Estimating derivative of difference quotient Suppose that $f: \mathbb D \to \mathbb C$ is a holomorphic function on the unit disk with $f(0)=0$, let $K \subset \mathbb D$ be compact and convex, and let $\Vert\cdot\Vert_\infty$ be the sup-norm over $K$.
It is not difficult to show that
$$ \left\Vert \frac{f(z)}{z} \right\Vert_\infty \leq \Vert f' \Vert_\infty, \qquad\qquad (*) $$
e.g. as demonstrated here.
I'm wondering if one can show an inequality of the form
$$ \left\Vert \frac{d^k}{dz^k} \frac{f(z)}{z} \right\Vert_\infty \leq c_k \Vert f^{(k+1)} \Vert_\infty. $$
My first intuition is to expand $f$ into its power series so
$$ \frac{f(z)}{z} = \sum_{j=0}^\infty \frac{f^{(j+1)}(0)}{(j+1)!} z^j $$
and then if one differentiates $k$ times,
$$ \frac{d^k}{dz^k} \frac{f(z)}{z} = \sum_{j=k}^\infty \frac{f^{(j+1)}(0)}{(j+1)!} k!\,z^{j-k}  = \frac{f^{(k+1)}(0)}{k+1} + k!\sum_{j=1}^\infty \frac{f^{(j+k+1)}(0)}{(j+k+1)!} z^{j}. $$
So I would guess that the inequality should hold with $c_k=1/(k+1)$ but I'm not really sure how to estimate the final sum in terms of the $(k+1)$-st derivative of $f$.
Alternatively, I thought one could calculate the derivative directly, so e.g. for $k=1$
$$ \frac{d}{dz} \frac{f(z)}{z} = \frac{zf'(z)-f(z)}{z^2},  $$
and then try to iterate $(*)$, but this seems to be lead to an endless loop.
Any brighter ideas?
 A: We must assume (see below) that $0 \in K$, so that $f'$ can be integrated along the straight line connecting $0$ and $z$ in $K$, as in the referenced Q&A:
$$
 f(z) = f(0) + \int_{[0, z]} f'(w) \, dw = z \int_0^1 f'(tz) \, dt \, .
$$
Then
$$
\frac{d^k}{dz^k} \frac{f(z)}{z} = \int_0^1 t^k f^{(k+1)}(tz) \, dt 
$$
and it follows that
$$
\left\Vert \frac{d^k}{dz^k} \frac{f(z)}{z} \right\Vert_\infty  \le
 \int_0^1 t^k \Vert f^{(k+1)} \Vert_\infty\, dt
= \frac{1}{k+1} \Vert f^{(k+1)} \Vert_\infty  \, ,
$$
confirming your conjecture that the inequality holds with $c_k = 1/(k+1)$.
Remark: The example $f(z) = z^{k+1}$ shows that this factor is best possible.

Without the assumption that $0 \in K$ already $(*)$ becomes wrong. As an example we can choose $F(z) = z-z^2$ and $K = \{ 1/2 \}$.
Then
$$
\left\Vert \frac{F(z)}{z} \right\Vert_\infty = \frac 12 > 0 =  \Vert F' \Vert_\infty \, .
$$
But we can pick any $z_0 \in K$ (and drop the condition $f(0) = 0$). Then we can integrate along the straight line from $z_0$ to $z$ and the same calculation as above gives
$$
\left\Vert \frac{d^k}{dz^k} \frac{f(z)-f(z_0)}{z-z_0} \right\Vert_\infty  \le
 \frac{1}{k+1} \Vert f^{(k+1)} \Vert_\infty  \, .
$$
