# Showing Lipschitz continuity for a particular distance functiom.

My friend and I have been working on trying to prove this inequality for awhile, however, I think there is some trick we are just not seeing.

Suppose $F$ is a closed set in $\mathbb{R}$, whose complement has finite measure, and let

$$\delta(x)=d(x,F)=\inf \{ \mid x - y \mid : y \in F \}$$

Prove $$\mid \delta(x) - \delta(y) \mid \le \mid x - y \mid.$$

Let $x,y\in \Bbb R$.

For all $z\in F$ we have $$\delta(x)\leq |x-z|\leq |x-y|+|y-z|$$ hence $\delta(x)-|x-y|\leq |y-z|$ for all $z\in F$ It follow that $\delta(x)-|x-y|\leq \inf_{z\in F}|y-z|=\delta(y)$

Now we get $\delta(x)-\delta(y)\leq |x-y|$

By the same argument we can show also that $\delta(y)-\delta(x)\leq |x-y|$

so $|\delta(x)-\delta(y)|\leq |x-y|$.

Remark: This result is true without the hypothesis whose complement has finite measure

• It seems you didn't even have to assume that $F$ is closed, or am I mistaken? – balu Oct 29 '15 at 23:20
• It is also true without this hypothesis (closed). – Hamou Nov 4 '15 at 9:09