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

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.

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!