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.

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

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 $\leq m$.

Now, I follow what the answer above 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 $

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    $\begingroup$ All derivatives of order higher than $m$ vanish in $0$, hence the Taylor series is a polynomial of degree $\leqslant m$. $\endgroup$ – Daniel Fischer Jul 7 '13 at 18:25
  • $\begingroup$ @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. $\endgroup$ – user64742 Jul 7 '13 at 23:42
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jul 7 '13 at 23:46

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 $m$.


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