Let $f:\mathbb D\rightarrow \mathbb C$ be a function in $\mathrm{H}^p$, i.e. $$\exists M>0,\text{ such that }\int_0^{2\pi}|f(re^{it})|^pdt\leq M<\infty,\forall r\in[o,1)$$ Consider a Möbius transform of the disk $\varphi :\mathbb D\rightarrow\mathbb D$, which generally does not fix $0$.
(or more generally consider a holomorphic function $g:\mathbb D \rightarrow \mathbb D$)
Is the composition $f\circ\varphi $ in $\mathrm H^p$?
If $\varphi $ fixes $0$ then we can apply Littlewood subordination theorem and derive the result. However, what can we say if $\varphi$ does not fix zero? Does it hold $f\circ\varphi $ in $\mathrm H^p$, or is there a counterexample?
Yes. If $\varphi$ is a holomorphic map of unit disk into itself, the composition operator $f\mapsto f\circ \varphi$ is bounded on $H^p$. In fact, $$\|f\circ \varphi\|_{H^p}\le \left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{1/p}\|f \|_{H^p} \tag1$$
Original source:
Sketch of proof. The function $|f|^p$ is subharmonic and has uniformly bounded circular means (i.e., mean values on $|z|=r$, $r<1$). Let $u$ be its the smallest harmonic majorant in $\mathbb D$. Then $u\circ \varphi$ is a harmonic majorant of $|f\circ \varphi|^p$. This implies that the circular means of $|f\circ \varphi|^p$ do not exceed $u(\varphi(0))$. In particular, $f\circ \varphi\in H^p$. Furthermore, the Harnack inequality for the disk yields $$u(\varphi(0))\le u(0)\frac{1+|\varphi(0)|}{1-|\varphi(0)|}$$ proving (1).