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Let $f$ be a holomorphic function such that $Im(f(z))$ is positive for all $z$. Prove that $f$ is constant.

Liouville's theorem states that if an entire function is bounded, then it must be constant. Let $f(z)=a(z)+ib(z)$ So I consider the function $e^{if(z)}=e^{ia-b}=e^{ia}e^{-b}$. Taking the absolute value $$|e^{if(z)}|=$$ $$|e^{ia}e^{-b}|=$$ $$|e^{ia}||e^{-b}|=$$ $$|e^{-b}|$$ $$\leq 1$$

From here one deduces $e^{if(z)}=z_0$, for $z_0 \in \mathbb C$.

I don't know how can I conclude from here that $f(z)$ must be constant. If all of these was taking place in $\mathbb R$, then $e^{f(x)}=k \implies f(x)=log(k)$. But in the complex case I have branches of logarithms and that confuses me a little bit.

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  • $\begingroup$ With branch cuts, you just have many options to define "$\text{log} z$". Once you choose one, you have a well-defined analytic function. $\endgroup$ – jbtimmer Jun 30 '14 at 6:41
  • $\begingroup$ For another approach, you could have found a Mobius transformation $\varphi$ that takes the upper half plane to the unit disc, and seen thus that $\varphi \circ f$ is constant, and thus that $f$ is constant. $\endgroup$ – user98602 Jun 30 '14 at 6:51
  • $\begingroup$ Here the function is given holomorphic therefore $e^{f(x)}$ will be differentiable.Can you conclude that $f(x)$ is constant?(there must be use of holomorphic function) $\endgroup$ – Siddhant Trivedi Jun 30 '14 at 7:04
  • $\begingroup$ Infact, if the function is given entire and it's imaginary or real part is bounded above or below than $f$ has to be constant. $\endgroup$ – Siddhant Trivedi Jun 30 '14 at 7:07
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Pick some $w_0$ so that $e^{i w_0} =z_0$.

Then for all $z \in \mathbb C$ you have $$e^{i f(z)}=e^{i w_0} \Rightarrow f(z)=w_0+2k_z \pi \,;\, k_z \in \mathbb Z\,.$$

Now use continuity of $f$ to deduce that $k_z$ is constant.

Here is a big hint for this: for each $k \in \mathbb Z$ you can easily prove that the set $$f^{-1}(w_0+2k \pi)$$ is open and closed.

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  • $\begingroup$ You can derive the relation $\exp(if(z))=z_0$; so $if^{\prime}(z)\exp(if(z))=0$, and we are done. $\endgroup$ – Kelenner Jun 30 '14 at 6:59
  • $\begingroup$ @Kelenner Is true that $f'(z)=0 \ \ \forall z$ implies $f(z)=0 \ \ \forall z$? $\endgroup$ – Siddhant Trivedi Jun 30 '14 at 7:10
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    $\begingroup$ No: $f$ entire and $f^{\prime}(z)=0$ is equivalent to $f$ is constant. $\endgroup$ – Kelenner Jun 30 '14 at 7:14
  • $\begingroup$ @N.S. I don't know how to show that $f^{-1}(w_0+2k\pi)$ is open and closed. If I could show that, then by connectedness of $\mathbb C$, $f^{-1}(w_0+2k\pi)$ is $\mathbb C$ or $\emptyset$. From here one can deduce $k$ is constant. Now, how can I prove that $f^{-1}(w_0+2k\pi)$ is clopen? $\endgroup$ – user100106 Jun 30 '14 at 15:33
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    $\begingroup$ @user100106 It is closed because it is the preimage of a single point. It is open, because if you pick an $\epsilon >0$ small enough, the ball of radius $\epsilon$ at $w_0+2k\pi$ doesnt contain any other point of this form. Therefore $f^{-1}(w_0+2k \pi)=f^{-1} (B_\epsilon w_0+2k \pi))$ is open as the preimage of an open set...... Note that Kelenner's solution is much simpler. $\endgroup$ – N. S. Jun 30 '14 at 16:20
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The Moebius transformation $$T:\quad w\mapsto{w-i\over w+i}$$ maps the upper half plane onto the unit disk $D$. Therefore the function $$g(z):=T\bigl(f(z)\bigr)$$ is entire and bounded. By Liouville's theorem $g$ has to be constant, and so is $f$.

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If $\Im(f(z)) \ge 0$, then $$ 1 = \Im(i) \le \Im(i)+\Im(f(z))=\Im(i+f(z))\le |\Im(i+f(z))|\le |i+f(z)|. $$ So $g(z)=1/(f(z)+i)$ is holomorphic because $i+f(z)$ is never $0$, and $$ 1\le |i+f(z)| \implies \frac{1}{|i+f(z)|} \le 1\implies |g(z)| \le 1. $$ That makes $g$ constant. So $g(z)=C$ for all $z$ and some $C\ne 0$. Solving for $f$ gives $$ f(z) = \frac{1}{C}-i. $$

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  • $\begingroup$ I don't understand why the inequality $|\Im (f(z)+i)| \ge \Im(f(z))+1$ holds. And also, you've proved $|\frac{1}{f(z)+i}| \leq 1$ but I don't see how it follows that from there that $f$ is bounded. Could you explain these things to me? $\endgroup$ – user100106 Jul 2 '14 at 20:45
  • $\begingroup$ @user100106 : I've rewritten the solution to be more verbose. $\endgroup$ – Disintegrating By Parts Jul 2 '14 at 21:01
  • $\begingroup$ Good answer!, thanks. $\endgroup$ – user100106 Jul 2 '14 at 21:48

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