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Assume $|f(z)|\leq 1/|y|$ for all $z\in\mathbb{C}$. Here $f$ is entire and we express $z=x+iy$. Then is $f$ constant ?

user74489
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2 Answers2

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Let $w\in\mathbb R$ and $r > 0$ be arbitrary. Let $C$ be the circle about $w$ with radius $r$. We have $$ \frac 1 {2\pi i}\int_C\frac{f(z)}{z-w}\,dz = f(w)\quad\text{and}\quad\frac 1 {2\pi i}\int_C(z-w)f(z)\,dz = 0. $$ Hence, $$ -f(w) = \frac 1 {2\pi i}\int_C\frac{r^{-2}(z-w)^2 - 1}{z-w}f(z)\,dz = \frac 1 {r^2}\cdot\frac 1 {2\pi i}\int_C\frac{(z-w)^2 - r^2}{z-w}f(z)\,dz. $$ Now, since $\left|\frac{(z-w)^2 - r^2}{z-w}\right| = 2|\operatorname{Im}z|$ for $z\in C$, we conclude that $$ |f(w)|\,\le\,\frac 1 {2\pi r^2}\cdot 2\pi r\cdot 2 = \frac 2 r. $$ Letting $r\to\infty$ shows $f(w) = 0$.

BTW: Here is another answer: Showing Entire Function is Bounded

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Friedrich Philipp
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  • I have two question further. How to check $|\tfrac{(z-w)^2-r^2}{z-w}|=2|\text{Im} z|$, (for if we write $z-w=re^{i\theta}$. Then this equals to $r|e^{2i\theta}-1|$). If f(w) is shown to be 0, how about other values near the real axis ? Thank you. – user74489 Mar 14 '16 at 13:58
  • Put $\xi = z-w$. Then$$|\xi^2-r^2|^2 = |\xi|^4 - 2\operatorname{Re}(r^2\xi^2) + r^4 = 2r^4 - 2r^2\operatorname{Re}(\xi^2) = 2r^2\left(r^2 - (\operatorname{Re}\xi)^2 - (\operatorname{Im}\xi)^2\right).$$Now use that $r^2 = |\xi|^2 = (\operatorname{Re}\xi)^2 + (\operatorname{Im}\xi)^2$. Your last question should be answered by the idetity theorem: http://en.wikipedia.org/wiki/Identity_theorem – Friedrich Philipp Mar 14 '16 at 14:10
  • Thanks a lot, but this is a little typing error: $2r^2 (r^2 - (\text{Re }\xi)^2 + (\text{Im }\xi)^2)$. – user74489 Mar 14 '16 at 14:17
  • Yes, indeed. Otherwise it would be zero. ;) – Friedrich Philipp Mar 14 '16 at 14:20
  • +1: This is a beautiful solution but a bit "out of the hat", at least for me. If you have time and will it would be nice if you explained how you came up with this idea. Especially, how did you take into account the integral $\int_C (z-w)f(z), dz=0$? – Giuseppe Negro Mar 14 '16 at 16:55
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    @GiuseppeNegro Hi, thanks. Actually, this is based on experience. ;o) I know this trick from a paper on spectral theory for operators in inner product spaces. Its name is "Locally definite operators in indefinite inner product spaces." You will find the relevant passage on page 6 (410 in journal). And I think they have it from a Gohberg-Krein book. – Friedrich Philipp Mar 14 '16 at 17:59
  • Is this the paper you mention? – Giuseppe Negro Mar 14 '16 at 18:10
  • Yes, that is it. :) – Friedrich Philipp Mar 14 '16 at 19:13
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A function like this must be zero. Indeed, let $g(z) := f(1/z)$, $z\neq 0$. Then $|g(z)|\le\frac{|z|^2}{|y|}$, i.e., $|g(z)|\le\frac 1 {|\operatorname{Im} z|}$ for $z\in\mathbb D\setminus\{0\}$. From this, it follows that $z=0$ is (at most) a pole of $g$. From here, it should be easy for you to deduce that $f = 0$.

Friedrich Philipp
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  • I am perhaps overlooking something, but why does $|g(z)|\le\frac 1 {|\operatorname{Im} z|}$ imply that $z=0$ is not an essential singularity of $g$? – Martin R Mar 14 '16 at 11:01
  • Well, essentially (;-)) because otherwise terms like $z^{-2}$ would let it grow faster. But I am about to make this more rigorous. – Friedrich Philipp Mar 14 '16 at 11:06
  • I couldn't get this straight. However, I am pretty sure it holds because $g$ is holomorphic outside zero. I posted another answer which is more comprehensible. – Friedrich Philipp Mar 14 '16 at 12:03
  • What I wrote about $g$ is true ($0$ must be a pole). One can use a similar argument as Yiorgos S. Smyrlis in http://math.stackexchange.com/questions/1174742/showing-entire-function-is-bounded?rq=1 – Friedrich Philipp Mar 14 '16 at 18:02