I'm not sure if Riemann map is standard terminology but, I mean the map guaranteed by Riemann Mapping Theorem.
I'm trying to prove the following theorem of Bohr-Pal:
If $f \in C(\mathbb{T})$ then there exists a homeomorphism $\phi:\mathbb{T} \to \mathbb{T}$ such that $S_N(f \circ \phi) \to f \circ \phi$ uniformly.
I've been suggested to prove this using the following facts, which I have succeeded in proving:
- If $f \in H^{1/2}(\mathbb{T})\cap C(\mathbb{T})$ then $S_Nf \to f$ uniformly.
- For $f \in H^{1/2}(\mathbb{T})$, $$4\sum_{n \neq 0} |n||\hat{f}(n)|^2 = \int_{\mathbb{T}} \int_{\mathbb{T}}\frac{|f(x) -f(y)|^2}{\sin^2(\pi(x-y))} \ dx \ dy.$$
So, it seems to me like the logical method would be to find a homeomorphism $\phi: \mathbb{T} \to \mathbb{T}$, such that
$$ \int_{\mathbb{T}} \int_{\mathbb{T}}\frac{|f(\phi(x)) -f(\phi((y))|^2}{\sin^2(\pi(x-y))} \ dx \ dy<\infty$$
which then implies $f \circ \phi \in H^{1/2}(\mathbb{T})$ , which then implies the result.
So, here's my candidate for the map.
Assuming $f>0$ (I've been told I can do this without loss of generality, but haven't seen why yet), we have the domain $\Omega = \{re^{2\pi i \theta} \ | \ 0 \le r \le f(\theta), 0 \le \theta <1\}$. Then $\Omega$ is simply connected, and its boundary $\partial \Omega$ is parametrized by $G(\theta) = f(\theta)e^{2\pi i \theta}$ for $0 \le \theta <1$. Note that $G$ is a homeomorphism.
By the Riemann mapping theorem, we have a conformal map $\psi: \mathbb{D} \to \Omega$, which extends to a homeomorphism between the boundaries $H: \mathbb{T} \to \partial \Omega$. So now I want to look at $\phi = G^{-1} \circ H: \mathbb{T} \to \mathbb{T}$. It is immediate that $f \circ \phi = |G \circ \phi| = |H|$. So now I want to look at
$$ \int_{\mathbb{T}} \int_{\mathbb{T}}\frac{||H|(x) - |H|(y)||^2}{\sin^2(\pi(x-y))} \ dx \ dy<\infty$$
and here I am lost.
Since $\psi$ is a conformal map onto a bounded domain, the integral of its Jacobian determinant is finite (it expresses the area of the image). The Jacobian is $|\psi'(z)|^2$. Plug the power series $\psi(z) = \sum_{n=0}^\infty \hat \psi(n) z^n$ in $|\psi'(z)|^2$ and integrate the result over the disk (using the orthogonality of exponentials, equivalently Parseval's identity):
$$ \int_0^1 \int_0^{2\pi} \left|\sum n \hat \psi(n)r^{n-1}e^{i(n-1)\theta}\right|^2\, r\,d\theta\,dr =2\pi \int_0^1 \sum n^2 |\hat \psi(n)|^2 r^{2n-2}\, r\,dr = \pi \sum n |\hat \psi(n)|^2 $$
Thus the boundary map induced by $\psi$, denoted $H$, belongs to $H^{1/2}$.
The description of $H^{1/2}$ norm by a double integral shows that $H\in H^{1/2}\implies |H| \in H^{1/2}$ (by the triangle inequality). This shows that $\phi$ chosen so that $f\circ \phi = |H|$ works.
By the way, the reason it suffices to consider $f>0$ is that the general case is reduced to this one by adding a constant to $f$.