Assume the representation formula \[u(z) = \operatorname{Re}\left[\frac{1}{2\pi i}\int_{|\eta|=R} \frac{\eta - z}{\eta + z} u(\eta)\frac{d\eta}{\eta}\right]\] for any harmonic function \(u\) defined on \(|z| < R\) and continuous up to \(|z| \leq R\).
If \(u(z)\) is a harmonic function defined on \(\mathbb{C}\) such that \(\lim_{z \to \infty} \frac{|u(z)|}{|z|} = 0\), show that \(u\) must be a constant.
Show that \[u(a) = \frac{1}{2\pi}\int_0^{2\pi} \frac{R^2 - |a|^2}{|Re^{i\theta} - a|^2} u(Re^{i\theta})\,d\theta.\]
Hint: The solution of (b) does not require (a).
Solution
1. From the representation, \(u(a) = \operatorname{Re}\frac{1}{2\pi i}\int_{|\eta|=R} \frac{\eta-a}{\eta+a}u(\eta)\frac{d\eta}{\eta}\).
hypothesis.
2. Bound \(|u(z)|=o(|z|)\), let \(R\to\infty\), then \(\frac{\eta-a}{\eta+a}=1+O(R^{-1})\), so \(u(a)=\operatorname{Re}\frac{1}{2\pi i}\int_{|\eta|=R} u(\eta)\frac{d\eta}{\eta}+o(1)\).
1 and growth.
3. The integral \(\frac{1}{2\pi i}\int_{|\eta|=R} u(\eta)\frac{d\eta}{\eta}\) is bounded independent of \(R\), so \(u\) is constant by letting \(R\to\infty\) for \(u(a)-u(0)\).
2 (difference tends to \(0\)).
4. Part (b) follows by writing \(\frac{\eta-a}{\eta+a}\frac{d\eta}{\eta}= \frac{R^2-|a|^2}{|Re^{i\theta}-a|^2}d\theta + i(\dots)\) and taking real part.
\(\eta=Re^{i\theta}\), \(d\eta=i\eta d\theta\).
5. Q.E.D.
3 and 4.