In the paper Extension operators for spaces of infinite differentiable Whitney jets (J. reine angew. Math. 602 (2007), 123—154, DOI:10.1515/crelle.2007.005) by Leonhard Frerick, a convenient condition for the existence of an extension operator (for Whitney jets on compact $K\subset \mathbb{R}^n$ to $\mathbb{R}^n$; if $K$ is the closure of its interior I gather this is the same as smooth functions on $K$) is given:

Assume there exist $\varepsilon_0,\ \rho\gt 0$, $r\geq 1$ such that for all $z$ in the boundary of $K$ and $0\lt \varepsilon \lt\varepsilon_0$ there is $x\in K$ with $|x-z|\lt\varepsilon$ and $\{y: |y-x| \leq \rho\varepsilon^r\} \subset K$. Then $\mathcal{E}(K)$ satisfies (DN) and therefore it admits an extension operator.

This implies the version proved much earlier by Stein that covers the case of $Lip_1$ or $C^1$ boundary, or convex sets with non-empty interior.

I'm interested in the case where I don't just take compact sets in Euclidean space, but some more general (in fact compact) manifold $M$, possibly with boundary (or even corners). I certainly am only interested in sets $K$ that are closures of open sets, so (as far as I can tell) I'm really working with smooth functions here.

For our general compact manifold-with-boundary $M$ pick a metric $g$, giving $M$ bounded geometry. (Aside: I believe one could even find a real-analytic structure on $M$, see e.g. K. Shiga, Some aspects of real-analytic manifolds and differentiable manifolds J. Math. Soc. Japan Volume 16, Number 2 (1964), 128-142. However I'm suspicious about the existence of a real analytic metric. Once one exists, then certainly one exists giving $M$ bounded geometry by ibid.)

So I am interested to know whether given some $K\subset M$ contained in a single chart, hence diffeomorphic to $K' \subset \mathbb{R}^n$, the two ways of defining a topology on what one might call $\mathcal{E}(K)$ coincide. That is, consider $\mathcal{E}(K')$ as defined in eg Frerick's paper linked above, or consider a variant of the topology defined directly from $K$ using the chosen metric $g$ for estimates. Clearly this reduces to the case where we have transported $g$ across the chart map, so that we are reduced to considering the case of $K'$, where now $\mathbb{R}^n$ is given some metric $g'$ (and one can choose it to be of bounded geometry, as the closure of the chart in $M$ is compact). Call this space $\mathcal{E}_{g'}(K')$.

Now I'm interested in compact sets that are closures of finite intersections of geodesically convex balls in general position, hence closures of sets that are geodesically convex. If I can believe that $\mathcal{E}(K') \simeq \mathcal{E}_{g'}(K')$ as Fréchet spaces, then I would like to apply an analogue of Frerick's result, using the metric induced by $g'$ instead of the usual norm. Then, relying on geodesic convexity of my set of interest, I hope to conclude that an extension operator exists.

Does this argument pan out?

  • $\begingroup$ Note that one can consider the inverse image of the compact set $K$ that is the closure of the intersection under the exponential mapping for a nearby point, assuming all the sets involved in the construction have radius smaller than the injectivity radius, and transfer the problem completely to there. This may simplify things a little, but I don't know how much it helps. $\endgroup$
    – David Roberts
    Apr 12, 2016 at 7:59
  • $\begingroup$ I find your question difficult to understand because much notation is undefined. It looks interesting, and there is a bunch of results about extensions of more or less smooth functions, including the case when wants the extension to depend linearly on the data, with Fefferman as a major actor. web.math.princeton.edu/facultypapers/Fefferman $\endgroup$ May 6, 2016 at 7:55
  • $\begingroup$ When I wrote the question I knew approximately zero about this area, so I think I was assuming I was using standard-ish notation I had learned from the references I was pointed to just prior. I really do need the smooth setting, but perhaps some of this can be adapted to other amounts of differentiability. $\endgroup$
    – David Roberts
    May 6, 2016 at 9:30
  • $\begingroup$ @BenoîtKloeckner, the notation $\mathcal{E}(K)$ means Whitney jets on the closed set $K$ - when $K = \overline{int(K)}$, then this is just smooth functions on $K$ as usual, which is all I care about. The $\mathcal{E}_g$ stuff I just made up! $\endgroup$
    – David Roberts
    May 6, 2016 at 12:00
  • 1
    $\begingroup$ Smooth functions is slightly ambiguous: do you mean that they are smoothly extendible up to an open set containing $K$? Or that they are smooth on the interior of $K$? Otherwise, what condition do you mean along the boundary? I think no notation is completely standard here, even if it is lengthy they should be really spelled out. $\endgroup$ May 6, 2016 at 12:06

1 Answer 1


A solution to this problem has now been accepted for publication:

  • David Michael Roberts, Alexander Schmeding, Extending Whitney's extension theorem: nonlinear function spaces, to appear, Annales de l'Institut Fourier, arXiv:1801.04126.

Original answer:

OK, here is a fairly general answer based on the sledgehammer theorem of Frerick mentioned in the question, going beyond what I asked above. I will state his result in the less general case of closed sets that are closures of their interior, and which satisfy the interior corkscrew condition (this is a nonlinear version of an interior cone condition). Call these closed interior corkscrew domains, for arguments' sake, or corkscrew domains for short (though this name has been used for sets with stronger conditions).

Theorem (Frerick) Let $K \subset \mathbb{R}^n$ be a closed interior corkscrew domain. Then the surjective map of Fréchet spaces $C^\infty(\mathbb{R}^n) \to C^\infty(K)$ has a continuous linear section.

Closed interior corkscrew domains make sense in any metric space, and we shall be considering flat $\mathbb{R}^n$, and a compact Riemannian manifold $(M,g)$ with the geodesic metric.

Note that for a closed Lipschitz map $M \to N$ between metric spaces, corkscrew domains (in my sense above) are sent to corkscrew domains. Since a compact Riemannian manifold is locally uniformly bi-Lipschitz to flat $\mathbb{R}^n$, for any corkscrew domain $K$ (in the geodesic metric) in $M$ that is contained in a single chart $M \supset U \xrightarrow{\phi} \mathbb{R}^n$, we know that $\phi(K)$ is a corkscrew domain. (Note that we also know that $C^\infty(K) \simeq C^\infty(\phi(K))$, where the Fréchet seminorms are defined using the metric $g$ and the flat metric respectively; the same holds for a ball surrounding K$ in the ambient chart and in its image).

Thus there is an extension operator $C^\infty(\phi(K)) \to C^\infty(\mathbb{R}^n)$ (or, if you like, to a large ball containing $\phi(K)$), and hence an extension operator for $K$ to a neighbourhood of $K$, and hence, by a smooth partition of unity, to $M$.

Now the case above, that of geodesically convex closed sets, follows since we can show that such sets satisfying an interior (geodesic) cone condition, hence the interior corkscrew condition.

We can do this all in greater generality, for instance by using the actual condition Frerick was using (weaker than the usual corkscrew condition, and in fact quite close to uniformly polynomially cuspidal), by considering non-compact manifolds with bounded geometry, and by considering closed subsets $K$ that are larger than any one chart (the case of compact boundary is easy, but I believe that is also not necessary, given enough uniform bounds in the geometry of $M$ in the neighbourhood of he boundary). A natural setting is that of homogeneously regular manifolds (i.e. complete, of bounded geometry and positive injectivity radius). Details will be forthcoming, but I would be interested to know of analogous results, or if this is all too obvious to bother writing down.

  • $\begingroup$ Oh, and letting $M$ be a manifold with corners is, I believe, possible too. $\endgroup$
    – David Roberts
    May 5, 2016 at 23:56

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