1

I've tried to use $e^{f(z)}$ like the question here: $f$ is an entire function with ${\rm Im}\ f\geq 0$.

However, I'm stuck here. I've learned Liouville's theorem; it's sufficient to show that $f$ is bounded, and it's not difficult to show that the constant is $0$, but how do I show $f$ is bounded?

RobPratt
  • 45,619
Meep
  • 55
  • Probably this has been asked before but one simple idea is to integrate the inequality after taking logarithms on large circles and show the integral of RHS goes to infinity so $\log |f(z)|=- \infty$ for all $z$ hence $f$ identically zero - – Conrad Feb 25 '23 at 02:48
  • @Conrad For the inequality $$\log |f(z)|\leq -\log |{\rm Im}(z)|$$ RHS becomes $+\infty$ on the large circle near the real axis, and I'm wondering why the integral will tend to $-\infty$ – Meep Feb 25 '23 at 03:11
  • The integral of $\log |\cos \theta|$ is finite for $\theta$ going on a circle and you get something like $-\log R+c$ – Conrad Feb 25 '23 at 03:33
  • 1
    There are a few solutions on this site - one uses a trick, another the idea above (since $\log$ is integrable near $0$ the idea above is very natural imho as $\log |f|$ is subharmonic so you can apply the submean property - or Jensen formula which is a quantitative form of the submean property - and then use the bound in the hypothesis on large circles): see https://math.stackexchange.com/questions/377782/application-of-liouvilles-theorem – Conrad Feb 25 '23 at 04:08
  • @Conrad Thanks a lot, I've been trouble by this for hours. – Meep Feb 25 '23 at 04:25
  • happy to be of help – Conrad Feb 25 '23 at 04:53

0 Answers0