Entire function bounded by a polynomial is a polynomial

Since $f$ is entire, it is equal to a power series centered at zero with radius of convergence $\infty$, which must match its Taylor series there.

$$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n$$

Since $|f(z)|\leq k|z|^m$, Cauchy's estimate gives

$$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$ for all $|z|=R$. For $n>m$, letting $R\rightarrow\infty$, we see that $|f^{(n)}|=0$. It follows that $f$ is a polynomial of degree $\leq m$.

Hints:

• We have by Cauchy's integral formula that $$|f^{(d)}(0)|=\frac{d!}{2\pi R}\left|\int_{C(0,R)}\frac{f(z)}{z^{d+1}}dz\right|.$$
• What about $f^{(d)}(0)$ if $d\geq n+1$?
• Use the fact that $f$ is analytic at $0$ to get that $f(z)=\sum_{j=0}^n\frac{f^{(j)}(0)}{j!}z^j$ in a neighborhood of $0$.
• Show that the last formula is in fact true for all $z\in\Bbb C$.