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I have a question regarding the proof for Theorem 1.2 in the following paper on page 8:

Link to Paper: https://pdfs.semanticscholar.org/a65d/6a11c4c9b810e3c467dce7a802e51aea6ba9.pdf?_ga=2.38893984.2019758983.1565736017-1780512687.1565553927



Preliminaries: Some Definitions and Results

Let $\mathbb{D}$ denote the open unit disk $\{z \in \mathbb{C} : |z| < 1\}$. Let $H^{\infty}_1(\mathbb{D})$ denote the set of all bounded, analytic functions defined on the unit disk whose first derivative vanishes at zero.

That is, if $\phi \in H^{\infty}_1(\mathbb{D})$, then $\phi$ is a bounded, analytic function on the unit disk such that $\phi^{\prime}(0) = 0$.

Although not in the proof, it is shown in another paper that the elements of $H^{\infty}_1(\mathbb{D})$ are power series where the first power is missing in the expansion. Thus if $\phi \in H^{\infty}_1(\mathbb{D})$, we have

$$\phi(z) = c_0 + c_2 z^2 + c_3 z^3 + c_4 z^4 + \cdots.$$



Question About Proof

In the very first paragraph in Theorem 1.2, the author shows that the function $g:\mathbb{D}\rightarrow\mathbb{D}$ satisfies $||g||_{\infty}\leq 1$ and belongs to $ H^{\infty}_1(\mathbb{D})$. Further, he writes \begin{align} g(z) &= z^2 (c_2 + c_3z + c_4 z^2 + c_5 z^3 \cdots )\\ &= z^2 h(z), \end{align} and then claims that $||h||_{\infty} \leq 1$.

My question is how he deduced that $||h||_{\infty} \leq 1$. He just says so, and I have no idea why. Any help regarding this question would be most appreciated.

If I haven't provided enough information or something about what I have written is unclear, please let me know.

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Aply MMP to $\{|z| \leq 1-\epsilon\}$. Since $|h(z)| \leq \frac 1 {(1-\epsilon)^{2}}$ when $|z|=1-\epsilon$ we get $|h(z)| \leq \frac 1 {(1-\epsilon)^{2}}$ when $|z|\leq 1-\epsilon$. For any fixed $z \in \mathbb D$ we get $|h(z)| \leq \frac 1 {(1-\epsilon)^{2}}$ for $0<\epsilon <1-|z|$. Since $\epsilon$ is arbitrary this gives $|h(z)| \leq 1$ for $|z| <1$.

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