Let $f\in H^p$ where $H^p$ denotes the Hardy space for $1\leq p<\infty$. That is, $f$ is a holomorphic function on the open unit disk, $\mathbb{D}$, such that $$\sup_{0<r<1}\int_0^{2\pi}\vert f(re^{i\theta})\vert^pd\theta<\infty.$$ Does it follow then that $f\in L^p(D(0,1))$, i.e. $$\int_{\mathbb{D}}\vert f(z)\vert^p dz<\infty?$$ I assume the answer ought to be no. If the integral in the above supremum equals $\frac{1}{\sqrt{1-r}}$, then the supremum is infinite, but the below intgral is finite. But I cannot find a holomorphic function with this property.
Edit: I corrected my title. I meant to ask the following inclusion: Suppose $f$ is a holomorphic function in $L^p(\mathbb{D})$. Then must $f\in H^p$?