Lipschitz maps on locally compact groups

120 Views Asked by At

Suppose $G$ is a locally compact second countable group. This means that there exists a proper (closed bounded sets are compact) left invariant ($d(gx,gy) = d(x,y) \ \forall g,x,y \in G$) metric on the group that generates the topology on $G$.

Let $A \subset G$ be a bounded set in $G$. For any point $x \in G$ define $d(x, A ) = \inf \{ d(x,y) : y \in A \}$. For all $\varepsilon > 0$ let $N_{\varepsilon}(A) = \{ x \in G : d(x,A) < \varepsilon \}$.

My question is, does there exist a Lipschitz map that is 1 on $A$ and 0 on $G \setminus N_{\varepsilon}(A)$ such that the Lipschitz constant does not depend on the size of $A$ and only depends on $\varepsilon$?

Here a function $f \colon G \to \mathbb{C}$ is Lipschitz if there exists $M > 0$ such that for all $g,h \in G$, $|f(g) - f(h)| \leq M d(g,h)$. Here $M$ is the Lipschitz constant.

My guess would be that the Lipschitz constant would be something like $1 / \varepsilon$.

1

There are 1 best solutions below

0
On BEST ANSWER

There is a Lipschitz function $\phi:[0,\infty)\to[0,1]$, with Lipschitz constant exactly $1/\epsilon$, such that $\phi(0)=1$ and $\phi(t)=0$ for $t\ge\epsilon$. Let $f(x)=\phi(d(x,A))$.