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Let $f: \mathbb{R}^2 \to \mathbb{R}^2$ be the function that swaps its arguments, e.g. $f(x,y) = (y,x)$. Is this function lipschitz continuous w.r.t the euclidean norm $\|\cdot\|_{2}$ ?

At first I was pretty sure that $f$ must be lipschitz, because it's almost like the identity function, but then I thought about it some more and now I'm not so sure anymore. In the past I only had to prove the lipschitz continuity of functions from $\mathbb{R}\to \mathbb{R}$, but this feels very different from those exercises... A hint is fine too :)

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  • $\begingroup$ What is your definition of "Lipschitz continuous"? What happens when you plug $f$ into it? $\endgroup$ – Milo Brandt Jul 5 '16 at 20:30
  • $\begingroup$ Hint : write $\|(x,y)-(x',y')\|_{2}$ :) $\endgroup$ – anonymus Jul 5 '16 at 20:30
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    $\begingroup$ $f$ is an isometry, so it is Lipschitz continuous. $\endgroup$ – Levent Jul 5 '16 at 20:31
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$$\|(x,y)-(z,w)\|=\|(x-z,y-w)\|=\sqrt{|x-z|^2+|y-w|^2}=\sqrt{|y-w|^2+|x-z|^2}=\|(y-w,x-z)\|=\|(y,x)-(w,z)\|$$

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  • $\begingroup$ I think in the last expression you mean $(y, x)$ instead of $(y,z)$. $\endgroup$ – jazzinsilhouette Jul 5 '16 at 20:52
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$f$ has the effect of transforming $(x,y)$ into its symmetric about the line $y = x$. Hence, it is an isometry: given $(x,y)$, we have that the triangle joining $(x,y), O$ and $f(x,y)$ is isosceles with vertex $O$ (in case of collinearity, $O$ is just the midpoint of the segment joining $(x,y)$ to $f(x,y)$). In particular, $f$ is Lipschitz.

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