Ive been tying to prove the following;
If $f$ is analytic with non-negative real part in $\mathbb{D}$. Then there is an unique non-negative Borel measure on $\mathbb{T}$ s.t $\mu(\mathbb{T})=Ref(0)$ and $f(z)=\int \frac{\zeta +z}{\zeta -z} d\mu$
Ive know that any positive harmonic function have these properites but when I try to extend it to the analytic completion it gets messy, Im aware of the fact that the Herlgotz kernal is the sum of the possion kernal and its conjugate which would be nice to use but I cant manage that either. Any hints on a proof involving this fact would be most appreciated.
As you noted, the real part of the Herglotz kernel is the Poisson kernel: $$ \Re \left(\frac{e^{i\theta}+z}{e^{i\theta}-z}\right) = \Re \left(\frac{(e^{i\theta}+z)(e^{-i\theta}-\overline{z})}{|e^{i\theta}-z|^2}\right) = \frac{1-|z|^2}{|e^{i\theta}-z|^2}. $$ Because $\Re f(z)$ is a non-negative harmonic function on $\mathbb{D}$, then there is a unique finite positive Borel measure $\mu$ on $\mathbb{T}$ such that $$ \Re f(z) = \int_{\mathbb{T}}\Re\left(\frac{w+z}{w-z}\right)d\mu(w),\;\;\; |z| < 1, \\ \Re f(0) = \mu(\mathbb{T}). $$ Then $$ g(z) = f(z) - \int_{\mathbb{T}}\frac{w+z}{w-z}d\mu(w) $$ is a holomorphic function on $\mathbb{D}$ with $0$ real part, which forces $g$ to equal $iC$ on $\mathbb{D}$, where $C$ is a real constant. Therefore, $$ f(z) = \int_{\mathbb{T}}\frac{w+z}{w-z}d\mu(w)+iC. $$ There's no way to get rid of the imaginary constant because $f(z)=i$ satisfies the hypotheses.