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This problem is taken off the 2008 problem set for Math 185 in the University of Chicago (PDF link via uchicago.edu) (problem 4 on page 3):

Suppose $f$ is analytic on a domain $\Omega \subset \mathbb{C}$ containing the unit square of the complex plane. The following are true;

$$f(z+1)-f(z)\geq0 \quad \forall z \, \, \text{such that} \, \, \Re(z)=0, \Im(z) \in [0,1]$$ $$f(z+i)-f(z)\geq0 \quad \forall z \, \, \text{such that} \, \, \Re(z)\in [0,1], \Im(z)=0$$

Prove that $f$ is constant.

My query:

What does an inequality in complex numbers mean here?

I took this to mean that if $f(x,y)=u(x,y)+iv(x,y)$ where $u,v$ are harmonic conjugates, then $$u(1,y)-u(0,y) \geq 0, \, \, y \in [0,1]$$ $$v(1,y)-v(0,y) \geq 0, \, \, y \in [0,1]$$ $$u(x,1)-u(x,0) \geq 0, \, \, x \in [0,1]$$ $$v(x,1)-v(x,0) \geq 0, \, \, x \in [0,1]$$

(I'm assuming that it means that the real and imaginary parts of complex number on the LHS must both be positive). Is that the correct interpretation and if so, is there any way for me to solve these functional inequalities in addition to the Cauchy Riemann equations to prove that $u=v=\text{const}$?

Blue
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Cognoscenti
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  • Instead of writing "$f(z+1)-f(z) \ge 0$" one could write something more pedantic like "$f(z+1)-f(z) = s + 0 i$ for some $s \ge 0$". I suspect, though, that the problem writer hoped or expected that common sense would lead one to the correct interpretation, in lieu of the more pedantic approach. – Lee Mosher Apr 16 '25 at 19:11

1 Answers1

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It is implicitly assumed that $f(z+1)-f(z)$ and $f(z+i)-f(z)$ are real. This is pretty standard when you see something like $z >0$ since inequalities are not defined for all complex numbers, as you state.

whpowell96
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