If f is a real function in $H^2\mathbb(D)$ then $f$ is constant.

Question

I want to prove that if $f$ is a REAL function in the Hardy space $H^2\mathbb(D)$ then $f$ is contant. We know that if $f\in H^2\mathbb(D)$ then $$f(z)=\sum_{n=0}^{\infty}a_nz^n$$ where $a_n=\frac{f^{(n)}(0)}{n!}$. Since $f$ is real then $f=\overline{f}$ hence: $$a_0+a_1z+a_2z^2+...=\sum_{n=0}^{\infty}a_nz^n=\overline{\sum_{n=0}^{\infty}a_nz^n}=\overline{a_0}+\overline{a_1z}+\overline{a_2z^2}+...$$ To prove that $f$ is constant i have to show that $f'(z)=0$ for all $z\in \mathbb{D}$ ie: $$\sum_{n=0}^{\infty}na_nz^{n-1}=0$$ which leads to prove that: $$a_1=a_2=...=0$$ Can someone help me please?

Answer 1

If $f \in H^2(D)$, then $f$ is holomorphic on the unit disc $D$. If $f$ is real, then $f(D)$ is not open, hence $f$ is constant.

Answer 2

A function $f \in L^2(T)$ ($T$=unit circle) is the boundary function of an element of $H^2(D)$ iff $$ \int_{T} f(z)z^{n}dz = \int_{0}^{2\pi}f(e^{i\theta})e^{i(n+1)\theta}d\theta = 0, \;\;\; n \ge 0. $$ If that boundary function is real a.e., then $$ \overline{\int_{T}f(e^{i\theta})e^{i(n+1)\theta}d\theta}=\int_{T}f(e^{i\theta})e^{-i(n+1)\theta}d\theta=0,\;\;\; n \ge 0. $$ Combining these two gives a function $f\in L^2(T)$ with Fourier coefficients that are all zero, except possibly for $\int_{T}f(e^{i\theta})d\theta$. So $f$ must be constant a.e.. on $T$.