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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.

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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 –  Une Femme Douce 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. –  Une Femme Douce Jun 14 '12 at 8:58

1 Answer 1

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 $u$.

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Thank you julian, specially for the second one.I did not know about BCT. –  Une Femme Douce Jun 14 '12 at 12:55

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