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).
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 $f:Sto mathbb{R}^n$ 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
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