The supremum of the $n$th derivative of a holomorphic function is bounded by the $L^1$ norm

Question

Let $U\subset \mathbb{C}$ be open, and $A\subset U$ compact. Suppose $n\geq 0$ where $n\in \mathbb{Z}$. Prove that there exists a constant $k$ (allowed to depend on $n$, $A$, and $U$) such that for any function $f$ which holomorphic is on $U$, we have $$\sup_{z\in A}|f^{(n)}(z)|\leq k\iint_{U}|f(z)|\text{d}x\text{d}y.$$

I have attempted to find points where $f^{(n)}$ might be maximized in $A$ and use Cauchy's Integral formula at disc neighorhoods, but this did not yield any fruit. Any and all help would be appreciated.

Answer 1

Let $4d>0$ be the distance from $A$ to $\partial U$. By compacity, there are finitely many closed discs of radius $d$ centered at the points of $A$ that cover it, so if we prove the relation required for the part of $A$ in each such disc we are done by taking for $k$ the maximum of all the $k's$ obtained at each disc and of course, it is enough to prove the required relation for the full closed discs of radius $d$ themselves

So wlog we can assume $A=\bar D(w,d)$. But then the disc $\bar D(w,3d) \subset U$, so we can apply Cauchy on each circle of radius $2d \le r \le 3d$ and get:

$f^{(n)}(z) = \frac{n!}{2\pi i}\int_{C_r}\frac{f(\zeta)}{(z-\zeta)^{n+1}}d\zeta$ But now using that $|z-\zeta| \ge d$, we get:

$|f^{(n)}(z)| \le \frac{n!}{2\pi d^{n+1}}\int_{C_r}|f(\zeta)||d\zeta|$ and using $|d\zeta|=rdt$, we can write this as:

$|f^{(n)}(z)| \le \frac{n!}{2\pi d^{n+1}}\int_{0}^{2\pi}|f(\zeta)|rdt$

Integrating this relation from $r=2d$ to $r=3d$ we get:

$|f^{(n)}(z)| \le \frac{n!}{2\pi d^{n+2}}\int_{2d}^{3d}\int_{0}^{2\pi}|f(\zeta)|rdtdr = \frac{n!}{2\pi d^{n+2}}\int_{A_d}|f(z)|dxdy$ where $A_d$ is the annulus centered at $w$ between the circles of radiuses $2d$ and $3d$. Since $A_d \subset U$, obviously $\int_{A_d}|f(z)|dxdy \le \int_U|f(z)|dxdy$ so we get the required relation and we are done!