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As mentioned in the comments, this is true if $K$ is compact and the complement of $K$ in the Riemann sphere is connected : it is the content of Mergelyan's Theorem on uniform polynomial approximation …

Let $g$ be analytic in $\mathbb{D}$, continuous on $\overline{\mathbb{D}}$ such that $\int \int_{\mathbb{D}} |g'|^2=\infty$. It is well-known that such functions exist (the disc algebra is not contain …

The family of averages of $|q|$ on $D(r)$ is indeed non-decreasing. To see this, let $A(r)$ be the average of $|q|$ on $D(r)$. Let $r<\rho$ Then $A(r)=v(r)$, where $$v(z)=\frac{1}{\pi}\int_{0}^{2\pi} …

Yes, there is a classical proof of the Riemann mapping theorem using harmonic maps and the Dirichlet problem. Riemann's original assumption of boundary smoothness can be removed using Perron's method …

I now know a lot more about this question than I did when I asked it, so for the sake of completness let me add : By applying the Riemann mapping theorem $n$ times, we can assume that $\Omega$ is bou …

Problems of this type are part of the so-called Brennan's conjecture. More precisely, suppose that $f:\mathbb{D} \to \mathbb{C}$ is univalent. Brennan's conjecture states that $$\int_{\mathbb{D}}|f'| …

With Misha's correction, the answer is no. The maximum principle imply that $f(\mathbb{D}) \subseteq \mathbb{D}$ (note that $f$ is not constant because of condition 3). Conditions 1 and 2 imply that $ …

I am also curious to know why you claim the inequality is true without knowing how to prove it. However, when I saw Fedja's challenge I wanted to try it myself :-) so here it goes : I can prove the i …

Of course (1) does not imply (2) : the functions $f_n:=z^n$ converge pointwisely to $0$ on the unit disk, but the convergence is not uniform. In fact, assuming (1), the convergence need not be local …

Hi, Let $f$ be an entire function. The spherical derivative $\rho(f)$ is defined by $$\rho(f)(z):= \frac{|f'(z)|}{1+|f(z)|^2}.$$ A result from Clunie and Hayman states that if $\rho(f)$ is bounded, …

The only example I know of a conformal map that "stretches" a point boundary component to a nondegenerate continuum was constructed by Gehring and Martio in Quasiextremal distance domains and extensio …

The following is too long for a comment. I will suppose here that your set of measure zero is compact. In this case, Your question is closely related to so-called continuous analytic capacity. Let $K …

Hi, Let $f$ be a non-identically zero bounded analytic function in the open unit disk $\mathbb{D}$. It is well-known that $f$ has radial limits almost everywhere on the unit circle $\mathbb{T}$. Let …

The answer seems to be no.The quantity $$\rho(f(z)):= \frac{|f'(z)|}{1+|f(z)|^2}$$ is called the spherical derivative of $f$. Since you're interested in the behaviour of $z\rho(f(z))$ near $\infty$, …

For a compact set $K$ in the complex plane, define the analytic capacity of $K$ by $$\gamma(K) := \sup |f'(\infty)|$$ where the supremum is taken over all functions $f$ holomorphic and bounded by $1$ …