Proof of Lipschitz Extension of a Real Function from a Subset of a Metric Space

The extension is given by an explicit infimum.

Fix $u_{0}\in D$, which exists as $D\neq\emptyset$. For $x\in X$ let

$$S_{x}=\{f(u)+L\,d(x,u):u\in D\},$$

a nonempty set of real numbers. It is bounded below: for $u\in D$ the hypothesis gives $f(u)\ge f(u_{0})-L\,d(u_{0},u)$ (with $d(u,u_{0})=d(u_{0},u)$ by Metric Space (iv)), and $d(u_{0},u)\le d(u_{0},x)+d(x,u)$ by Metric Space (v), so, as $L\ge0$,

$$f(u)+L\,d(x,u)\ge f(u_{0})-L\,d(u_{0},u)+L\,d(x,u)\ge f(u_{0})-L\,d(u_{0},x).$$

Hence the infimum

$$F(x)=\inf S_{x}$$

exists by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (viii), and this defines a function $F:X\to\mathbb{R}$.

The function $F$ extends $f$. Let $u\in D$. Taking $u$ itself as the point of $D$ in $S_{u}$ gives $F(u)\le f(u)+L\,d(u,u)=f(u)$, since $d(u,u)=0$ by Metric Space (iii). Conversely, the hypothesis gives $f(u)\le f(v)+L\,d(u,v)$ for every $v\in D$, so $f(u)$ is a lower bound of $S_{u}$, and $f(u)\le F(u)$ because the infimum is the greatest lower bound (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv)). Hence $F(u)=f(u)$.

Finally, $F$ is $L$-Lipschitz. Let $x,y\in X$. For every $v\in D$, by Metric Space (v),

$$F(x)\le f(v)+L\,d(x,v)\le f(v)+L\,d(y,v)+L\,d(x,y).$$

So $F(x)-L\,d(x,y)$ is a lower bound of $S_{y}$, whence $F(x)-L\,d(x,y)\le F(y)$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order (iv). Exchanging $x$ and $y$, and using $d(y,x)=d(x,y)$ (Metric Space (iv)), gives $F(y)-L\,d(x,y)\le F(x)$; together, $|F(x)-F(y)|\le L\,d(x,y)$. This proves (i).