In some papers and books I read they used that the polynomials are dense in $H^p$ for $1\le p<\infty$, but I could not find a proof for this stament. I tried to proof it myself but failed. My idea was to use the theorem of Stone Weierstrass, but I did not come any futher. So my question is, how can I proof it?
Edit: For $1\le p<\infty$ and $H(\mathbb{D})$ the set of all holomorphic functions on $\mathbb{D}=\{z\in \mathbb{C}: \lvert z\rvert<1\}$ we define $$ H^p=H^p(\mathbb{D})=\{f\in H(\mathbb{D}):\lVert f\rVert_{H^p}<\infty\} $$ with $$ \lVert f\rVert_{H^p}=\sup \limits_{0\le r<1}\left(\frac{1}{2\pi} \int \limits_0^{2\pi}\lvert f(re^{i\theta})\rvert^pd\theta\right)^{\frac{1}p}. $$
Let $f_r(z):=f(rz)$. Recall that, for $f\in H^p$ with $p\ge 1$, there exists the radial limit $f^*\in L^p(\partial D)$ and $||f^*||_{L^p}=||f||_{H^p}$. It is easy to see that $\lim_{r\to 1}||f_r^*- f^*||=\lim_{r\to 1}||f(re^{i\vartheta})-f^*(e^{i\vartheta})||=0$, by definition of radial limit. Thus $\lim_{r\to 1} ||f_r-f||_{H^p}=\lim_{r\to 1} ||f_r^*-f^*||_{L^p}=0$.
Now let $0<r_n<1$ be an increasing sequence such that $r_n\to 1$. By elementary results, there exists a polynomial $p_n:\sup_{D}|f_{r_n}-p_n|<\frac1{2^n}$. Since $||f_{r_n}-p_n||_{H^p}\le \sup_D|f_{r_n}-p_n|<\frac{1}{2^n}$. Since by the previous result we have $||f-f_{r_n}||_{H^p}\to 0$, the triangular inequality implies $$\lim_{n\to \infty}||f-p_n||_{H_p}\le\lim_{n\to \infty} ||f-f_{r_n}||_{H^p}+||f_{r_n}-p_n||=0$$