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Let $f$ be an entire function. Suppose that $f$ satisfies $$ |f(x+iy)|\leq\frac{1}{|y|}. $$ for all $x,y\in\mathbb{R}$. Prove that $f$ is identically zero.

I'm having some trouble with this, but I'm probably just overthinking it. The first instinct as usual is to try and make something happen with Liouville's theorem (and then since $f$ tends to zero on the imaginary axis we can conclude that $f\equiv 0$), but since $f$ is potentially unbounded on the real axis it's inapplicable. I'm pretty sure it's still impossible for $|f|\to\infty$ as $x\to\pm\infty$ while the function is bounded everywhere else, as it seems there would be some topological obstruction, but I could be wrong. I've also thought of getting an estimate on the $|f'|$ as in the proof of Liouville's theorem, but it fails for the same reasons. Any hints or tips are greatly appreciated. Thank you

Blake
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2 Answers2

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First approach: The first thing that comes to mind is Jensen's Formula.

For any point $\omega \in \mathbb{R}$ and $r > 0$, we have:

$$\log |f(\omega)| \le \frac{1}{2\pi}\int_0^{2\pi} \log |f(\omega + re^{i\theta})|\,d\theta$$

Since, $\displaystyle |f(\omega + re^{i\theta})| \le \frac{1}{|r\sin \theta|}$, we have:

$$\log |f(\omega)| \le -\frac{1}{2\pi}\int_0^{2\pi} \log |r\sin \theta|\,d\theta = - \log r - \frac{1}{2\pi}\int_0^{2\pi} \log |\sin \theta|\,d\theta$$

Since, the integral on the RHS is a constant,

$$|f(\omega)| \le \frac{K}{r}$$

Hence, $f(\omega)$ must be identically zero on $\mathbb{R}$ and hence on $\mathbb{C}$.

Second approach: Let us first fix a $r > 0$ and $z = re^{i\theta}$, then

$$|(z^2 - r^2)f(z)| = 2r|\Im(z).f(z)| \le 2r$$

Then, $g(\omega) = (\omega^2 - r^2)f(\omega)$ is holomorphic in the unit disc $D_r = \{\omega : |\omega| \le r \}$

Hence, by Maximum Modulus Principle:

$$|g(\omega)| \le 2r \textrm{ in the disc } D_r$$

and hence, $$|f(\omega)| < \frac{2r}{|\omega^2 -r^2|} \textrm{ in the interior of the disc }D_r$$

Letting $r \to \infty$, it follows that $f$ must be identically $0$.

r9m
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    The first proof looks good, the second proof doesn't look right. We don't know $2iyf(z)$ is holomorphic, so how can you apply Cauchy to it? Also $a_{n-1} - |z|^2a_{n+1}$ is not a coefficient of a power series. However, you should be able to integrate $|2iyf(z)|^2$ on circles of radius $r$ to get the result. – zhw. Jul 06 '15 at 07:16
  • @zhw. you are right, the second approach needs tinkering. I'll improve it later. Thanks! :-) – r9m Jul 06 '15 at 07:54
  • For the first proof, is there any reasonable way to show the integral is finite? Also is using Lindelöf's theorem a possible method of proving this? – Blake Jul 06 '15 at 09:15
  • @Blake $\displaystyle \int_0^{2\pi} \log |\sin x| ,dx = -2\pi\log2$ :-) and as for Lindeloff Thm .. I don't see what motivations should I have for using that (although I admit the working principle are quite similar) – r9m Jul 06 '15 at 09:20
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    @r9m +1 for the first development. Just curious .... although it does not change the result, you left of the sum $$\sum_{n=1}^N\log \left(\frac{|z_n|}{r}\right)$$ where ${z_n}$ is the set of zeros of $f$ from Jensen. Was this tacitly omitted inasmuch as it simply has no impact on the result? – Mark Viola Jul 06 '15 at 17:05
  • @Dr.MV I used the inequality form of Jensen's Formula, since $\log \frac{|z_n|}{r} < 0$ (as, $|z_n| < r$), we get a negative contribution from that part. – r9m Jul 06 '15 at 17:10
  • Apology for missing the inequality. Interestingly, one could have retained that component and expressed the equality in the first step, applied the bound to the integral afforded by the presumed bound on $f$ as $|f|\le\frac{1}{|\text{Im}(z)|}$ and reached the same conclusion. But, really very well done!!!! – Mark Viola Jul 06 '15 at 17:23
  • The new second proof seems incorrect. All you've shown is that $|f(z)| \le \frac{2|z|}{|z^2-|z|^2|}.$ The expression on the right can be $\infty,$ so how could this help? Also, you can't divorce $z$ from $r$ as you seem to do by writing $|z|\le r$ etc. – zhw. Jul 07 '15 at 02:48
  • @zhw. First we fix a $r > 0$. The function $g(\omega) = (\omega^2 - r^2)f(\omega)$ is holomorphic in the disc $D_r = {\omega | |\omega| \le r}$. Hence, I applied Max mod on $g$, and argued the max happens in the boundary of $D_r$, which in turn gives us $|g(\omega)| < \frac{2r}{|\omega^2 - r^2|}$ in the interior of the disc $D_r$. – r9m Jul 07 '15 at 03:34
  • Of course! I should have seen that. Nice proof, both of them. – zhw. Jul 07 '15 at 04:30
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Here's another approach: Note that $z-\bar z=2iy.$ Thus

$$\tag 1 |(z-\bar z)f(z)| \le 2$$

for all $z.$ Let's write $f(z) = \sum_{n=0}^{\infty}a_nz^n.$ Setting $z=re^{it},$ we get

$$\tag 2 (z-\bar z)f(z) = (re^{it} - re^{-it})(\sum_{n=0}^{\infty}a_nr^ne^{int}).$$

Play with $(2)$ a little to see it equals

$$\tag 3 \sum_{n=0}^{\infty}(a_nr^{n+1} - a_{n+2}r^{n+3})e^{i(n+1)t} -a_0re^{-it} - a_1r^2.$$

Let's now integrate the modulus squared of $(3)$ over $[0,2\pi]$using the orthogonality of the exponentials. We get, recalling $(1),$

$$\sum_{n=0}^{\infty}|a_nr^{n+1} - a_{n+2}r^{n+3}|^2 +|a_0r|^2 +|a_1r^2|^2 \le 4.$$

This inequality holds for all $r>0.$ Verify that this holds iff $a_n=0$ for all $n,$ i.e., iff $f\equiv 0.$

zhw.
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