Let $M \subset \mathbb{R}^p$ be a Riemannian submanifold. In what follows, when we talk about a Lipschitz function $f$ on $M$, namely $f: M \to \mathbb{R}$, we will assume there is a $L > 0$ so that:

$$ \lvert f(x) - f(y)\rvert \leq Ld_M(x,y)\quad \forall x, y \in M. $$

My question is: What is either a necessary and sufficient condition (preferred) OR if not possible, then at least a sufficient condition that every Lipschitz function on $M$ can be extended to a Lipschitz function on $\mathbb{R}^p$? So to write mathematically, what is or are sufficient condition(s) on $M$ so that for every Lipschitz $f: M \to \mathbb{R}$, there exists an extension $\tilde{f}$ of $f$ so that $\tilde{f}$ is also Lipschitz (possibly with a bigger Lipschitz constant)?

On the other side, what is a necessary and sufficient condition on $M$ so that every Lipschitz function on $\mathbb{R}^p$ restricts to a Lipschitz function on $M$? I feel like for a compact manifold, this will be true, as there we will have:

$$C_1\lVert x- y\rVert \leq d_M(x,y) \leq C_2\lVert x-y\rVert,$$

because the exponential map is smooth, so the norm of its gradient will have an upper and a lower bound on compact $M$, resulting in the above inequality. Is it true though?

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    $\begingroup$ For future reference, it's "Lipschitz"; I have edited accordingly. I also changed "sufficient or necessary and sufficient" to just "sufficient". $\endgroup$
    – LSpice
    Mar 3, 2020 at 20:06
  • $\begingroup$ No problme with Lipscitz, thanks. May I ask why are you editing the necessary and sufficient part? It'd be nice to have an equivalent condition, no? $\endgroup$ Mar 3, 2020 at 20:08
  • $\begingroup$ I agree, but asking for a condition that is sufficient or necessary and sufficient is the same as asking for a condition that is sufficient (and may also be necessary). If you think it makes it clearer, then please feel free to put it back in. $\endgroup$
    – LSpice
    Mar 3, 2020 at 20:10
  • $\begingroup$ @LSpice OKay. Sorry but I'm most happy with necessary and sufficient, slighly less happy with sufficient. So I'm putting it back in. $\endgroup$ Mar 3, 2020 at 20:11

1 Answer 1


Say $M$ admits universal Lipschitz extension if, for any Lipschitz $f : M \to \mathbb R$, there exists Lipschitz$F : \mathbb R^n \to \mathbb R$ such that $F|_M = f$. $M$ admits universal Lipschitz extension if and only if there exists $C$ such that $d_M(x,y) \leq C \|x-y\|$ for all $x,y \in M$ (note that it always holds that $\|x-y\| \leq d_M(x,y)$).

In the forward direction, suppose $d_M(x,y) \leq C \|x-y\|$ for all $x,y \in M$. Let $f : M \to \mathbb R$ be $L$-Lipschitz with respect to $d_M$. Then $f$ is $CL$-Lipschitz with respect to the Euclidean metric on $M$, and therefore by McShane's theorem, the function $$F(x) = \sup\{f(y) + CL \|y-x\| : y \in M\}$$ is $CL$ Lipschitz and extends $f$.

For the other direction, suppose without loss of generality that $0 \in M$. Let $\mathcal L_M$ be the Banach space of Lipschitz functions $f$ on $M$ (Lipschitz with respect to $d_M$) such that $f(0) = 0$, equipped with the norm $\|f\|_{\mathcal L_M} = \text{Lip}(f)$. Let $\mathcal L_{\mathbb R^n}$ be the Banach space o\f Lipschitz functions on $\mathbb R^n$ equipped with the Lipschitz norm $\|f\|_{\mathcal L_{\mathbb R^n}}$. The restriction map $R : \mathcal L_{\mathbb R^n} \to \mathcal L_M$, where $R(F) = F|_M$, is $1$-Lipschitz. Provided that $M$ admits universal Lipschitz extension, $R$ is surjective, and so by the open mapping theorem there exists $C \in (0,\infty)$ such that for any $f \in \mathcal L_M$, there exists $F \in \mathcal L_{\mathbb R^n}$ such that $RF = f$ and $\|F\|_{\mathcal L_{\mathbb R^n}} \leq C \|f\|_{\mathcal L_M}$. Now, suppose for contradiction that there exist $x_n,y_n \in M$ such that $d_M(x_n,y_n) \geq n \|x_n - y_n\|$. The function $f_n(y) = d_M(x_n,y) - d(x_n,0)$ is in $\mathcal L_M$ with $\|f_n\|_{\mathcal L_m} = 1$, but for any $F \in \mathcal L_{\mathbb R^n}$ with $R F = f_n$, $$\|F\|_{\mathcal L_{\mathbb R^n}} \geq \frac{|f_n(y_n) - f_n(x_n)|}{\|x_n - y_n\|} = \frac{d_M(x_n,y_n)}{\|x_n - y_n\|}\geq n.$$

  • $\begingroup$ Open mapping theorem guarantees boundedness from below only for injections. I think you should instead of $\mathcal{L}_{\mathbf{R}^n}$ consider $\mathcal{L}_{(M,\|\cdot\|)}$. $\endgroup$
    – erz
    Mar 4, 2020 at 3:08
  • $\begingroup$ Thanks, you are right of course. Edited. $\endgroup$ Mar 4, 2020 at 3:34
  • $\begingroup$ @Justthisguy Thank you for the answer! Quick question tio brush up my geometry: in the last line of the first paragraph of your answer, you wrote: "note that it always holds that $∥x−y∥≤d_M(x,y)$". Why is it true? $\endgroup$ Mar 4, 2020 at 8:02
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    $\begingroup$ The distance on $M$ is the inf of the lengths of paths in $M$ connecting $x$ and $y$. The distance in $\mathbb R^n$ between $x$ and $y$ is the inf over the lengths of paths in $\mathbb R^n$ connecting $x$ and $y$, so it is in the inf over a larger set and therefore bounded above by $d_M(x,y)$. $\endgroup$ Mar 4, 2020 at 14:37
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ Mar 4, 2020 at 19:13

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