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

Question

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 $D(0,1).$

Answer 1

Hint: use the Cauchy formula.

Answer 2

Here is a proof using the Schwarz lemma and maximum modulus principle. Let $D$ denote the open unit disc.

A minor point: Suppose $h$ is a non-constant holomorphic function on $D$ such that $|h(z)| \le 1$ for $z \in D$. Then $|h(z)| < 1$ for $z \in D$. To see this, suppose $|h(z_0)| =1$ for some $z_0 \in D$, then since $|h(z)| \le 1$, this would imply that $h$ is constant, a contradiction.

Since $f$ has a zero of order 3 at zero, we can write $f(z) = z^3 g(z)$, where $g$ is holomorphic on $D$.

On the following, I am assuming that $z \in D$:

We have $|z^3 g(z)| \le 1$, hence $|z^3 g(z)| < 1$, and so $|z^3 g(z) | \le |z|$, and so $|z^2 g(z)| \le 1$.

We have $|z^2 g(z)| \le 1$, hence $|z^2 g(z)| < 1$, and so $|z^2 g(z) | \le |z|$, and so $|z g(z)| \le 1$.

We have $|z g(z)| \le 1$, hence $|z g(z)| < 1$, and so $|z g(z) | \le |z|$, and so $|g(z)| \le 1$.

If $g(z) = \sum_k g_k z^k$, then we have $|g_0| \le 1$. Since $f(z) = \sum_k g_k z^{k+3}$, we see that $f^{(3)}(0) = 3! g_0$, and so we have $|f^{(3)}(0)| \le 6$.

If $|f^{(3)}(0)|= 6$, then we have $|g_0| = 1$, hence $|g(0)| =1$ and so $g$ is a constant. In that case, we have $f(z)=z^3 g_0$.