Prove the Jensen's formula
$$\int_{T}f(z+re^{2\pi i\theta})d\theta-f(z)=\iint_{D(z,r)}\log{\frac{r}{|w-z|}}\Delta f(w)dm(w)$$ where $w$ is in $D(z,r)$ and $f$ is a two-dimensional $C^2$ function.
You may notice that if $f$ is a harmonic function, the equality is the maximum principle. For the simplicity, put $r=1$, $z=0$. Then it is equal to
$$\int_{T}f(e^{2\pi i\theta})d\theta-f(0)=-\iint_{D(0,1)}(\log{|w|})\Delta f(w)dm(w)$$
By the Green's second identity, $$\iint_{D(0,1)}(\log{|w|})\Delta f(w)dm=\int_{D(0,1)}f(w)\Delta \log{|w|}dm+\int_{\partial D}\log{|w|}\frac{d}{dn}f(w)d\sigma-\int_{\partial D}f(w)\frac{d}{dn}\log{|w|}d\sigma$$
Since $\log{|w|}$ is the harmonic function, the first term on the right side is always $0$. The second term is also $0$ because $\log{|w|}=0$ on the unit disc. Therefore, $$\iint_{D(0,1)}(\log{|w|})\Delta f(w)dm=-\int_{\partial D}f(w)\frac{d}{dn}\log{|w|}d\sigma$$ Let's calculate the left term. $$\int_{\partial D}f(w)\frac{d}{dn}\log{|w|}d\sigma=\int_{\partial D}f(w)(\triangledown \log{|w|} \cdot n )d\sigma =\int_{0}^{1} f(e^{2\pi i\theta})[(\cos(2\pi \theta),\sin(2\pi \theta)) \cdot (\cos(2\pi \theta),\sin(2\pi \theta))]=\int_{0}^{1}f(e^{2\pi i\theta})d\theta$$ In conclusion, we get $$\int_{T}f(e^{2\pi i \theta})d\theta =-\iint_{D(0,1)}(\log{|w}|)\Delta f(w)dm(w)$$ I need your sharp eyes. I fail to find my mistake.