This is a problem from Rudin's Real and complex analysis.
If $\varphi$ is a non-constant, non zero inner function in $\mathbb{D}$, then $1/\varphi \notin H^p$ for $p>0$.
Inner function is a function $f\in H^{\infty}$ for which $|f^*|=1$ a.e on $\mathbb{T}$ ($f^*$ is the radial limit of $f)$. Few things I know is that $ \log|\varphi|$ is negative harmonic and I tried if I can get a contradiction assuming $1/\varphi \in H^p$ and then we get that $|1/\varphi^*|=1$ a.e. and therefore $\log|1/\varphi^*|=0$ a.e. Also, if $1/\varphi \in H^p$, we also get that $\log|1/\varphi^*|\in L^1(\mathbb{T})$. I thought I might get a contradiction from this but I couldn't. Any help is appreciated. Let me know if you need more details. Thanks!
If $1/\varphi$ is in $H^p$, when $f = 1/\varphi^{p/2}$ is in $H^2$ (the power $\varphi^{p/2}$ is holomorphic because $\varphi$ has no zeros). Expanding $f$ into a power series $\sum_{n=0}^\infty c_n z^n$ and using Parseval, we get $$ \sum_{n = 0}^\infty |c_n|^2 = \frac{1}{2\pi}\int_{0}^{2\pi}|f(e^{it})|^2\,dt = 1 $$ since $|\varphi|=1$ a.e. on the boundary. This implies $|f(0)|=|c_0|\le 1$, hence $|\varphi(0)|\ge 1$, which by the maximum principle is possible only if $\varphi$ is constant.
One can do a similar argument based on $1/\varphi^p\in H^1$, using the integral representation of $H^1$ functions.