- Let u be a harmonic function in the unit disc that is continuous on its closure. Deduce Poisson’s integral formula
\[ u (z _ {0}) = \frac {1}{2 \pi} \int_ {0} ^ {2 \pi} \frac {1 - | z _ {0} | ^ {2}}{| e ^ {i \theta} - z _ {0} | ^ {2}} u (e ^ {i \theta}) d \theta \quad \mathrm{for} | z _ {0} | < 1 \]
from the special case \(z _ { 0 } = 0\) (the mean value theorem). Show that if \(z _ { 0 } = r e ^ { i \varphi }\) ， then
\[ \frac {1 - | z _ {0} | ^ {2}}{| e ^ {i \theta} - z _ {0} | ^ {2}} = \frac {1 - r ^ {2}}{1 - 2 r \cos (\theta - \varphi) + r ^ {2}} = P _ {r} (\theta - \varphi), \]
and we recover the expression for the Poisson kernel derived in the exercises of the previous chapter.
[Hint: Set \(u _ { 0 } ( z ) = u ( T ( z ) )\) where
\[ T (z) = \frac {z _ {0} - z}{1 - \overline {{z _ {0}}} z}. \]
Prove that \(u _ { 0 }\) is harmonic. Then apply the mean value theorem to \(u _ { 0 }\) , and make a change of variables in the integral.]
Solution
1. Let \(T(z) = \frac{z_0 - z}{1 - \overline{z_0} z}\) and define \(u_0(z) = u(T(z))\).
the hint.
2. \(T\) is a holomorphic automorphism of the disk with \(T(0) = z_0\) and \(T(z_0) = 0\).
\(T\) is a Blaschke factor (an involution of \(\mathbb{D}\)).
3. \(u_0\) is harmonic.
\(u_0 = u \circ T\) is the composition of a harmonic function with a holomorphic map, which is harmonic.
4. By the mean value theorem (the special case \(z_0 = 0\)), \(u_0(0) = \frac{1}{2\pi}\int_0^{2\pi} u_0(e^{i\theta})\,d\theta\).
the mean value property of harmonic functions at the origin.
5. \(u_0(0) = u(T(0)) = u(z_0)\).
2.
6. Hence \(u(z_0) = \frac{1}{2\pi}\int_0^{2\pi} u(T(e^{i\theta}))\,d\theta\).
4 and 5.
7. Make the change of variables \(e^{i\varphi} = T(e^{i\theta})\); then \(d\theta = \frac{1 - |z_0|^2}{|e^{i\theta} - z_0|^2}\,d\varphi\) (the Jacobian of the Blaschke factor on the circle).
the derivative of \(T\) on the unit circle has modulus \(\frac{1 - |z_0|^2}{|e^{i\theta} - z_0|^2}\).
8. Hence\[u(z_0) = \frac{1}{2\pi}\int_0^{2\pi} \frac{1 - |z_0|^2}{|e^{i\theta} - z_0|^2} u(e^{i\theta})\,d\theta.\]
6 and 7.
9. For \(z_0 = re^{i\varphi}\), \(\frac{1 - |z_0|^2}{|e^{i\theta} - z_0|^2} = \frac{1 - r^2}{1 - 2r\cos(\theta - \varphi) + r^2} = P_r(\theta - \varphi)\).
\(|e^{i\theta} - re^{i\varphi}|^2 = 1 - 2r\cos(\theta - \varphi) + r^2\).
10. Q.E.D.
8 and 9.