# Given $g:\mathbb{D}\rightarrow\mathbb{D}$ is $g(z)=z^2h(z)$ and $|g|_{\infty}\leq 1$, why is $|h(z)|_{\infty}\leq 1$

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

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

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

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.
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 $$|z| <1$$.