Let \(f\) be analytic on the unit disk \(\Delta\), with \(f(0)=0\) and \[ \int_\Delta |f(z)|^2\,|dz|^2<\infty. \] Prove that \[ \int_\Delta |\operatorname{Re}f|^2\,|dz|^2 = \int_\Delta |\operatorname{Im}f|^2\,|dz|^2. \]
Solution
Write \[ f=u+iv, \qquad u=\operatorname{Re}f, \quad v=\operatorname{Im}f. \] Since \[ u^2-v^2=\operatorname{Re}(f^2), \] it suffices to show that \[ \int_\Delta f(z)^2\,|dz|^2=0. \]
1. The function \(f^2\) is integrable on \(\Delta\).
By hypothesis, \[ \int_\Delta |f(z)|^2\,|dz|^2<\infty. \] Since \[ |f(z)^2|=|f(z)|^2, \] the function \(f^2\) is absolutely integrable. Thus polar-coordinate integration and Fubini’s theorem may be applied to \(f^2\).
2. For every \(0<r<1\),\[ \int_0^{2\pi} f(re^{i\theta})^2\,d\theta=0. \]
The function \(f^2\) is analytic on \(\Delta\), and \[ f(0)^2=0. \] By the mean-value formula for analytic functions, \[ \frac1{2\pi}\int_0^{2\pi} f(re^{i\theta})^2\,d\theta =f(0)^2 =0. \]
3. One has\[ \int_\Delta f(z)^2\,|dz|^2=0. \]
1. The inner integral vanishes forevery \(r\) by 2, so the whole integral is zero.
4. Therefore\[ \int_\Delta u^2\,|dz|^2 = \int_\Delta v^2\,|dz|^2. \]
Taking real parts in 3 gives \[ 0 =\operatorname{Re}\int_\Delta f^2\,|dz|^2 =\int_\Delta \operatorname{Re}(f^2)\,|dz|^2 =\int_\Delta (u^2-v^2)\,|dz|^2. \] Hence \[ \int_\Delta |\operatorname{Re}f|^2\,|dz|^2 = \int_\Delta |\operatorname{Im}f|^2\,|dz|^2. \]