How to prove that a function $f\colon c_0\to c_0$ is not Lipschitz continuous?

Question

I wonder if the following function $f\colon c_0\to c_0$ ( $c_0$ is a space of real sequences convergent to 0 with supremum norm) $$ f(x)=(f_{n}(x)),$$ where $f_n(x)=\sqrt{|x_n|}+\frac{1}{n+1}$ is Lipschitz? I want to prove that it isn't. So, I pick $x,y\in c_0$ and consider $$\|f(x)-f(y)\|=\sup_{n\in\mathbb{N}}|\sqrt{|x_n|}-\sqrt{|y_n|}|.$$ At first I thought that it is enough to ask whether $g(x)=\sqrt{|x|}$ is Lipschitz continuous, but on the other hand there is the supremum. Maybe there are sequences $(x_{n}), (y_n)$ such that Lipschitz codition does not hold?

Answer 1

That should be enough. Because you can take the the subset $c_0' \subset c_0$ defined by $$c_0' := \{ (x_n)_n \ | \ x_k = 0 \text{ for } k \geq 2\}$$ Then take $x,y \in c_0'$ so that $$f(x) - f(y) = (\sqrt{|x_n|} - \sqrt{|y_n|})_n = (\sqrt{|x_1|} - \sqrt{|y_1|}, 0,0,...)$$ Let $g(x) = \sqrt{|x|}$, then $$|g(x_1) - g(y_1)| = ||f(x) - f(y)|| = \big|\sqrt{|x_1|} - \sqrt{|y_1|}\big|$$ Which is not Lipschitz.