Why is $g(z)=f(z)-\overline{f(\overline{z})}$ analytic, if $f(z)$ is analytic? If a complex function $f(z)$ is analytic in some domain $D$, then why is the following function $g(z)$ also analytic in that same domain?
$$g(z)=f(z)-\overline{f(\;\overline{z}\;)}$$
My doubt arises from the fact that $\overline{z}$ is not analytic.

I've looked at this related question but it didn't help provide a clear answer.
 A: You're correct.  $\bar z$ is not analytic.  But $\overline{\bar z}=z$ is analytic.
Suppose $f(z)=u(x,y)+iv(x,y)$ is analytic for $z\in D$ and is symmetric about the $x$-axis.  Clearly, $u$ and $v$ satisfy the Cauchy-Riemann equations.
Then, we have
$$\overline {f(\bar z)}=u(x,-y)-iv(x,-y)=U(x,y)+iV(x,y)$$
Now show that $U$ and $V$ satisfy the Cauchy-Riemann equations.
A: $g(z)=f(z)-\overline{f(\;\overline{z}\;)}$
Then,
$\begin{equation}
\displaystyle{\lim_{h \to 0}}\frac{g(z+h)-g(z)}{h}
\end{equation}$
${=\displaystyle{{\lim_{h \to 0}}\frac{[{{f(z+h)-\overline{f(\;\overline{z+h}\;)}-f(z)+\overline{f(\;\overline{z}\;)}
}}] }{h}}}$
${{=\displaystyle{\lim_{h \to 0}}\frac{[{f(z+h)-f(z)]-[\overline{f(\;\overline{z+h}\;)}-
\overline{f(\;\overline{z}\;)}]
} }{h}}}$
${=\displaystyle{\lim_{h \to 0}}[{\frac{f(z+h)-f(z)}{h}}-{\frac{\overline{f(\;\overline{z+h}\;)}-\overline{f(\;\overline{z}\;)}}{h}}]}$
${=\displaystyle{
f'(z)-
\lim_{h \to 0}}[{\frac{\overline{f(\;\overline{z+h}\;)}-\overline{f(\;\overline{z}\;)}}{h}}]}$
${=\displaystyle{
f'(z)-
\lim_{h \to 0}}{\frac{\overline{[{f(\;\overline{z+h}\;)}-{f(\;\overline{z}\;)}]}}{h}}}$
${=\displaystyle{
f'(z)-
\lim_{h \to 0}}[{\frac{\overline{{f(\;\overline{z+h}\;)}-{f(\;\overline{z}\;)}}}{\overline {h}}}]}$
${=\displaystyle{
f'(z)-
\overline{{ [\lim_{\overline {h} \to 0}}{\frac{{{f(\;{\overline z+\overline h}\;)}-{f(\;\overline{z}\;)}}}{\overline {h}}}]}}}$
$= f'(z)- \overline{f'(\overline z)}$
Hence, $ g'(z)=f'(z)- \overline{f'(\overline z)}$
[ Note :

*

*$ f: z\to {\overline z}$ is continuous.


*It is enough to prove $\overline{f(\;\overline{z}\;)}$ is Holomorphic.]
