A Lipschitz function on $\mathbb{Q}$ extends uniquely to a Lipschitz function on $\mathbb{R}$ with the same constant

Regard $\mathbb{Q}$ as a subspace of $\mathbb{R}$ with the metric inherited from the usual metric (The rationals embed densely in the reals, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset), let $L \in \mathbb{R}$ with $L \ge 0$, and let $f : \mathbb{Q} \to \mathbb{R}$ be Lipschitz with constant $L$ (Lipschitz map, $\alpha$-Hölder map for rational $0 < \alpha \le 1$, and contraction), that is

$|f(p) - f(q)| \;\le\; L\,|p - q| \qquad \text{for all } p,q \in \mathbb{Q}.$

Then there is exactly one continuous $F : \mathbb{R} \to \mathbb{R}$ with $F(q) = f(q)$ for every $q \in \mathbb{Q}$, and that $F$ is again Lipschitz with the same constant $L$.

• The extension theorem gives uniform continuity; the constant is recovered separately. A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space transports a modulus of continuity, not a Lipschitz constant, so step 4.1 is genuinely needed. The mechanism is generic: a non-strict inequality that holds on a dense set and whose two sides are continuous holds everywhere, by Limits preserve non-strict inequalities.
• Nothing here is special to $\mathbb{Q}$ and $\mathbb{R}$. The same argument extends a Lipschitz map from any dense subspace of any metric space into any complete metric space, with the same constant; $\mathbb{Q} \subseteq \mathbb{R}$ is used only because it is the density statement this library already has (The rationals embed densely in the reals).
• Uniqueness is what makes "the" extension meaningful, and it needs only continuity, not the Lipschitz condition: two continuous maps agreeing on a dense set agree everywhere (A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed, For a map of metric spaces the following agree: $\varepsilon$-$\delta$ continuity everywhere, preimages of open sets are open, preimages of closed sets are closed, sequential continuity, and $f(\overline{A}) \subseteq \overline{f(A)}$).
• The hypothesis cannot be relaxed to continuity. A continuous $f : \mathbb{Q} \to \mathbb{R}$ need not extend continuously to $\mathbb{R}$ at all, and the reason is the same one that defeats every argument of this kind: continuity does not control a function along a Cauchy sequence (A uniformly continuous map sends Cauchy sequences to Cauchy sequences, $x \mapsto 1/x$ is continuous on $(0,1)$ and sends the Cauchy sequence $(1/(k+2))_{k \ge 0}$ to an unbounded one).