Corollary from Maximum Modulus Principle and Schwarz's Lemma

Question

Need to prove this implication derived from the maximum principle, but have no clue how.

$$\forall k=0,...,N. \ f^{(k)}(0)=0 \implies\exists M=const . \ \forall z.|z|\lt1:|f(z)|\le M|z|^{N+1}$$

Answer 1

As the question is stated, you don't need the maximum modulis principle.

The assumption that $f^{(k)}(0) = 0$ for $k = 0, \dots, N$ implies that $f(z)$ has a zero of order $N+1$ at the origin. Therefor, the function $$g(z) = \frac{f(z)}{z^{N+1}}$$ is analytic in the closed unit disc (assuming that $f$ is analytic in the closed unit disc that is). By compactness of the closed unit disc, there is a constant $M$ such that $|g(z)| \leq M$ in $|z|\leq 1$. That is, $|f(z)| \leq M |z|^{N+1}$ in $|z|\leq 1$.

If you use the maximum modulis principle, then you can deduce that $|f(z)| < M |z|^{N+1}$ in $|z|< 1$.

Notice that the assumption that $f$ is analytic in the closed unit disc is essential. The function $f(z) = \frac{z^{N+1}}{1-z}$ fulfilles your assumption, but tends to $\infty$ as $z\rightarrow 1^-$.