[proofplan] Unlike the scalar case, applying [McShane's Extension Theorem](/theorems/3067) componentwise gives an extension that is only $\sqrt{m}\, L$-Lipschitz, not $L$-Lipschitz, because each component picks up its own constant. The standard approach (due to Valentine) extends $f$ one point at a time: by Zorn's lemma it suffices to show that for any $L$-Lipschitz $f: E \to \mathbb{R}^m$ and any $z \in \mathbb{R}^n \setminus E$, there exists $w \in \mathbb{R}^m$ such that setting $\tilde{f}(z) := w$ produces an $L$-Lipschitz extension to $E \cup \{z\}$. The existence of $w$ reduces to a geometric statement: the closed balls $\overline{B}(f(y), L|z-y|)$ for $y \in E$ have non-empty intersection. This is proved via Helly's theorem in $\mathbb{R}^m$ (see standard convex-geometry texts for Helly's theorem in $\mathbb{R}^m$) (any finite intersection is non-empty) combined with a compactness argument. The finite-intersection step is the technical core: it reduces to a quadratic-form analysis on a configuration of $m+1$ points in $\mathbb{R}^n$ and their images in $\mathbb{R}^m$. [/proofplan] [step:Dispense with the trivial case $L = 0$] If $L = 0$, then $f$ is constant: pick any $y_0 \in E$ and observe that for every $y \in E$, $|f(y) - f(y_0)| \le 0 \cdot |y - y_0| = 0$, so $f(y) = f(y_0)$. Define $F: \mathbb{R}^n \to \mathbb{R}^m$ by $F(x) := f(y_0)$ for all $x$. Then $F$ is constant, hence $0$-Lipschitz, and $F|_E = f$. This achieves $\operatorname{Lip}(F) = 0 = L$. In what follows, assume $L > 0$. [/step] [step:Reduce the global extension problem to one-point extensions via Zorn's lemma] Define the partially ordered set \begin{align*} \mathcal{P} := \{(F, A) : E \subseteq A \subseteq \mathbb{R}^n,\ F: A \to \mathbb{R}^m,\ F|_E = f,\ \operatorname{Lip}(F) \le L\}, \end{align*} ordered by extension: $(F_1, A_1) \le (F_2, A_2)$ iff $A_1 \subseteq A_2$ and $F_2|_{A_1} = F_1$. The set $\mathcal{P}$ is non-empty since it contains $(f, E)$. Any chain $\{(F_\alpha, A_\alpha)\}_{\alpha \in I}$ has an upper bound given by $A := \bigcup_\alpha A_\alpha$ and $F: A \to \mathbb{R}^m$ defined as $F(x) := F_\alpha(x)$ for any $\alpha$ with $x \in A_\alpha$. Well-definedness uses the chain condition: if $x \in A_\alpha \cap A_\beta$, then by the total ordering of the chain either $A_\alpha \subseteq A_\beta$ or $A_\beta \subseteq A_\alpha$, so $F_\alpha(x) = F_\beta(x)$. The map $F$ is $L$-Lipschitz: for $x, y \in A$ choose $\alpha, \beta$ with $x \in A_\alpha$, $y \in A_\beta$; by total order one of $A_\alpha \subseteq A_\beta$ or $A_\beta \subseteq A_\alpha$ holds, so $x, y \in A_\gamma$ for some $\gamma$, and $|F(x) - F(y)| = |F_\gamma(x) - F_\gamma(y)| \le L|x - y|$. By Zorn's lemma, $\mathcal{P}$ has a maximal element $(F^*, A^*)$. We claim $A^* = \mathbb{R}^n$. Suppose for contradiction $z \in \mathbb{R}^n \setminus A^*$. The forthcoming Steps 3 and 4 produce a value $w \in \mathbb{R}^m$ such that $|w - F^*(y)| \le L|z - y|$ for every $y \in A^*$. Defining $\tilde{F}: A^* \cup \{z\} \to \mathbb{R}^m$ by $\tilde{F}|_{A^*} := F^*$ and $\tilde{F}(z) := w$ produces an $L$-Lipschitz extension strictly larger than $(F^*, A^*)$, contradicting maximality. Hence $A^* = \mathbb{R}^n$ and $F^*$ is the desired extension. [/step] [step:Reformulate the one-point extension as a ball-intersection problem] Fix $z \in \mathbb{R}^n \setminus A^*$. Note $z \ne y$ for every $y \in A^*$, so $|z - y| > 0$. We seek $w \in \mathbb{R}^m$ such that \begin{align*} |w - F^*(y)| \le L|z - y| \quad \text{for every } y \in A^*. \end{align*} Equivalently, defining the closed balls \begin{align*} B_y := \overline{B}\bigl(F^*(y),\, L|z - y|\bigr) \subseteq \mathbb{R}^m \quad (y \in A^*), \end{align*} we seek $w \in \bigcap_{y \in A^*} B_y$. Since $L > 0$ (Step 1) and $|z - y| > 0$, every $B_y$ has strictly positive radius and hence is a non-degenerate compact convex set in $\mathbb{R}^m$. The one-point extension exists iff this infinite intersection is non-empty. [/step] [step:Reduce the infinite intersection to finite intersections via Helly's theorem (see standard convex-geometry texts for Helly's theorem in $\mathbb{R}^m$)] Fix any $y_0 \in A^*$; then $B_{y_0}$ is a compact convex subset of $\mathbb{R}^m$. The intersection $\bigcap_{y \in A^*} B_y$ is non-empty iff every finite subfamily has non-empty intersection. This is the **finite intersection property**: a family of closed subsets of a compact space has non-empty intersection iff every finite subcollection does. We apply this in the compact ambient space $B_{y_0} \subseteq \mathbb{R}^m$ to the family $\{B_y \cap B_{y_0}\}_{y \in A^*}$ of closed subsets of $B_{y_0}$. Therefore it suffices to prove \begin{align*} \bigcap_{i=1}^N B_{y_i} \ne \varnothing \end{align*} for every finite collection $y_1, \dots, y_N \in A^*$. We further reduce to $N = m+1$ by Helly's theorem (see standard convex-geometry texts for Helly's theorem in $\mathbb{R}^m$). We invoke Helly's theorem: if $\mathcal{C}_1, \dots, \mathcal{C}_N$ are convex subsets of $\mathbb{R}^m$ with $N \ge m+1$, and every $m+1$ of them have a common point, then all of them have a common point. The hypotheses are met: each $B_{y_i}$ is a closed ball, hence convex, and lives in $\mathbb{R}^m$. Therefore it suffices to verify \begin{align*} \bigcap_{i=1}^{m+1} B_{y_i} \ne \varnothing \end{align*} for every choice of $m+1$ points $y_1, \dots, y_{m+1} \in A^*$. [/step] [step:Set up the $(m+1)$-point configuration and the auxiliary extremal problem] Fix $y_1, \dots, y_{m+1} \in A^*$ (necessarily distinct from $z$). Set $a_i := y_i \in \mathbb{R}^n$, $b_i := F^*(y_i) \in \mathbb{R}^m$, and $r_i := L|z - a_i| > 0$ for $i = 1, \dots, m+1$. We must produce $w \in \mathbb{R}^m$ with $|w - b_i| \le r_i$ for $i = 1, \dots, m+1$. Define \begin{align*} \Psi: \mathbb{R}^m &\to \mathbb{R} \\ w &\mapsto \max_{1 \le i \le m+1} \frac{|w - b_i|^2}{r_i^2}. \end{align*} $\Psi$ is continuous as a finite max of continuous functions. Each function $w \mapsto |w - b_i|^2 / r_i^2$ is coercive (tends to $+\infty$ as $|w| \to \infty$), and the maximum of coercive functions is coercive, so $\Psi$ is coercive. Hence $\Psi$ attains its infimum at some $w^* \in \mathbb{R}^m$; set $\lambda^2 := \Psi(w^*) \ge 0$. Our goal is to show $\lambda \le 1$, for then $|w^* - b_i| \le r_i$ for all $i$ and $w := w^*$ is the desired common point. [/step] [step:Derive the convex-combination optimality condition via Danskin's theorem] Suppose for contradiction $\lambda > 1$. Let $I := \{i \in \{1, \dots, m+1\} : |w^* - b_i|^2 / r_i^2 = \lambda^2\}$ be the set of active indices. Each function $\varphi_i : \mathbb{R}^m \to \mathbb{R}$, $\varphi_i(w) := |w - b_i|^2 / r_i^2$, is $C^1$ and convex, with gradient $\nabla \varphi_i(w) = 2(w - b_i)/r_i^2$. The objective $\Psi = \max_i \varphi_i$ is the pointwise maximum of finitely many convex $C^1$ functions, hence convex. Apply Danskin's theorem (equivalently, the subdifferential calculus for the max of finitely many smooth convex functions): the subdifferential of $\Psi$ at $w^*$ is the convex hull of the gradients of the active functions, \begin{align*} \partial \Psi(w^*) = \operatorname{conv}\bigl\{\nabla \varphi_i(w^*) : i \in I\bigr\}. \end{align*} Since $\Psi$ is convex with finite values, the minimizer satisfies Fermat's rule for convex subdifferentials: $0 \in \partial\Psi(w^*)$. Hence there exist $\theta_i \ge 0$ for $i \in I$ with $\sum_{i \in I} \theta_i = 1$ and \begin{align*} 0 = \sum_{i \in I} \theta_i \nabla \varphi_i(w^*) = \sum_{i \in I} \theta_i \cdot \frac{2(w^* - b_i)}{r_i^2}. \end{align*} Dividing by 2 and discarding indices with $\theta_i = 0$ (we may assume $\theta_i > 0$ for $i \in I$ by restricting $I$ to the support of $\theta$), \begin{align*} \sum_{i \in I} \frac{\theta_i}{r_i^2}\,(w^* - b_i) = 0. \end{align*} Set $\mu_i := \theta_i / r_i^2 > 0$ for $i \in I$ and $T := \sum_{i \in I} \mu_i > 0$. Then \begin{align*} w^* = \frac{1}{T}\sum_{i \in I} \mu_i\, b_i, \end{align*} i.e., $w^*$ is the $\mu$-weighted centroid of $\{b_i\}_{i \in I}$. [/step] [step:Compute the target-side energy $S_b = \lambda^2$] Define \begin{align*} S_b := \sum_{i \in I} \mu_i\, |w^* - b_i|^2. \end{align*} By the active-set condition $|w^* - b_i|^2 = \lambda^2 r_i^2$ for $i \in I$, and using $\mu_i r_i^2 = (\theta_i / r_i^2) r_i^2 = \theta_i$, \begin{align*} S_b = \sum_{i \in I} \mu_i \cdot \lambda^2 r_i^2 = \lambda^2 \sum_{i \in I} \theta_i = \lambda^2 \cdot 1 = \lambda^2. \end{align*} [/step] [step:Derive $S_b \le L^2 S_a - L^2 T |z - z^*|^2$ via the centroid identity and Lipschitz bound] Define the $\mu$-weighted centroid in the source space, \begin{align*} z^* := \frac{1}{T}\sum_{i \in I} \mu_i\, a_i \in \mathbb{R}^n, \end{align*} and the source-side energy \begin{align*} S_a := \sum_{i \in I} \mu_i\, |z - a_i|^2. \end{align*} We use the **centroid identity**: for any vectors $\{x_i\}_{i \in I} \subseteq \mathbb{R}^d$ and positive weights $\{\mu_i\}_{i \in I}$ with $T = \sum_i \mu_i$, \begin{align*} \sum_{i, j \in I} \mu_i \mu_j\, |x_i - x_j|^2 = 2T \sum_{i \in I} \mu_i\, |x_i|^2 - 2\,\Bigl|\sum_{i \in I} \mu_i\, x_i\Bigr|^2. \end{align*} This follows from the expansion $|x_i - x_j|^2 = |x_i|^2 - 2 x_i \cdot x_j + |x_j|^2$, summing against $\mu_i \mu_j$, and recognising $\sum_{i,j} \mu_i \mu_j (x_i \cdot x_j) = |\sum_i \mu_i x_i|^2$. **Apply to $x_i := w^* - b_i \in \mathbb{R}^m$ for $i \in I$.** Since $w^*$ is the $\mu$-centroid of $\{b_i\}$, we have $\sum_{i \in I} \mu_i (w^* - b_i) = T w^* - T w^* = 0$. The centroid identity gives \begin{align*} \sum_{i, j \in I} \mu_i \mu_j\, |b_i - b_j|^2 = \sum_{i,j \in I} \mu_i \mu_j\, |(w^* - b_j) - (w^* - b_i)|^2 = 2T \sum_{i \in I} \mu_i\, |w^* - b_i|^2 - 0 = 2T S_b. \end{align*} **Apply to $x_i := z - a_i \in \mathbb{R}^n$ for $i \in I$.** Since $z^*$ is the $\mu$-centroid of $\{a_i\}$, we have $\sum_{i \in I} \mu_i (z - a_i) = T z - T z^* = T(z - z^*)$. The centroid identity gives \begin{align*} \sum_{i, j \in I} \mu_i \mu_j\, |a_i - a_j|^2 = \sum_{i,j \in I} \mu_i \mu_j\, |(z - a_j) - (z - a_i)|^2 = 2T \sum_{i \in I} \mu_i\, |z - a_i|^2 - 2|T(z - z^*)|^2 = 2T S_a - 2T^2 |z - z^*|^2. \end{align*} **Combine via the Lipschitz bound.** Since $F^*$ is $L$-Lipschitz on $A^*$, for all $i, j \in I$, \begin{align*} |b_i - b_j|^2 = |F^*(a_i) - F^*(a_j)|^2 \le L^2\, |a_i - a_j|^2. \end{align*} Multiplying by $\mu_i \mu_j \ge 0$ and summing over $i, j \in I$ preserves the inequality: \begin{align*} \sum_{i,j \in I} \mu_i \mu_j\, |b_i - b_j|^2 \le L^2 \sum_{i,j \in I} \mu_i \mu_j\, |a_i - a_j|^2. \end{align*} Substituting the two centroid-identity evaluations, \begin{align*} 2T S_b \le L^2 \bigl( 2T S_a - 2T^2 |z - z^*|^2 \bigr). \end{align*} Dividing by $2T > 0$: \begin{align*} S_b \le L^2 S_a - L^2 T\, |z - z^*|^2. \end{align*} [/step] [step:Compute $L^2 S_a = 1$ and obtain the contradiction $S_b \le 1$] We compute $L^2 S_a$ exactly. By definition $r_i^2 = L^2 |z - a_i|^2$, so $|z - a_i|^2 = r_i^2 / L^2$ (using $L > 0$ from Step 1). Substituting and using $\mu_i r_i^2 = \theta_i$: \begin{align*} L^2 S_a = L^2 \sum_{i \in I} \mu_i\, |z - a_i|^2 = L^2 \sum_{i \in I} \mu_i\, \frac{r_i^2}{L^2} = \sum_{i \in I} \mu_i\, r_i^2 = \sum_{i \in I} \theta_i = 1. \end{align*} Substituting into the bound from Step 7, \begin{align*} S_b \le L^2 S_a - L^2 T\, |z - z^*|^2 = 1 - L^2 T\, |z - z^*|^2 \le 1, \end{align*} where the final inequality uses $L^2 T |z - z^*|^2 \ge 0$. But Step 6 gave $S_b = \lambda^2 > 1$ (under our standing assumption $\lambda > 1$). This contradicts $S_b \le 1$. Therefore $\lambda \le 1$, which means $|w^* - b_i| \le r_i$ for all $i = 1, \dots, m+1$, so $w := w^*$ lies in $\bigcap_{i=1}^{m+1} B_{y_i}$. [/step] [step:Combine the steps to conclude the extension exists with $\operatorname{Lip}(F) = L$] Steps 4 and 5–8 verify the $(m+1)$-fold finite intersection $\bigcap_{i=1}^{m+1} B_{y_i} \ne \varnothing$ for every choice of $m+1$ points in $A^*$. By Helly's theorem (see standard convex-geometry texts for Helly's theorem in $\mathbb{R}^m$) (Step 4) every finite subfamily of $\{B_y\}_{y \in A^*}$ has non-empty intersection, and by the finite intersection property in the compact set $B_{y_0}$ (Step 4) the full intersection $\bigcap_{y \in A^*} B_y$ is non-empty. Any element $w$ of this intersection satisfies the required inequality, contradicting the maximality of $(F^*, A^*)$ in Step 2. Hence $A^* = \mathbb{R}^n$, so $F := F^*: \mathbb{R}^n \to \mathbb{R}^m$ is an $L$-Lipschitz extension of $f$. Restriction can only decrease the Lipschitz constant, so $\operatorname{Lip}(F) \ge \operatorname{Lip}(F|_E) = \operatorname{Lip}(f) = L$. Combined with $\operatorname{Lip}(F) \le L$, we conclude $\operatorname{Lip}(F) = L$. [/step]