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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?

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  • $\begingroup$ I guess you mean $f(\mathbf{x})_n=f_n(x_n)$? $\endgroup$ – Ian Apr 15 at 13:45
  • $\begingroup$ I mean, for $x\in c_0$, $f(x)=y$, where $y\in c_0$ and $y=(y_n)$, $y_n=\sqrt{|x_n|}+\frac{1}{n+1}$. $\endgroup$ – zorro47 Apr 15 at 13:47
  • $\begingroup$ Yes, that is what I said. $\endgroup$ – Ian Apr 15 at 13:48
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    $\begingroup$ Anyway, if $x_n=y_n$ for all $n \neq k$, then $\| f(\mathbf{x})-f(\mathbf{y}) \| =|\sqrt{x_k}-\sqrt{y_k}|$, and now you've reduced to the one dimensional case. $\endgroup$ – Ian Apr 15 at 13:49
  • $\begingroup$ ok, and the square root is not a Lipschitz function, at least if I recall corectly (on $[0,1]$). Could you explain, what do you mean by $x_n=y_n$ if $n\ne k$? $\endgroup$ – zorro47 Apr 15 at 13:57
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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.

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  • $\begingroup$ as @ Ian mentioned, you can take the kth element instead of the first. $\endgroup$ – Dayton Apr 15 at 14:05
  • $\begingroup$ Exactly, thank you! $\endgroup$ – zorro47 Apr 15 at 14:06

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