Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In a question paper I got the following two questions.

  1. $u(z)=u(x,y)$ be a harmonic function in $\mathbb{C}$ satisfying $u(z)\le a|\ln|z||+b$ for some positive constants $a,b$ and for all complex numbers. Show that $u$ is constant.
  2. If for all complex numbers, $u(z)\le |z|^n$ for some $n\in \mathbb{N}$, then $u$ is a polynomial in $x,y$.

I am completely stuck with this one.

share|improve this question
Do you know the Greens function for $\mathbb{R}^2$? This might help with 1 (did not check the details, though). –  user20266 Jun 14 '12 at 5:39
Dear Sir, I dont know about greens function –  Bunuelian Trick Jun 14 '12 at 5:40
What results do you know about harmonic functions in general? –  Did Jun 14 '12 at 8:43
since $u$ is harmonic in $\mathbb{C}$ then there exist a harmonic conjugate $v$ such that $g(z)=u+iv$ is entire. –  Bunuelian Trick Jun 14 '12 at 8:58
add comment

1 Answer

up vote 2 down vote accepted

Let $f(z)$ be an entire function such that $\Re(f(z))=u(z)$.

First question. Let $g(z)=e^{f(z)}$. Then $$ 0<|g(z)|\le e^{a|\ln|z|\,|+b}=e^b|z|^a\qquad\forall z\in\mathbb{C}. $$ It follows that $g$ is a polynomial of degree $\le a$. Since it never vanishes, it must be a constant, and so must $f$ and $u$.

Second question We need to bound the modulus of a holomorphic function in terms of its real part. For this we use the Borel-Carathéodory theorem. From it it is easy to deduce that $$ |f(z)|\le C\,|z|^n $$ for some constant $C>0$. It follows that $f$ is a polynomial of degree $\le n$ and hence so is $u$.

share|improve this answer
Thank you julian, specially for the second one.I did not know about BCT. –  Bunuelian Trick Jun 14 '12 at 12:55
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.