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Problem of the Week #279 - Oct 02, 2018

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Show that if $f$ is an entire function with $\lim\limits_{z\to \infty} \dfrac{\operatorname{Re}f(z)}{z} = 0$, then $f$ is constant.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Congratulations to Janssens for his correct solution, which is posted below.


Instead of the Cauchy integral formula (used for the Liouville Theorem), we invoke (a corollary of) the Poisson integral formula for harmonic functions (Bak and Newman, Complex Analysis, 2010, Theorem 16.9). Namely, writing $f = u + i v$ with $u := \Re{f}$ and $v := \Im{f}$, we have
$$
f(z) = \frac{1}{2\pi}\int_0^{2\pi} u(R e^{i\theta})\frac{R e^{i\theta} + z}{R e^{i\theta} - z}\,d\theta + i v(0),
$$
for all $z \in \mathbb{C}$ and $R > |z|$. This is useful, because it enables reconstruction of $f$ from its real part. So, for any fixed $z \in \mathbb{C}$ and arbitrary $R > |z|$,
\begin{align*}
|f(z) - f(0)| &\le \frac{1}{2\pi}\int_0^{2\pi} |u(R e^{i\theta})| \frac{2|z|}{|R e^{i\theta} - z|}\,d\theta\\
&= \frac{1}{\pi} \int_0^{2\pi} \frac{|u(R e^{i\theta})|}{|R e^{i\theta}|} \frac{|z|}{|1 - \frac{z}{R}e^{-i\theta}|}\,d\theta.
\end{align*}
By the reverse triangle inequality,
$$
\left|1 - \frac{z}{R}e^{-i\theta}\right| \ge \left|1 - \frac{|z|}{R}\right|,
$$
so
$$
|f(z) - f(0)| \le \frac{|z|}{\pi|1 - \frac{|z|}{R}|}\int_0^{2\pi} \frac{|u(R e^{i\theta})|}{|R e^{i\theta}|} \, d\theta.
$$
Now, by the definition of the complex limit, the remaining integrand tends to zero as $R \to \infty$, uniformly for $\theta \in [0,2\pi]$. Hence the integral tends to zero as well. Also, the factor in front of the integral remains bounded, so this gives the result.
 
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