In the book "Geometric Nonlinear Functional Analysis" by Benyamini and Lindenstrauss, I came across an example (Example 1.5) in which the authors construct a retraction from $l_\infty$ to $c_0$ (both equipped with the supremum metric) and claim it is a Lipschitz mapping with constant 2. I give the definition below
For a sequence $x=(x_n)_{n\in\mathbb N}\in l_\infty$, denote $d(x)=\limsup|x_n|$. We define $r:l_\infty\to c_0$ as follows. For $x=(x_n)_{n\in\mathbb N}$ and $n\in\mathbb N$
$$r(x)_n = \begin{cases} 0 &, |x_n|<d(x) \\ (|x_n|-d(x))sign(x_n)&, |x_n|\geq d(x).\end{cases}$$
I can see why it is a well-defined retraction. However I cannot obtain the Lipschitzness at constant 2. Could you please give a solution or a hint?
Let $x=(x_n)$ and $y=(y_n)$ belong to $\ell_{\infty}$ with $\|x-y\| <\epsilon$. Note that $d(x)+\epsilon > |x_n|$ for all $n$ sufficiently large. This gives $|0-(|x_n|-d(x))sign(x_n)| <\epsilon$ if $|x_n| \geq d(x)$ and $n$ is sufficiently large. Similarly, $|0-(|y_n|-d(y))sign(y_n)| <\epsilon$ if $|y_n| \geq d(y)$ and $n$ is sufficiently large. In the case $|x_n| \geq d(x)$ and $|y_n| \geq d(y)$ use the fact that $|d(x)-d(y)| \leq \epsilon$ and triangle inequality to show that $|(r(x)_n-r(y)_n| \leq 2 \epsilon$ I hope this helps you to complete the proof. [Let $\epsilon$ decrease to $\|x-y\|$ to end the proof].