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Suppose we have a map $f: S \rightarrow \mathbb{R}^{n}$, where $S \subset \mathbb{R}^{m}$, such that for each $a \in S$ there exists an $m$ by $n$ matrix $A$ such that

$\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)-Ah}{|h|} = 0.$

What conditions must be satisfied so that $f$ can be extended to a differentiable function defined on an open set containing $S$? I know that if $f$ can be locally extended to a differentiable function, then $f$ can be extended in the desired way. However, is there a more general result...possibly an if and only if condition?

1 Answers1

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If $S$ is closed, and $A$ is suitably continuous, you can apply Whitney extension theorem to extend $f$ to be real analytic outside $S$.

If $S$ is not closed, then in a neighborhood of $\bar{S}\setminus S$ you can easily construct an example of a function that is $C^\infty$ on $S$ but cannot be continuously extended to $\bar{S}$ (think $1/|x|$ on the punctured disk).

Since there are some confusion about Whitney's theorem in the comments, let me add a few lines about it here.

In the context of the current problem posed, Whitney's theorem says:

Preamble. Let $S$ be a closed subset of $\mathbb{R}^m$. Let $f:S\to \mathbb{R}^n$ be a continuous function. Let $A$ be a continuous function on $S$ taking values in the space of $n\times m$ matrices. Define the function $R: S\times S\to \mathbb{R}^n$ by $$ R(x,y) = f(y) - f(x) - A(x) \cdot (y-x). $$ Theorem. If for every $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $|x-y| < \delta$ we have $|R(x,y)| < \epsilon |x-y|$, then there exists a function $F:\mathbb{R}^m \to \mathbb{R}^n$ that is continuously differentiable, such that

  • $F = f$ on the set $S$
  • $\nabla F = A$ on the set $S$
  • $F$ is real analytic on $\mathbb{R}^m \setminus S$.

Basically, the theorem says that if you have a continuous $f$, and a continuous "putative derivative" $A$ (since $S$ is only assumed to be closed, given a point $x\in S$ it may not be possible to define the derivative of $f$ [this is the case when $x$ is not in the interior of $S$]), then provided the "putative remainder term" $R$ (remainder in the sense of Taylor polynomials) behaves like a remainder term (that when $x,y$ approaches each other it should go to zero fast enough), then you can extend $f$ to a continuously differentiable function $F$.

Willie Wong
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    For various versions of Whitney type extension theorems, a decent reference is E. Stein's Singular Integrals and Differentiability Properties of Functions, Chapter VI. – Willie Wong Jul 13 '11 at 16:23
  • It is not clear to me how to apply the linked result to the question that was asked, which wants to know if a function can be extended to a differentiable function on an open domain. I also observe that the counter-example when the domain is the already-open set $\mathbb{R}^n \setminus 0$ seems to be a mismatch for the question. – Michael Jul 24 '20 at 15:04
  • @Michael: when $S$ is closed, Whitney's extension theorem literally says that there is an extension to all of $\mathbb{R}^n$, and hence in particular there is an extension to an open neighborhood. // For the second part, I believe I understood OP's "open set containing $S$" to mean that the open set has to be strictly larger than $S$, that is to say, contains $\bar{S}$. Otherwise when $S$ is open this isn't an "extension" problem. – Willie Wong Jul 24 '20 at 15:41
  • Of course, this being from nine years ago and the OP no longer has an account on this site, it is a bit hard to ask for clarifications. – Willie Wong Jul 24 '20 at 15:44
  • For example, the Whitney extension theorem seems to talk about the extension being differentiable on $\mathbb{R}^n \setminus S$, which does not include the boundary points of $S$ when $S$ is closed. [The overall question is important because one wants to know when we can say a function $f:S\rightarrow\mathbb{R}$ is "differentiable" even when $S$ is not open. There is no issue if $S$ is already open.] – Michael Jul 24 '20 at 15:46
  • @Michael: the Whitney extension theorem's conclusion usually states that there exists a differentiable (or $n$-times differentiable) function $F$ that agrees with $f$ on $S$ with derivative agreeing with $A$ on $S$ such that $F$ is real analytic outside of $S$. That is, $F$ is differentiable everywhere with improved regularity away from $S$. (This is how the theorem is stated on Wikipedia too, by the way...) – Willie Wong Jul 24 '20 at 15:50
  • The statement on wikipedia seems to start by assuming $f$ is defined already on $\mathbb{R}^n$ (equation 1 there), though that may just be "motivation." In equation 2, it seems to define $D^{\alpha}f$, possibly at all points on the closed set of interest, and it is not clear what this means at boundary points. The statement also seems to require some form of uniform error bound. The statement is complex enough for me to not be confident in applying to the basic question of extending a real-valued function (on a closed domain) to a differentiable function on an open domain. – Michael Jul 24 '20 at 15:57
  • @Michael: the extension problem is anything but "basic"! Sure, it is basic if you know precisely what $S$ looks like, especially if $S$ is one of the basic shapes (closed ball, closed hypercube, etc.) But the geometry of $S$ can look quite horrible in general, if you only know that $S$ is a closed set. The power of Whitney's theorem is that the horrible geometry doesn't really matter. The way that the statement looks complex is precisely because at the boundary points (which could be all of $S$) one cannot just say $A = \nabla f$. See the edits above. – Willie Wong Jul 24 '20 at 17:17
  • Your edited answer is much appreciated! (+1) I also notice you added the structural assumptions that $A(x)$ is continuous and there is a uniform error bound (which were big parts of my concerns about applying the Whitney theorem to the problem). By "basic" I just meant 1-derivative problems not multi-derivative problems. – Michael Jul 24 '20 at 19:03