# extension of a function to a differentiable function

Mathematics Asked by user13255 on January 1, 2022

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?

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).

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

Preamble. Let $$S$$ be a closed subset of $$mathbb{R}^m$$. Let $$f:Sto mathbb{R}^n$$ be a continuous function. Let $$A$$ be a continuous function on $$S$$ taking values in the space of $$ntimes m$$ matrices. Define the function $$R: Stimes Sto mathbb{R}^n$$ 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 $$xin S$$ 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 $$F$$.

Answered by Willie Wong on January 1, 2022