Let $f(z)$ be an entire function such that $$|f(z)|<\frac{1}{|\text{Im}(z)|},\qquad z\in\Bbb C-\Bbb R.$$ The question asked me to prove that $f(z)=0$. At least looking at it, it really seems to have an application of Liouville's theorem lurking around somewhere, but I haven't found it.

My thoughts first led me to think about doing this by contradiction and using Picard's little theorem. So, I've considered a strip containing the real axis (say of width $2$ for simplicity). This will imply the complement maps into the unit disk, which implies that this strip has to map almost everywhere else on the complex plane. Now, on the imaginary axis, I know that $f(z)$ will vanish as $z\to\infty$, and this seems like it might be useful. I'm sort of under the impression $f(z)$ might have an essential singularity at $\infty$, which would mean $f(1/z)$ has an essential singularity at $0$. Either way, it of course can't have a pole at $\infty$ because of $f(z)$ vanishing on the imaginary axis, and if it is a removable singularity, it must be $0$, which still gives the solution. Therefore, I think it must have something to do with $f(z)$ having an essential singularity at $\infty$.

I was hoping to combine this with the above inequality and deduce a contradiction, but I haven't thought of one yet.

Any hints or suggestions? I'm studying for a prelim. exam, so this isn't homework.

Update: So, in the spirit of searching methods using the essential singularity at $z=0$, we have $$\left|f\left(\frac{1}{z}\right)\right|\leq\frac{1}{\left|\text{Im}\left(\frac{1}{z}\right)\right|}=\frac{|z|^2}{|y|}.$$ Thus, for $|y|>1$, we have $$\left|f\left(\frac{1}{z}\right)\right|<|y|\left|f\left(\frac{1}{z}\right)\right|<|z|^2.$$ Thus, if I can show that $|f(1/z)|$ is a polynomial, we'll know it can't have an essential singularity at the origin, concluding my proof.

  • $\begingroup$ Do you know Jensen's inequality, the one for holomorphic functions (not for convex ones)? It's a corollary of Jensen's formula if you drop the sum over zeros. $$\log |f(a)|\le \frac{1}{2\pi}\int_0^{2\pi} \log|f(a+re^{it})|\,dt$$ $\endgroup$ – 75064 May 1 '13 at 3:52
  • $\begingroup$ @user75064: Yes I know the inequality, but I am not quite sure how to apply it here. $\endgroup$ – Clayton May 1 '13 at 3:57
  • $\begingroup$ Fix a real number $a$ (arbitrary) and let $r\to \infty$. $\endgroup$ – 75064 May 1 '13 at 4:06
  • $\begingroup$ @user75064, no matter how large your circle is, there is always an arc with small imaginary part and there is no uniform bound on that arc. $\endgroup$ – user27126 May 1 '13 at 4:14
  • $\begingroup$ @Sanchez Correct, but I don't need a uniform bound. See my answer below. $\endgroup$ – 75064 May 1 '13 at 4:23

Let $w \in \mathbb{C}$ and let $R > |w|$. We first bound the analytic function $(z^2 - R^2)f(z)$ on the circle $|z| = R$.

If $|z| = R$ and $\operatorname{Re}z \ge 0$, then

$$|(z- R)f(z)| \le \frac{|z - R|}{\operatorname{Im}z} = \sec \theta$$

where $\theta \in [0, \dfrac{\pi}{4}]$ so $|(z - R)f(z)| \le \sqrt{2}$.

For $\operatorname{Re} z \le 0$ and $|z| = R$, we have

$$|(z + R)f(z)| \le \sqrt{2}$$


$$|(z^2 - R^2)f(z)| \le 3R$$

on $|z| = R$. By the maximum modulus principle the same bound holds in the interior of the disk $|z| \le R$ and so

$$|f(w)| \le \frac{3R}{|w^2 - R^2|}$$ Letting $R \to \infty$, we get the desired result.

  • $\begingroup$ Where does the $3R$ come from? $\endgroup$ – Clayton May 1 '13 at 4:38
  • $\begingroup$ @Clayton, I was just wondering that myself. A sharper bound is $2\sqrt{2} R$, since the point on $|z| = R$ farthest from $\pm R$ is $\mp R$ at a distance of $2R$. It looks like this was rounded up to $3R$. $\endgroup$ – Antonio Vargas May 1 '13 at 4:41
  • $\begingroup$ @Clayton $|z - R||z + R||f(z)| \le 2\sqrt{2}R$ $\endgroup$ – Ink May 1 '13 at 4:42
  • $\begingroup$ Okay, I was wondering about that. Last question: Why is $\theta\in[0,\pi/4]$? I think I have a vague idea, just not 100% sure. $\endgroup$ – Clayton May 1 '13 at 4:50
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    $\begingroup$ Nice answer! A little simplification of your argument: when $|z|=R$, $|z^2-R^2|=|z^2-z\bar{z}|=|z||z-\bar{z}|=2R|\mathrm{Im}(z)|$. @AntonioVargas, so $2R$ is a better bound. $\endgroup$ – Hu Zhengtang May 1 '13 at 5:57

Let $\sum_{n=0}^\infty a_n z^n$ be the Taylor expansion of $f(z)$. Also denote $a_{-1}=a_{-2}=0$. By definition, $$a_n=\frac{1}{2\pi i}\int_{|z|=R}\frac{f(z)}{z^{n+1}}dz,\quad\forall n\ge -2,\, \forall R>0.\tag{1}$$

Note that $\mathrm{Im} (z)=\frac{z-\bar{z}}{2i}$, so when $|z|=R$, $$\mathrm{Im} (z)=\frac{z-\frac{R^2}{z}}{2i}.\tag{2}$$ Substituting $(2)$ into $(1)$, we have $$a_{n-1}-R^2 a_{n+1}=\frac{1}{\pi}\int_{|z|=R}\frac{\mathrm{Im} (z)\cdot f(z)}{z^{n+1}}dz,\quad\forall n\ge -1,\, \forall R>0.\tag{3}$$ By the assumption of $f$ and continuity, $|\mathrm{Im} (z)\cdot f(z)|\le 1$ on $\mathbb{C}$. Then from $(3)$ we know $$|a_{n-1}-R^2a_{n+1}|\le \frac{2}{R^n},\quad\forall n\ge -1,\, \forall R>0.\tag{4}$$

Given $n\ge -1$, dividing both sides of $(4)$ by $R^2$ and letting $R\to \infty$, we have $a_{n+1}=0$. Therefore, all the coefficients of the Taylor expansion of $f$ are $0$, i.e. $f\equiv 0$.

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    $\begingroup$ Nice, but do you really need induction? Letting $R\to\infty$ in (4) immediately gives $a_{n+1}=0$ and $a_{n-1}=0$. $\endgroup$ – 75064 May 1 '13 at 4:16
  • $\begingroup$ @user75064: Yes, you are right. Thank you! $\endgroup$ – 23rd May 1 '13 at 4:19
  • $\begingroup$ @user75064: I have edited my answer based on your suggestion. Thank you again. $\endgroup$ – 23rd May 1 '13 at 4:24
  • $\begingroup$ I'm really not sure I follow the steps to $(3)$. Could you elaborate just a little? Thanks. $\endgroup$ – Clayton May 1 '13 at 4:26
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    $\begingroup$ @Landscape: I know your proof didn't use my ideas exactly, but I didn't see any reason for the bounty to go to waste. Your approach seemed most direct, so I liked it. I'm still thinking on how to solve it using my original thoughts. I went to a professor and his thoughts matched mine identically up to this point, at which he got stuck as well. $\endgroup$ – Clayton May 11 '13 at 17:52

"Me too!" Fix an arbitrary $a\in \mathbb R$. By Jensen's inequality for every $r>0$ we have $$\begin{split} 2\pi \log|f(a)| &\le \int_0^{2\pi} \log|f(a+re^{it})|\,dt \le \int_0^{2\pi} \log(|r\sin t|^{-1})\,dt \\ &= -2\pi\log r- \int_0^{2\pi} \log(|\sin t|)\,dt \end{split}$$ The integral on the right is finite, because $ \log |x|$ is integrable near $0$. Let $r\to \infty$ to conclude $f(a)=0$. Since $f$ vanishes on $\mathbb R$, it vanishes identically.

By the way, this works equally well for bounds like $|f(z)|\le 1/|\mathrm{Im}\,z|^{N}$ or even $|f(z)|\le \exp( 1/|\mathrm{Im}\,z|^{1-\epsilon})-1$. Fails with the assumption $|f(z)|\le \exp( 1/|\mathrm{Im}\,z|)-1$, probably for a good reason.

  • $\begingroup$ Probably the next thing I'll attempt will be showing it for the exponential bound. Thanks! $\endgroup$ – Clayton May 1 '13 at 4:34
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    $\begingroup$ This is nice, yet the "...because $\,\log|x|\,$ is integrable near zero" part is a little shaky, unless the OP knows this: $$\lim_{\epsilon\to 0^+}\int\limits_e^a\log x\,dx=\lim_{\epsilon\to 0}\left.\left(x\log x-x\right)\right|_\epsilon^a=\text{finite}$$ since $\,x\log x\xrightarrow[x\to 0^+]{}0\,$ $\endgroup$ – DonAntonio May 1 '13 at 11:55
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    $\begingroup$ @DonAntonio: Thanks, but it seems pretty obvious. I think most calculus II students should be able to show something like that. $\endgroup$ – Clayton May 1 '13 at 14:29
  • $\begingroup$ I think you're right, @Clayton . Thanks. $\endgroup$ – DonAntonio May 1 '13 at 18:16
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    $\begingroup$ Nice application of Jensen's. Two small remarks. The first integral should have $a + r e^{it}$ instead of $re^{it}$ and in the last expression it should be $-2\pi \log(r)$. $\endgroup$ – WimC May 1 '13 at 19:08

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