Does a homeomorphism on the unit disk, that is the identity on the boundary have bounded displacement? I want to show a connection between the hyperbolic metric and the boundary values of a homeomorphism. Assuming there is a homeomorphism $f: \mathbb{D} \rightarrow \mathbb{D}$ on the unit disk that can be extended continuously $\partial{\mathbb{D}}$ and that $f$ is the identity on the boundary, is there a way to show that for $z \in \mathbb{D}$ it is true that
$$
p(z,f(z)) := \frac{|z -f(z)|^2}{(1-|z|^2)(1-|f(z)|^2)} = \sinh(\frac{\rho(z,(f(z))}{2})/4
$$
is bounded? Here $\rho$ is the hyperbolic metric on the unit disk.
I tried using a sequence $z_n \rightarrow z$ with $z\in \partial \mathbb{D}$ via
\begin{align*}
&\frac{|z_n -f(z_n)|^2}{(1-|z_n|^2)(1-|f(z_n)|^2)} \\
=& \frac{|z_n -f(z_n)| \ |z_n -f(z_n)|}{(1-|z_n|)(1+|z_n|) \ (1-|f(z_n)|)(1+|f(z_n)|)} \\
\leq& \frac{|z_n -f(z_n)|}{(1-|z_n|)} \ \frac{|z_n -f(z_n)|}{(1-|f(z_n)|)} \\
=& \frac{|z_n -f(z_n)|}{(|z|-|z_n|)} \ \frac{|z_n -f(z_n)|}{(|f(z)|-|f(z_n)|)},
\end{align*}
but sadly holds only
$$\geq \frac{|z_n -f(z_n)|}{|z-z_n|} \ \frac{|z_n -f(z_n)|}{|f(z)-f(z_n)|}.
$$
Another way I tried was with
$$
|z_n - f(z_n)| = |z_n - z + f(z) - f(z_n)| \leq |z_n - z| + |f(z) - f(z_n)|,
$$
but $(\epsilon + \delta)^2 = \epsilon^2 + 2\epsilon \delta + \delta^2$ and I don't know why we should know anything about $\epsilon/\delta$.
So does this not hold or is there a way to prove the statement?
Edit for clarification: $\rho$ is the hyperbolic metric while $p$ is a different function defined by the equation above.
 A: Here is a reasonably simple construction of a counterexamples.
Associated to any homeomorphism $d: [0,1] \to [0,1]$ fixing the endpoints there is a homeomorphism $f_d: \overline{\mathbb D} \to \overline{\mathbb D}$ defined in polar coordinates on the closed disc $\overline{\mathbb D}$ by the formula
$$f_d(r,\theta) = (d(r),\theta)
$$
This homeomorphism fixes each point on $\overline{\mathbb D}^2$. Now we just have to choose $d$ so that the restriction of $f_d$ to $\mathbb D$ does not have bounded displacement with respect to the Poincaré metric. I'll construct $d$ so that, as a point in $\mathbb D$ gets closer and closer to $\partial \mathbb D^2$ in the closed disc $\overline{\mathbb D}^2$, the distance that point gets moved by the map $f_d$ diverges to $+\infty$.
Let $\alpha : [0,\infty) \to \mathbb D$ be the unique geodesic ray with respect to the Poincaré metric which can be written in polar coordinates as $\alpha(t)=(\alpha_R(t),\alpha_\Theta(t))$ such that $\alpha_R(0)=0$ and $\alpha_\Theta(t)=0$ for all $t \in [0,1)$. It doesn't matter what the formula for $\alpha_R$ is; one can work that formula out at one's leisure. The key thing to notice is that $\alpha_R : [0,\infty) \to [0,1)$ is a homeomorphism.
Define
$$\rho(r) = 
\begin{cases}
\alpha_R\bigl((\alpha_R^{-1}(r))^2\bigr) &\quad r \in [0,1) \\
1 &\quad r=1
\end{cases}
$$
It follows that for each $t \in [0,\infty)$ the function $f_\rho$ moves the point $\alpha(t)$ by a distance equal to $t^2-t$.
One can construct many other counterexamples by replacing the "squaring function" $t \mapsto t^2$ with any homeomorphism of $[0,\infty)$ that does not have bounded displacement.
