McShane Extension of a Real-Valued Lipschitz Function on a Metric Space

Let $(X,d)$ be a metric space, let $A\subseteq X$ be nonempty, and let $d_{A}$ denote the restriction of $d$ to $A$, which is a metric on $A$ by claim 1 of The Restriction of a Metric to a Subset Induces the Subspace Topology. Write $\mathbb{R}$ for the real numbers, $|\cdot|$ for the absolute value, and $d_{\mathbb{R}}$ for the absolute value metric $d_{\mathbb{R}}(s,t)=|s-t|$ on $\mathbb{R}$.

Let $L\in\mathbb{R}$ with $0\le L$ and let $f:A\to\mathbb{R}$ be Lipschitz with constant $L$ from $(A,d_{A})$ to $(\mathbb{R},d_{\mathbb{R}})$, that is

$$|f(a)-f(b)|\le L\,d(a,b)\qquad\text{for all }a,b\in A .$$

For $x\in X$ put

$$S(x)=\bigl\{\,f(a)+L\,d(x,a)\;:\;a\in A\,\bigr\}\subseteq\mathbb{R}.$$

Then the following hold.

1. (The McShane extension is well defined) (i) For every $x\in X$ the set $S(x)$ is nonempty and bounded below, so that by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below it has a greatest lower bound in $\mathbb{R}$. We may therefore define a function $F:X\to\mathbb{R}$, the McShane extension of $f$ with constant $L$, by

$$F(x)=\inf S(x)\qquad(x\in X).$$

2. (It extends $f$) (ii) $F(a)=f(a)$ for every $a\in A$.

3. (It keeps the Lipschitz constant) (iii) $F$ is Lipschitz with constant $L$ from $(X,d)$ to $(\mathbb{R},d_{\mathbb{R}})$; that is, $|F(x)-F(y)|\le L\,d(x,y)$ for all $x,y\in X$.