Denote the unit disk by $\mathbb{D}$. Let $w: \mathbb{D} \rightarrow \mathbb{D}$ be a Möbius transformed defined by $w(z) = \frac{z-\lambda}{1-\overline{\lambda}z}$ where $\lambda \in \mathbb{D}$ is fixed.
Suppose $f$ is a function defined on $\mathbb{D}$ such that $||f||_{\infty} \leq 1$.
If we consider the composition $w\circ f$, I do not understand why $||w(f(z))||_{\infty} \leq 1$.
Since $w$ is bijective, and $f(z) \in \mathbb{D}$, I think $||w||_{\infty}= 1$ is the only possibility. In my mind, even if $\text{im}(f)$ is properly contained in $\mathbb{D}$, $w$ still has to map $\mathbb{D}$ to $\mathbb{D}$. So the only least upper bound in any case is $1$.
Any clarification or insight is appreciated. Thank you.
Edit Since the range of $w$ is $\mathbb{D}$ and the range of the composition $w \circ f$ is contained the range of $w$, it follows that $w(f(z)) < 1.$ Therefore, the least upper bound of $|w(f(z))|$ is at most $1$; that is, $||w(f(z))||_{\infty} \leq 1$.
Of course $\Vert w \circ f \Vert_\infty \le 1$ holds, but the inequality can be strict. Concretely: If $$ \Vert f \Vert_\infty = K < 1 $$ then $$ \Vert w \circ f \Vert_\infty \le L $$ with $$ L = \max \{ |w(z)| : |z| \le K \} < 1 $$