Source: Marshall's Harmonic Measure Chapter 2 Problem 8
I have been working on this problem the entire day, and I could not figure out a way to crack it.
Question: Let $u(z)$ be the Poisson integral of $f \in L^2(\overline{\mathbb{D}})$. Use Green's theorem to show$$\frac{1}{2\pi} \int |f - f(0)|^2 = \frac{2}{\pi} \int \int_{\mathbb{D}} |\nabla u(z)|^2 \log\frac{1}{|z|} \, dx dy.$$
Hints(copy from the book): Replace $f$ by $u(re^{i \theta})$, use the identity $\Delta |u|^2 = 4 |\nabla u|^2$ and Green's theorem, and send $r \to 1$.
What I have tried: I tried to start with an easy case where $f(0) = 0$. Then, by Green's identity ($\int_U \Delta u \, dx = \int_{\partial U}\frac{\partial u}{\partial n} \, dS$), I could see that $$\frac{d}{dr} \frac{1}{2\pi} \int_{0}^{2\pi} |u(re^{i \theta})|^2 \, d\theta =\frac{2}{\pi r} \int \int_{r \mathbb{D}} |\nabla u|^2 \, dx dy,$$and we could take the limit $r \to 1$. However, I am stuck after this step, nor was I entirely sure the above observation is helpful.