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I try to solve this problem by using Liouville's theorem but I could not find any way to prove that $f$ is bounded.

Edit: I have noticed that my question seems like a duplicate of the question "An entire function which has a bounded real part is constant". Well, the major difference here is that while the other question works with the real part of the function, my question is dealing with the real part of z itself. So unless we can prove that Re($f(z)$) is equivalent to Re($z$) (which will never happen), these two question are completely different from each other.

Another edit (after being enlightened): Mine is a duplicate of this problem. An entire function whose real part is bounded must be constant.

José Carlos Santos
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Ngân Trần Triệu Thanh
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    What does it mean for a function not to receive any value from a set? –  Jul 19 '20 at 16:03
  • Do you mean $\text{Re}\big(f(z)\big)\leq 0$ for all $z\in\mathbb{C}$? If that is the case, consider $$g(z):=\dfrac{1}{1-f(z)}\text{ for all }z\in\mathbb{C},.$$ Show that $g$ is a bounded entire function. – Batominovski Jul 19 '20 at 16:04
  • I don't think my question is the same version of the question "An entire function whose real part is bounded", or at least, I haven't seen the similarity yet. I think it just simply states that an entire function that is not defined on the set contains z, which has positive real part, that function is a constant. Please correct me if I am wrong. – Ngân Trần Triệu Thanh Jul 19 '20 at 17:21
  • Then you should answer @Gae.S.'s question: What does it mean that “$f$ does not take any value from $z \in \mathbb{C}: \text{Re } z>0$”? I would understand it as $\operatorname{Re} f(z) \le 0$ for all $z \in \Bbb C$, and that is exactly what is used in https://math.stackexchange.com/a/229319/42969. If you mean something different then please clarify the question. – Martin R Jul 19 '20 at 17:37
  • Sorry for causing all the misunderstanding, and yes, this is a duplicate. I was completely ignoring the fact that $f$ is an entire function and so believing in my false interpretation. I will mark it as a duplicate right now. Thank you for enlighten me. – Ngân Trần Triệu Thanh Jul 19 '20 at 18:51

2 Answers2

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First note that unless it is constant, the image of $f$ is an open subset of the left half plane. Now find a bijective holomorphic function $\varphi$ from the left half plane to the unit disc. Now $\varphi\circ f$ is entire and bounded, and thus constant, and so is $f=\varphi^{-1}\circ\varphi\circ f$.

Vercassivelaunos
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Define $g(z)=e^{f(z)}$, this is an entire function which satisfies:

$|g(z)|=|e^{f(z)}|=e^{Re(f(z))}\leq e^0=1$

So $g$ is entire and bounded, hence constant by Liouville's theorem. Now we have to show that this implies $f$ is also constant.

Define a new function $h(z)=\frac{f(z)-f(0)}{2\pi i}$. This function is clearly continuous. We will show it is integer valued. Let $z\in\mathbb{C}$. We already know that $e^{f(z)}=e^{f(0)}$, hence $e^{f(z)-f(0)}=1$, and thus there is some $k\in\mathbb{Z}$ such that $f(z)-f(0)=2\pi ik$, which implies $h(z)=k\in\mathbb{Z}$.

So $h$ is continuous in $\mathbb{C}$ (which is a connected domain) and integer valued, hence constant. And this clearly implies that $f$ is constant.

Mark
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  • Could you please explain why Re($z$) $\leq 0$ implies Re($f(z)$) $\leq 0$, as it seems like you are using it in the second line? – Ngân Trần Triệu Thanh Jul 19 '20 at 17:26
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    I believe that the meaning of "$f$ does not receive any value from ${z: Re(z)>0}$" is that for each $z\in\mathbb{C}$ the number $f(z)$ does not belong to this set, i.e $Re(f(z))\leq 0$. I see that in the comments above you wrote that you think that the meaning is different: that $f$ is not defined on the set ${z: Re(z)>0}$. But it can't be true, because an entire function is by definition a function which is holomorphic in the whole complex plane, so in particular it must be defined everywhere. – Mark Jul 19 '20 at 17:33
  • Thank you so much. I completely ignore the fact that $f$ is an entire on $\mathbb{C}$. So, correct me if I am wrong (again...). So as I understand, in this problem, we can prove that unless f is constant, Re($f(z)$) must be equal or less than 0. Then with the later claim, we can simply prove $f$ is once again must be constant, by using the function $g=e^f$. If $g(z)=c, \forall z$ then $e^{f(z)}=c$, which implies $f(z)=\ln c, \forall z$. Is that right? Thank you. – Ngân Trần Triệu Thanh Jul 19 '20 at 18:41
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    We know that $Re(f(z))\leq 0$ for all $z\in\mathbb{C}$, this is the information given in the exercise. We use this fact to prove that the function $g=e^f$ is constant. We want to show this implies $f$ is also constant. However, we can't take logarithms here. There is no branch of logarithm in the whole complex plane, so there is no way to define a logarithm as a continuous function. So this is not the way to deduce $f$ is constant. A correct way is like I did in my answer, using the function $h$. – Mark Jul 19 '20 at 20:27
  • Thank you so much. – Ngân Trần Triệu Thanh Jul 20 '20 at 01:14