = F(z)+G(z^\prime)$ with $F$ and $G$ complex analytic. Then restrict them back to $\mathbb{R}^2$.

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