TITLE: Is it true or not : if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic.
QUESTION [5 upvotes]: Is it true or not :  if $u(z)$ is harmonic, $u(\overline{z})$ is also harmonic.
My try :
$u(z)=u(x,y)$ is harmonic
Define $s=-y$
Let $U := u(\overline{z})=u(x,-y)=u(x,s)$ : $$\frac{\partial U}{\partial x}=\frac{\partial u}{\partial x} \Rightarrow \frac{\partial^2 U}{\partial x^2}=\frac{\partial^2 u}{\partial x^2}$$
And $$\frac{\partial U}{\partial y}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial y} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial y}  = - \frac{\partial u}{\partial s}  $$
Similarly $$\frac{\partial^2 U}{\partial y^2}=   \left[ \frac{\partial }{\partial y} ( \frac{\partial U}{\partial y}) \right] =- \left[ \frac{\partial }{\partial y} ( \frac{\partial u}{\partial s}) \right] = ... = - \left[ - \frac{\partial }{\partial s} ( \frac{\partial u}{\partial s}) \right] = \frac{\partial^2 u}{\partial s^2}  $$ Hence
$$\frac{\partial^2 U}{\partial x^2} + \frac{\partial^2 U}{\partial y^2} = \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial s^2}= 0 \ \ \ (*)$$
Therefore $u(\overline{z})$ is also harmonic.
*My Questions is : Is my try problematic?Does $(*)$ needs justification? *

REPLY [6 votes]: Let $f(z)$ be a function which is analytic on the given domain and such that $\mbox{Re}(f)=u$. Then $\overline{f(\bar{z})}$ is Analytic and  $\mbox{Re}(\overline{f(\bar{z})})=u(\bar{z})$. Hence, $u(\bar{z})$ is harmonic.
The proof that $\overline{f(\bar{z})}$ is Analytic is simple: it follows immediately from the definition of the derivative that if $f$ is differentiable at $z_0$ then $\overline{f(\bar{z})}$ is differentiable at $\bar{z_0}$.