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Suppose that $f$ is entire and that for each $z$, either $|f(z)|≤1$ or $|f′(z)|≤1$. Prove that f is a polynomial of degree $\leq 1$.

Hint: Use a line integral to show that $|f(z)|≤A+|z|$ where$ A = \max\{1,|f(0)|\}$.

This is an exercise of complex analysis (author : newman, bak)

Actually, I found similar question but I don't understand answer.

Suppose that $ f $ is entire and that for each $ z $, either $ |f(z)| \leq 1 $ or $ |f^\prime (z) |\leq 1 $. Prove that $ f $ is a linear polynomial.

Is there someone explaining easily?

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jakeoung
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  • That's a pretty good answer over there. What is it that you don't you understand? – Julien Apr 19 '13 at 11:22
  • I don't follow any logic follow. E.g. I don't understand take line [0,z] and why suppose |f(z)|>1 – jakeoung Apr 19 '13 at 11:35
  • $[d,z]$ is the line (segment) joining $d$ and $z$. It can be parametrized by $\gamma:t\mapsto (1-t)d+tz$ over $[0,1]$. By definition, $\int_\gamma g(z)dz=\int_0^1g(\gamma(t))\gamma'(t)dt=(z-d)\int_0^1g((1-t)d+tz))dt$. – Julien Apr 19 '13 at 11:40
  • Thank you very much, I very appreciate your comment! But I lost my way again.. Is there any meaning about |f(w)| or only mathematical expression? I don't know why 3rd line is deducted. (Then |f′(z)|≤1 on [d,z].) It is very hard for me.. – jakeoung Apr 19 '13 at 12:48
  • $|g(w)|$ is the modulus of $g(w)$. There is a little problem in the definition of $d$. It should be the sup of $w$ such that $|f(w)|\leq 1$. Then for $w$ in $[d,z]$, except possibly at $d$, we have $|f(w)|>1$ hence $|f'(w)|\leq 1$. – Julien Apr 19 '13 at 12:54
  • @julien, I think it may be a good idea if you post that as an answer and, perhaps, giving a link to the other question's answer, as there seem to be quite a few unclear things for topy in that proof (also for me, btw). – DonAntonio Apr 19 '13 at 14:05
  • @DonAntonio Ok, I wrote something. Let me know if it is clear enough or if you have suggestions for improvement. – Julien Apr 19 '13 at 15:00
  • Bak & Newmans problems are often diabolical in that they appear simple at first, and then not so much later on... – Emily Apr 19 '13 at 17:54

1 Answers1

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Inequality: I'll repeat the argument given by muzzlator here, modulo a minor modification which, I think, is necessary. I'll try also to give a few more details. Of course, muzzlator deserves all the credit. I will also show how to conclude that $f$ is a polynomial of degree $\leq 1$ to really add something to his answer.

If $z\in\mathbb{C}$ and denote $[0,z]$ the line segment with endpoints $0$ and $z$, parametrized by $\{tz\;;\;t\in[0,1]\}$. Then the set $$ C:=\{t\in[0,1]\;;\;|f(tz)|\leq 1\} $$ is closed by continuity of $f$. It is also bounded, so $C$ is compact.

If $C$ is empty, set $t_0:=0$ so $|f(t_0z)|=|f(0)|>1$ hence $|f(t_0z)|=|f(0)|=\max\{|f(0)|,1\}=A$.

If not, we set $t_0:=\sup C=\max C$. Note that $|f(t_0z)|\leq 1\leq \max\{|f(0)|,1\}=A$.

If $t_0=1$, then $|f(z)|=|f(t_0z)|\leq 1 \leq \max\{|f(0)|,1\}=A$. And the inequality you seek follows.

So we have to prove the inequality when $0\leq t_0<1$, in which case $|f(t_0z)|\leq A=\max\{|f(0)|,1\}$. By definition of $t_0$, we have $|f(tz)|>1$ for every $t_0<t\leq 1$. By assumption, this forces $|f'(tz)|\leq 1$ on this interval. And by continuity of $f'$, this holds also at $t_0$. Hence $$ |f'(tz)|\leq 1\qquad\forall t_0\leq t\leq 1. $$ Now consider the funtion $g(t):=f(tz)$. It is $C^1$ on $\mathbb{R}$ with derivative $g'(t)=f'(tz)z$. Therefore, by the Fundamental Theorem of Calculus $$ f(z)=g(1)=g(t_0)+\int_{t_0}^1g'(t)dt=f(t_0z)+z\int_{t_0}^1f'(tz)dt $$ whence $$ |f(z)|\leq |f(t_0z)|+|z|\int_{t_0}^1|f'(tz)|dt\leq A+|z|\int_{t_0}^11dt=A+|z|(1-t_0)\leq A+|z|. $$

Line integral: the integral $\int_{t_0}^1g'(t)dt=\int_{t_0}^1f'(tz)zdt$ is the line integral of the analytic function $f'$ over the line segment $[t_0z,z]$. Up to orientation, this does not depend on the parametrization. Here we took the parametrization $t\mapsto tz$ over $[t_0,1]$, oriented from $t_0z$ to $z$. It is denoted $$ \int_{[t_0z,z]}f'(w)dw=\int_{t_0}^1f'(tz)zdt. $$ Be careful, the lhs is not the Lebesgue integral of $f'$ over the zero measure set $[t_0z,z]$ in $\mathbb{C}$, otherwise it would be $0$.

Conclusion: we can deduce that $f$ is a polynomial of degree $\leq 1$ from the estimate $|f(z)|\leq A+|z|$, knowing that $f$ is entire.

You can either adapt the proof of Liouville's theorem by proving that the coefficients of the Taylor series representation of $f$ at $0$ are zero for $n\geq 2$ from Cauchy's integral formula.

Or you can observe that the inequality entails that the entire function $$h(z):=\frac{f(z)-f(0)}{z}\quad \forall z\neq 0\qquad h(0):=f'(0)$$ is bounded, whence constant by Liouville.

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Julien
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  • +1 Very nice, yet there's still that $,|f(0)|>1,$ that you use in your 6th line. I also was confused by this, yet the argument by Muzzlator seems to be that "even if the module of the function is greater than one, it is enough to assume its derivative's module is less than one or equal to one". – DonAntonio Apr 19 '13 at 15:09
  • @DonAntonio Thank you. Not sure I understand you. If $C=\emptyset$, this means in particular that $|f(0)|>1$. And this could happen. – Julien Apr 19 '13 at 15:43
  • @DonAntonio Regarding your "even if ...: this is by assumption: either $|f(z)|\leq 1$ or $|f'(z)|\leq 1$, so $|f(z)|>1$ implies $|f'(z)|\leq 1$. Of course you saw that, so I'm not sure qhat you meant. – Julien Apr 19 '13 at 15:49
  • It's awesome. I understand it and now I am trying to understand it more deeply. Thank you very much!!!!!! – jakeoung Apr 20 '13 at 04:54