$u$ is harmonic iff $u(z)=f(z)+g(\bar z) $, for entire $f,g$ Prove: $u: \Bbb C \to \Bbb R$ is a harmonic function if and only if $u(z)=f(z)+g(\bar z),$ for entire functions $f,g$
For the first direction, assume $u(z) = f(z)+g(\bar z).$ Then we have: 
$$\Delta u= \Delta f+ \Delta g = 0$$
Is this correct? also, I'm pretty stuck with the second diretcion. Any suggestions?
Edit: $u $ is defined as follows: $u:\Bbb C \to \Bbb R$
 A: Hint:
We know the following 

Lemma $f: \Bbb C \to \Bbb R$ iff there exists $h: \Bbb C \to \Bbb C$ analytic such that $$f =Re(h).$$

Assume $u$ harmonic so are $Re(u)$ and $Im(u)$ since,  $$ u(z) = Re(u) +iIm(u) $$
Since   $Re(u), Im(u): \Bbb C \to \Bbb R$ form the Lemma above there exist $f,g:\Bbb C \to \Bbb C$ entire such that, 
$$  Re(u) = Re(f) =\frac{1}{2}(f(z)+\overline{f(z)}) =\frac{1}{2}(f(z)+\overline{f}(\bar z))$$
and $$Im(u) = Re(g) = \frac{1}{2}(g(z)+\overline{g(z)}) = \frac{1}{2}(g(z)+\overline{g}(\bar z))$$
Hence, $$ u(z) = Re(u) +iIm(u) = \color{red}{\frac{1}{2}(f(z)+ig(z))} + \color{blue}{\frac{1}{2}(\overline{f}(\bar z)+i\overline{g}(\bar z))}  $$

**Note **for an entire function  $h$ it is easy to check   by using the Taylor series that $\overline{h( z)} =\overline{h}(\bar z))$
  

A: Let $u$ be harmonic, then $\dfrac{\partial^2u}{\partial\bar{z}\partial z}=0$ shows $\dfrac{\partial u}{\partial z}$ is analytic. Let 
$\dfrac{\partial u}{\partial z}=f'$ for an analytic function $f$, if we let $g=\overline{u-f}$ so $u=f+\bar{g}$ and also we can conclude $g$ is analytic:
$$\dfrac{\partial g}{\partial\bar{z}}=\dfrac{\partial\overline{u-f}}{\partial\bar{z}}=\overline{\dfrac{\partial(u-f)}{\partial z}}=\overline{u_z-f'}=0$$
