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Let $f:\mathbb{C}\to \mathbb{C}$ be an entire function such that $Re(f)\le |p(z)|$ for some polynomial, can we derive that $f(z)$ is a polynomial.

If $p(z)$ is constant, then this can be shown by considering $e^f$. If we instead consider $|u(z)|\le |p(z)|$, then it can also be shown. But if we do not establish the lowerbound, then I cannot figure out how to generlize the proof.

Pax
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  • What is the absolute value of $\exp if(z)$? – Mariano Suárez-Álvarez Feb 09 '16 at 07:29
  • you didn't say on what set $u(z) \le p(z)$ it is not the same if it is on a line, on a disk, on the whole complex plane – reuns Feb 09 '16 at 07:33
  • @user1952009 I meant on the entire complex. The question has been edited. Thanks! – Pax Feb 09 '16 at 07:34
  • @MarianoSuárez-Alvarez |exp(if(z))|=exp(-Im(f))$? – Pax Feb 09 '16 at 07:35
  • if it is analytic on the whole complex plane it is entire $f(z) = \sum_{k=0}^\infty c_k z^k$ everywhere so it is easy to see that $Re(\sum_{k=0}^\infty c_k z^k)$ as $|\sum_{k=0}^\infty c_k z^k|$ cannot be bounded by a polynomial unless $c_k = 0$ for $k > D$ – reuns Feb 09 '16 at 07:37
  • @user1952009 But I'm only thinking about the Real part being bounded, so maybe I'm missing something simple, but how do we see $|f|$ being bounded? – Pax Feb 09 '16 at 07:39
  • for example by noticing that as Mariano said $e^{i f(z)-p(z)}$ is a everywhere $< 1$ entire function, or understanding the maximum modulus principle – reuns Feb 09 '16 at 07:42
  • but the most important point is that $\sum_{k=0}^\infty c_k z^k$ is a sort of degree $\infty$ polynomial : the more is the degree of the polynomial to approximate it, the more you can extend the radius on which the approximation is good. you can also count the zero : it has an infinite number of zeros or it is a polynomial multiplied by $e^{\sum_m a_m z^m}$ etc... – reuns Feb 09 '16 at 07:51
  • @user1952009 But if we take $f(z)=-iz$ and $p(z)=iz$, don't we get $e^{i(-iz)-z}=e^{z-iz}=e^ze^{iz}$ which has norm unbounded? I'm sorry, but I'm a bit confused... – Pax Feb 09 '16 at 07:51
  • @user1952009 but what if, $p(z)=iz$ and $f(z)=-z$, so that $e^{p(z)-f(z)}=e^{iz+z}=e^{z}e^{iz}$. Am I making a mistake, it has been a while since I have dealed with complex numbers... – Pax Feb 09 '16 at 07:57
  • Since you are referring to $u$ in $u+iv,$ maybe the title should refer to the real part of an entire function. – coffeemath Feb 09 '16 at 08:08
  • @coffeemath Yes! Thank you for catching that mistake! – Pax Feb 09 '16 at 08:10
  • Why do you focus on the real part ? If $f$ is an entire function, so is $if$, and that swaps the real and imaginary parts. –  Feb 09 '16 at 08:31
  • @Yves Daoust I don't care which, I just an trying to determine if an upper bound on just the real or imaginary part will force a bound on the entire function. – Pax Feb 09 '16 at 08:37
  • @Chanler: for the given reason the upper bound must be on the modulus. –  Feb 09 '16 at 08:52

1 Answers1

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The condition is equivalent to $\operatorname{Re} f\le K|z|^m$ for some $m$ and $K$.
Under this condition we can conclude that $f(z)$ is a polynomial of order less than or equal to $m$.

Let $f(z)=u+iv=\sum_{k=0}^\infty a_kz^k$ and $A(r)=\max _{|z|=r} u(z)$.
It is well-known that for $k\ge 1$ $$ a_kr^k=\frac{1}{\pi}\int_0^{2\pi} u(re^{i\theta })e^{-ik\theta }d\theta . $$ This leads $$ |a_k|r^k+2u(0)\le \frac{1}{\pi}\int_0^{2\pi} \left(|u(re^{i\theta })|+u(re^{i\theta })\right)d\theta \tag{1} $$ since $u(0)=\frac{1}{2\pi}\int_0^{2\pi} u(re^{i\theta })d\theta .$

If $A(r)\le 0$, then $|u|+u=0$ and we have $|a_k|r^k+2u(0)\le 0$ from $(1)$. If $A(r)>0$, then we have $$ |a_k|r^k+2u(0)\le 4A(r),$$ since $|u|+u\le 2A(r)$. In both cases we have $|a_k|r^k\le \max\{4A(r),\, 0\}-2u(0).$

Now suppose that $\operatorname{Re} f \le K|z|^m$. Then we have $$ |a_n|\le 4Kr^{m-n}-\frac{2u(0)}{r^n}\to 0 \quad (r\to \infty) $$ for $n>m$.

ts375_zk26
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  • How can we know that $$a_k r^k = \frac{1}{\pi} \int_0^{2\pi} u(re^{ix})e^{-ikx} dx$$? – Sato Apr 15 '21 at 17:16
  • @Sato We have $$a_k=\frac{1}{2\pi i}\int _{|\zeta|=r} \frac{f(\zeta)}{\zeta^{k+1}}d\zeta=\frac{1}{2\pi r^k}\int_0^{2\pi} f(re^{i\theta})e^{-ik\theta }d\theta .$$

    On the other hand we have $$0=\int_{|\zeta |=r} f(\zeta )\zeta^{k-1}d\zeta=ir^k\int_0^{2\pi}f(re^{i\theta })e^{ik\theta }d\theta .$$ So we have $$a_k=\frac{1}{2\pi r^k}\int_0^{2\pi} (f(re^{i\theta})e^{-ik\theta }+\overline{f(re^{i\theta})e^{ik\theta }})d\theta =\frac{1}{\pi r^k}\int_0^{2\pi} e^{-ik\theta }u(re^{i\theta })d\theta .$$

     – ts375_zk26 Apr 16 '21 at 05:07