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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.

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.

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 complex conjugate $w^*$ 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 $c$ are real constants. (And your original inequality is in fact always equality). Or in other words, the general form of such $f$ is $f=g+biz$ where $g$ is real on the real line.

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.

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 complex conjugate $w^*$ 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 $c$ are real constants. (And your original inequality is in fact always equality).

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 complex conjugate $w^*$ 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 $c$ are real constants. (And your original inequality is in fact always equality). Or in other words, the general form of such $f$ is $f=g+biz$ where $g$ is real on the real line.