Is there any characterization of the set of entire functions $f(z) $ such that $\Re(f(z)) \geq \Re(\overline{f(\bar{z})})$ for all $z\in \mathbb{C}^{+} $?
($\Re$ stands for the real part)
Edit: I was thinking in writing $f(z) $ in the form
$$f(z) =f(x+iy) =u(x, y) +iv(x, y) $$
and
$$\overline{f(\bar{z})}=\overline{f(x-iy)} =u(x, - y) - iv(x, - y) $$
so, it is like finding all such continuous functions $u$ for which $$u(x, y) \geq u(x, - y), \;\;\forall\; y>0, x\in\mathbb {R} $$
Your last inequality implies that $v(x,y):=u(x,y)-u(x,-y)\geq 0,\; y>0$, and similarly $v(x,y)\leq 0,\; y<0$. Since $v$ is harmonic, this easily implies that $v(x,y)=cy$ for some real constant $c$. Now $w(x,y):=u(x,y)-cy/2$ will satisfy $w(x,y)=w(x,-y)$ This implies $(\partial w/\partial y)(0,y)=0$ and by Cauchy-Riemann, the conjugate function $w^*$ to $w$ is constant on the real line. Since $(u+iu^*)(z)=w+iw^*(z)+ciz/2$, we obtain a characterization: these are exactly those entire functions whose imaginary part is of the form $a+bx$ on the real line, where $a$ and $b$ are constants, $b$ is real. (And your original inequality is in fact always equality). Or in other words, the general form of such $f$ is $f=g+a+biz$ where $g$ is real on the real line.
Remark. I used the fact that a harmonic function which is positive for $y>0$ and negative for $y<0$ must be $cy$. This is easy to prove, but is also a special case of a general theorem describing meromorphic functions with positive real part in the upper half-plane and negative real part in the lower half-plane, see for example Levin, Distribution of zeros of entire functions, Ch. VI, sect. 1, Them 1.
