Derivatives of holomorphic function Let $f:U\rightarrow \mathbb{C}$ be a holomorphic function on a open and bounded subset $U$ of $\mathbb{C}$. Then suppose $|\frac{df}{dz}|&lt;C$ for a certain constant $C&gt;0$. Does this condition imply that all the norms of the other derivatives of $f$, $(|\frac{d^2f}{dz^2}|,|\frac{d^3f}{dz^3}|,\cdots$) are bounded?

I think the answer is yes: around $z_0\in U$ the derivative of the function is $\frac{df}{dz}(z)=a_1+2a_2(z-z_0)+3a_2(z-z_0)^2+\dots$ and if this sum is bounded in every point then I think it's safe to say that also the sum $\frac{d^2f}{dz^2}(z)=2a_2+6a_2(z-z_0)+\dots$ is bounded.. But this reasoning is very rough, so I was hoping someone could help me with a better argument
The claim is not true. A counter example: look at $f(z) = z^{3/2} + z$ over an open set $U = B_{2/3}(0) - (-\infty, 0] \subset \mathbb{C}$, where $B_{2/3} = \{z \in \mathbb{C}||z| &lt; 2/3\}$. Then $f$ is holomorphic on $U$ with $df/dz = (3/2)z^{1/2} + 1/2$ therefore $1/2 &lt; |df/dz|\leq 3/2|z| + 1/2 &lt; 3/2(2/3) + 1 = 2$. But it is clear that $d^2f/dz^2 = 3/4z^{-1/2}$ would be unbounded as $z \rightarrow 0$ in $U$.
No, take for $U=\{|z|&lt; 1\}$ the unit disc and let $f'(z)=\sum_{n\geq 1} \frac{1}{n^2} z^n$ so that $|f'(z)|&lt;2$ uniformly on $U$. Then  $f''(z)=\sum_{n\geq 1}\frac{1}{n}z^{n-1}\rightarrow \infty$ as $z\rightarrow 1^-$. However, if you restrict yourself to $|x|\leq r&lt;1$ then you may write down bounds for every higher order derivative on this smaller disc e.g. by using Cauchy's formula for the derivatives.