# Riemannian submanifolds of Euclidean space admitting Lipschitz extension of Lipschitz functions, and converse statement

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?

• For future reference, it's "Lipschitz"; I have edited accordingly. I also changed "sufficient or necessary and sufficient" to just "sufficient". Mar 3, 2020 at 20:06
• 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? Mar 3, 2020 at 20:08
• 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. Mar 3, 2020 at 20:10
• @LSpice OKay. Sorry but I'm most happy with necessary and sufficient, slighly less happy with sufficient. So I'm putting it back in. Mar 3, 2020 at 20:11

## 1 Answer

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 $$R F = f_n$$, $$\|F\|_{\mathcal L_{\mathbb R^n}} \geq \frac{|f_n(y_n) - f_n(x_n)|}{\|x_n - y_n\|} = \frac{d_M(x_n,y_n)}{\|x_n - y_n\|}\geq n.$$

• Open mapping theorem guarantees boundedness from below only for injections. I think you should instead of $\mathcal{L}_{\mathbf{R}^n}$ consider $\mathcal{L}_{(M,\|\cdot\|)}$.
– erz
Mar 4, 2020 at 3:08
• Thanks, you are right of course. Edited. Mar 4, 2020 at 3:34
• @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? Mar 4, 2020 at 8:02
• 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 $d_M(x,y)$. Mar 4, 2020 at 14:37
• Mar 4, 2020 at 19:13