Sol. verfication: Prove that if $f$ is entire and bounded on the open disk $|z|$f$ is entire. 
Let $D= \{z \in \mathbb C| |z|<R \}$, $f(0)=f'(0)=0$ and there is a real $K$ so that $|f(z)|\leq K$ for every $z \in D$. Show that for every $z \in D$, $|f(z)| \leq \frac{K|z|^2}{R^2} $
My attempt to solve: Define $g(z)= \frac {f(z)}{z^2}$, and $g(0)=0$. So $g$ is entire. So let there be a circle $C_r$ with radius $r \lt R$, centered at $z=0$. So by Cauchy's Integral Formula we get:
$|g(z)| \leq \frac {1}{2\pi} \int_{C_r}| \frac {g(z)}{z}||dz| \leq \frac {1}{2\pi} \cdot 2 \pi r \cdot \max _{z \in C_r}| \frac {1}{z}|\cdot \max _{z \in C_r}|\frac{f(z)}{z^2}|=r \cdot \frac {K}{r^3}=\frac {K}{r^2}$
So now set $r \rightarrow R$, and we get $|g(z)| \leq \frac {K}{R^2}$. Multiply both side with $|z|^2$ and we get: $|g(z)||z|^2=\frac {K|z|^2 }{R^2} \rightarrow \frac {|f(z)|}{|z|^2}\cdot|z|^2=\frac {K|z|^2 }{R^2}$ and then we get the desired result: $|f(z)| \leq \frac{K|z|^2}{R^2} $
Is this solution correct?
 A: I don't see how your definition of $g$ creates even a holomorphic function on $D$. It is true that $f/z^2$ will have a removable singularity on $D$, but it is not necessarily true that the value of the extended function at $z=0$ will be $0$. For example, $z^2$ is entire and $0^2=2\cdot 0=0$ however $z^2/z^2=1$ does not equal $0$ anywhere, as it is constant (I might be misunderstanding something here however). Also I don't see why $f$ needs to be entire, wouldn't just a function holomorphic on $D$ be sufficient? There might be some inconsistencies with our terminology, but I can give you a modest attempt simply assuming $f$ is a holomorphic function on $D$.
Here's my attempt:
Notice that $g(z):=f(Rz)/K$ maps $\mathbb{D}$ into $\mathbb{D}$ ($\mathbb{D}$ being the open unit disk) and $g(0)=0$, so Schwarz lemma says that
$$
|g(z)|=\frac{|f(Rz)|}{K}\leq |z|
$$
for $z\in\mathbb{D}$. It follows that (by replacing "$z$" with "$z/R$") 
$$
|f(z)|\leq \frac{K|z|}{R}.
$$
for $z\in D$, where $D$ is your set $\{z\in\mathbb{C}\,|\,|z|< R\}$. Using this information, we deduce that $h(z):=\frac{Rf(Rz)}{Kz}$ is a holomorphic (because $f(0)=0$) function mapping $\mathbb{D}$ into $\mathbb{D}$ where $h(0)=0$ (because $f'(0)=0$). Reapplying Schwarz lemma gives
$$
|h(z)|=\frac{R|f(Rz)|}{K|z|}\leq |z|
$$
for $z\in\mathbb{D}$ and (as before)
$$
|f(z)|\leq\frac{K|z|^2}{R^2}.
$$
Let me know if any of these steps were unclear. I sincerely tried to follow your attempt more closely but I feel as though are terminology many be different or something of the sort. By entire do you mean holomorphic? I don't believe that we even need $f$ to be defined outside of $D$.
