# Show that f is a polynomial of degree at most N,

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.

Thanks,

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

Briefly: $M(2^k)\le 2^{kN}M(1)$ by induction. Fix $j>N.$ Then for any $k,$
$$|f^{(j)}(0)| \le j!\frac{M(2^k)}{(2^k)^j}\le \frac{2^{kN}M(1)}{2^{kj}} \to 0.$$
• It's actually easier: $f(z) = \sum_{n=0}^\infty [D^nf(0)/n!]z^n = \sum_{n=0}^N[D^nf(0)/n!]z^n.$