Suppose $f(z)$ is holomorphic on $D(0,1)$ such that $|f(z)| \leq 1, \forall z \in D(0,1)$ and $f$ has zero of order $3$ at $z=0.$ 

Prove that $|f^{\prime\prime\prime}(0)| \leq 6$ and determine the forms of $f(z)$ if $|f^{\prime\prime\prime}(0)|=6. \ $ 

May you verify if my proof is correct? Thank you.

Let $\epsilon >0$ such that $D(0,1-\epsilon) \subset D(0,1). $ By Schwarz's lemma, $|f(z)|\leq |z|,$ so $\left|\dfrac{f(z)}{z^4}\right| \leq \left|\dfrac{1-\epsilon}{(1-\epsilon)^4}\right|, \forall z$ on $ C(0,1-\epsilon).$

By Cauchy Integral Formula, $|f^{\prime\prime\prime}(0)|=\begin{align}\Biggl|\dfrac{3}{\pi i}\int\_{C(0,1-\epsilon)} \end{align}\dfrac{f(z)}{z^4}dz \Biggl| \leq ...\leq 6,$ where $\epsilon \to 0$

Suppose $|f^{\prime\prime\prime}(0)|=6. \ $ Since $f^{\prime\prime}(0)=0,$ there exists $r<1$ such that $|f^{\prime\prime}(z)|<1, \forall z\in D(0,r).$ Let $h(z)=\dfrac{f''(z)}{3!}, \forall z\in D(0,r). \ $ Since $|f^{\prime\prime\prime}(0)|=6, $ so $ |h^{\prime}(0)|=1,$ and by Schwarz's Lemma, $h(z)=cz,$ for some constant $c, $ in $D(0,r). $ Hence, $ f(z)=cz^3+dz+f,$ for some constants $d,f. \ $ Since $f^{\prime}(0)=f^{\prime\prime}(0)=0, \ f(z)=cz^3$ in $D(0,r).$ By Identity theorem, $f(z)=cz^3$ in $D(0,1).$