# Show that an entire function is a polynomial

There is a question in the book that asks me to show that if f is an entire function such that $$|f(z)| \le L|z|^m$$ where $$|z| \ge R$$, then $$f$$ is a polynomial of degree of at most $$m$$.

The problem gives me a hint that I should use the Cauchy estimates for n>m and $$r \to \infty$$

The below is from a post https://math.stackexchange.com/a/143881/64742

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 us

$$|f^{(n)}(0)|\leq \frac{n!k|z|^m}{R^n}$$ 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 $$\leq m$$.

Now, I follow what the above answer says except where "It follows that f is a polynomial of degree $$\le m$$. Why do we arrive at that conclusion? The preceding statement merely says that $$|f^{(n)}|=0$$

• All derivatives of order higher than $m$ vanish in $0$, hence the Taylor series is a polynomial of degree $\leqslant m$. Commented Jul 7, 2013 at 18:25
• @DanielFisher I can see that since derivatives of order higher than m vanish to 0, if f is a polynomial, the degree is no greater than m. However, how do I know f is a polynomial and not another function? I am sorry if that is a dumb question but I am just having such difficult time with complex analysis. Commented Jul 7, 2013 at 23:42
• Ah, perhaps you don't know that a holomorphic function is the sum of its Taylor series? So by the vanishing of the higher derivatives, we know the Taylor series in $0$ is $f(z) = \sum_{k = 0}^m c_kz^k$. But the Taylor series converges to the function, so $f$ is a polynomial, because its Taylor series is. Commented Jul 7, 2013 at 23:46
• Does this answer your question? Show that if $|f(z)| \leq M |z|^n$ then $f$ is a polynomial max degree n Commented Dec 10, 2021 at 12:30
• Does this answer your question? Entire function bounded by a polynomial is a polynomial Commented Feb 4, 2022 at 21:21

Since $f$ is entire, for every $z\in\mathbb C$ it holds that $$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^n \tag1$$ Once we know that $f^{(n)}(0)=0$ for $n>m$, the formula (1) simplifies to $$f(z)=\sum_{n=0}^m \frac{f^{(n)}(0)}{n!}z^n \tag2$$ where the expression on the right is evidently a polynomial of degree at most $m$.