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Given analytic function $f(z)$ on $\mathbb{H}:=\{x>0\}$ satisfying $$0\leq \Re{f(z)}\leq M\Re{z}$$ for some $M>0$ and $z \in \mathbb{H}$ I want to show that $f$ takes form $$f(z)=mz+ic$$ where $m\in[0,M],c\in\mathbb{R}$.

[Observation] If $f$ can be extended to $\partial{\mathbb{H}}$, then the condition implies that $f$ must takes purely imaginary number on $\{x=0\}$. By proper rotation we can extend $f$ to the whole plane by Reflection Principle and thus the entire function $e^{f(z)}$ have at most growth order of 1, whence by Hadamard's factorization theorem with some detailed argument we get the conclusion. Here's the only obstacle that left within the argument:

Can $f(z)$ be continuously extended to the boundary $\{x=0\}$ from the assumptions?

Roy Han
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2 Answers2

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Yes, $f$ can be extended to the entire plane by reflection.

Consider $u = \Re f$. The function

$$U(z) = \begin{cases}0 &, \Re z = 0\\ u(z) &, \Re z > 0\\ -u(-\overline{z}) &, \Re z < 0\end{cases}$$

is clearly continuous (by the bound on $\Re f$) and has the mean value property (for $\Re z \neq 0$, because it is known to be harmonic, for $\Re z = 0$ by the symmetry), hence it is an entire harmonic function.

$\mathbb{C}$ is simply connected, hence $U$ has a conjugate harmonic $V$, i.e. a real-valued entire harmonic function such that $F = U + iV$ is holomorphic. $\Re F = \Re f$ in the right half plane, so $f$ and $F$ differ by a purely imaginary constant there, which can be absorbed into $V$ to get the entire extension of $f$.

Daniel Fischer
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  • question: can you explain why f and F differ by a purely imaginary constant on x > 0? I see that they differ by a purely imaginary function, but why is it a constant? Thanks. – Betty Mock Aug 30 '13 at 22:57
  • Because a non-constant holomorphic function has open range. Alternatively, you can get it from the Cauchy-Riemann equations, if the real part is constant, its partial derivatives are zero, then by CR the partial derivatives of the imaginary part are zero, hence the imaginary part is constant. – Daniel Fischer Aug 30 '13 at 23:00
  • Very illuminating by consider harmonic extension! I was thinking about to define the function pointwise but stuck with how to deal with the continuity of the imaginary part! Thank you very much! – Roy Han Aug 31 '13 at 03:16
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I've got an attempted solution that goes in a different direction. I can't think of an easy way to show that $f(z)$ can be extended to the boundary. Instead, I'll use Runge's Theorem.

Let $K_n$ be the compact set $[1/n, n]\times[-in, in]$. Then $f(z)$ is analytic in a neighborhood of $K_n$, so there exists a rational function $p_n(z)$ with a pole in each component of $\mathbb{C}\backslash K_n$ such that: $$\sup_{z\in K_n}|f(z)-p_n(z)|<n^{-1}$$ Furthermore, we can specify the location of each pole. Since $\mathbb{C}\backslash K_n$ has only one component, we can make $p_n(z)$ to have a pole only at $\infty$, so that $p_n(z)$ is a polynomial. On $K_n$, $\Re p_n(z)$ is bounded by $M\Re z+n^{-1}$. This tells us that $p^{(k)}_n(z)\to 0$ for $k\ge 2$, but $p_n^{(k)}(z)\to f^{(k)}(z)$, so $f$ is linear and the result follows. (This last argument is not rigorous at all; I'd appreciate some help in filling it out).

Connor
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