Estimating derivative of difference quotient

Question

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?

Answer 1

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.

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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 $$ \left\Vert \frac{d^k}{dz^k} \frac{f(z)-f(z_0)}{z-z_0} \right\Vert_\infty \le \frac{1}{k+1} \Vert f^{(k+1)} \Vert_\infty \, . $$