Let f(z) be an entire function, and let $M(R)=max_{∣∣z∣∣=R}∣∣f(z)∣∣$, for R>0. Suppose that $M(2R)≤2^NM(R)$ for all R>0 and for some positive integer N. Show that f(z) is a polynomial of degree not exceeding N.

I used the standard Cauchy estimate to bound the derivatives of f, getting

$$|f^{N+k}(0)|=|\frac{N+k!}{2\pi i}\int \frac {f(z)}{z^{N+k+1}}dz|$$ $$\le \frac {(N+k)!}{R^{N+k}} M(R)$$

for k $\ge$ 1.

I want to show that the right hand-side goes to zero, as R grows to infinity, however we need to know how M(R) is growing and whether we can bound it or perhaps get rid of its dependence on R.

Any hints or solutions are welcome.


  • $\begingroup$ Does bounding the derivatives depend on a certain $N$ ? $\endgroup$
    – Zophikel
    Commented Jan 8, 2018 at 19:07

1 Answer 1


Briefly: $M(2^k)\le 2^{kN}M(1)$ by induction. Fix $j>N.$ Then for any $k,$

$$|f^{(j)}(0)| \le j!\frac{M(2^k)}{(2^k)^j}\le \frac{2^{kN}M(1)}{2^{kj}} \to 0.$$

  • $\begingroup$ Hi @zhw., I finally got it now - I finally understood the log_2 argument in the solution. High school math that I have not used in awhile (or never understood deeply enough) is always a weakness of mine, when it shows up again in upper-level coursework, e.g., plane geometry stuff in complex analysis. Can I ask you a quick follow-up question? So, I am able to show all derivatives of order N+k, for k greater than or equal to 1, are 0 at zero. How do I now say that these derivatives are in fact identically zero on the whole complex plane -- so that f is a polynomial of degree at most N? $\endgroup$
    – User001
    Commented Sep 29, 2015 at 0:02
  • $\begingroup$ My attempt would be this: the derivative of order N+k is also entire, and Taylor expansion about the origin z=0 has an infinite radius of convergence, and the derivative function, being entire, agrees with its Taylor series everywhere on C. But the expansion at the origin z=0 ... has Taylor coefficients all equal to zero, and so the derivative must be zero everywhere on C - i.e., it is identically zero. What do you think? I don't want to just cite the "Identity Theorem" since I don't feel like I'd have shown it rigorously enough, say, for an exam. I'd love to hear your thoughts - thanks, $\endgroup$
    – User001
    Commented Sep 29, 2015 at 0:08
  • 1
    $\begingroup$ It's actually easier: $f(z) = \sum_{n=0}^\infty [D^nf(0)/n!]z^n = \sum_{n=0}^N[D^nf(0)/n!]z^n.$ $\endgroup$
    – zhw.
    Commented Sep 29, 2015 at 2:35
  • $\begingroup$ That's so cool, @zhw. Thanks so much and have a great night :-) $\endgroup$
    – User001
    Commented Sep 29, 2015 at 2:41

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