Let $p\in [1, \infty]$. Suppose $g \in L^p(T)$ where $T$ is the unit circle.
I want to prove g is boundary value of $f \in H^p$ if and only if $\int_{-\pi}^{\pi}g(e^{it})e^{-int} dt = 0$ for all $n \in \{-1, -2, \cdots \}$.
I can prove one direction ($\Rightarrow$): Since $f \in H^p$, so it is a Poisson integral of boundary value. Therefore $f(z) = \int_{-\pi}^{\pi} P_r(\theta - t)g(e^{it}) dt$ where $P_r(\theta - t) = \sum_{-\infty}^{\infty} r^{|n|}e^{in\theta}e^{-int}$ and $z = re^{i\theta}$. Therefore, $f(z) = \sum r^{|n|}e^{in\theta}\int g(e^{it})e^{-int}dt = \sum_{-\infty}^{\infty} \hat{g}(n)z^n$ where $\hat{g}(n)$ is the $n$th Fourier coefficient. Since $f$ is holomorphic, therefore the negative Fourier coefficients are zero.
Can anybody help me to prove the other direction? I have no clear idea to prove it.