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Let $a\in \mathbb{R}$ and $f, g$ two entire functions such that $\operatorname{Re}(f(z))\le a·\operatorname{Re}(g(z))$.

Prove that there exists a constant $c\in \mathbb{C}$ such that $f(z)=a·g(z)+c$

Is there a way to prove this that doesn't involve Liouville's theorem or any of its corollaries?

Bernard
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NFC
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1 Answers1

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Well, $f, g$ are entire and hence so is $f-ag$. For all $z\in\mathbb{C}$, it is given that $\operatorname{Re}(f-ag)\ge 0$. A known corollary of Liouville's theorem states that if the real part of an entire function is bounded, it is constant. Therefore $f-ag=c$ or $f(z)=ag(z)+c$ for all $z\in\mathbb{C}$.

Yuval Gat
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  • While this does seem like the fastest approach to solve the problem, we haven't done Liouville's heorem in class yet, so I hoped to find another way to solve it (if there is one) – NFC Mar 22 '19 at 21:01
  • Oh, I missed that, sorry. This seems weird to me, as the subject of dominance of entire functions is usually handled with Liouville's theorem. – Yuval Gat Mar 22 '19 at 21:02
  • First of all, $Re(f-ag)\leq 0$. The second point is that you show it is bounded below. How do you know that it is bounded above? – Ri-Li Mar 06 '21 at 21:26