Does the Lipschitz constant depend on the used norm?
In my courses I know that the Lipschitz constant does not depend on the norm in $\mathbb{R^n}$ because it is complete so all the norms are equivalent.
I found nothing in the online litterature concerning the correspondance norm - Lipschitz constant in the general case, when the studied function $f$ is not defined on a complete space.
Many thanks
The Lipschitz constant depend of the norm: Imagine an $f:\mathbb{R}\rightarrow \mathbb{R}$ such that $$ |f(x)-f(y)|\leq k|x-y|.$$ If I consider the norm $|\cdot|_2 := 2\times |\cdot|$ for the arrival space, I will have $$ |f(x)-f(y)|\leq \frac{k}{2}|x-y|_2.$$
Remark: The norms on $\mathbb{R}^n$ are not equivalent because $\mathbb{R}^n$ is complete! But because $\mathbb{R}^n$ is a finite vectoriel space... $L^2(\mathbb{R}^n)$ is complete but the norms are not equivalent on it.
Lipschitz constants depend on the norm on $\Bbb R^n$. For instance, consider $n=2$ and the norms $\lVert \bullet\rVert_1,\ \lVert \bullet\rVert_2$. Consider $g(x)=\frac{\sqrt2}2(x_1-x_2,x_1+x_2)$. This is an isometry of $\left(\Bbb R^2,\lVert\bullet\rVert_2\right)$, but $$\sup_{x\ne y} \frac{\lVert g(x)-g(y)\rVert_1}{\lVert x-y\rVert_1}=\sup_{\lVert z\rVert_1=1} \lVert g(z)\rVert_1=\frac{\sqrt2}{2}\sup_{\lvert z_1\rvert+\lvert z_2\rvert=1} \lvert z_1-z_2\rvert+\lvert z_1+z_2\rvert\ge \sqrt2$$
so $g$ has Lipschitz constant $\ge \sqrt2$ as a function $g:(\Bbb R^2,\lVert\bullet\rVert_1)\to(\Bbb R^2,\lVert \bullet \rVert_1)$.