# Function $u(x,y) \in C^2$ that admits harmonic conjugate

Problem

1) Prove that if the real and imaginary part of a holomorphic function are of class $C^2$, then they are harmonic.

2)Deduce from 1) that if $u(x,y) \in C^2$ is a function that admits a harmonic conjugate, then $u$ is harmonic.

My attempt

I could prove part 1) using the Cauchy-Riemann equations. I am having problems with 2): I am not suppose to use that holomorphic functions are analytic (which also means $u,v \in C^{\infty}$).

I want to show that $u_{xx}+u_{yy}=0.$

By hypothesis, $u$ is the real part of a holomorphic function $f(x+iy)=u(x,y)+iv(x,y).$

Using the Cauchy-Riemann equations, I have $$u_{xx}=v_{xy} \space;\space u_{yy}=-v_{yx}$$

I don't know how one can deduce from here that $\Delta u=0$. I would appreciate some help.

• Just add the two equations you obtained and use Schwarz Theorem $v_{xy}=v_{yx}$. – Dario Jun 26 '14 at 15:48

The Laplace operator $\nabla^2$ or $\Delta$ is simply $$\nabla^2 = \Delta = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}$$ so we want to show that $\Delta u = u_{xx} + u_{yy} = 0$. From your identities shown, $u_{xx} = v_{xy}$ and $u_{yy} = -v_{yx}$, we have $$\Delta u = u_{xx} + u_{yy} = v_{xy} - v_{yx}.$$ Since $u\in C^2$, $v_{xy}$ and $-v_{yx}$ exist and are continuous. Schwarz's Theorem states that
If $f:\mathbb{R}^n\to\mathbb{R}$ has continuous second partial derivatives at any point in $\mathbb{R}^n$, then $$\frac{\partial^2 f}{\partial x \, \partial y} = \frac{\partial^2 f}{\partial y \, \partial x}.$$
Now the complex plane can be viewed as the two-dimensional Euclidean space $\mathbb{R}^2$. Therefore, $$\Delta u = u_{xx} + u_{yy} = v_{xy} - v_{yx} = v_{xy} - v_{xy} = 0.$$