In the book of Analysis on Manifolds, by Munkres, at page 144, question 3, it is asked that
Questions:
First of all, I was writing this question to ask whether my proof was correct or not, but down the road, I have realised that I have made a mistake, so now I have 2 questions:
1-) How to proceed from where I have left ?
2-) Is the part of the proof that I have already established correct ?
Note: I'm aware of this question, but it has no answer.
Edit:
I just want to point out that, as you can guess, I'm studying real analysis.
Unfinished Proof:
We first give a short lemma, and then prove the part a-), and then part b-)
Lemma: If $\phi: U -> \mathbb{R} $, where $U$ is open in $\mathbb{R}^n$, is continuous, s.t $S_\phi = Support \phi \subset U$, then $\phi(x) = 0$ for $x \in \partial S_\phi.$
Assume the opposite, i.e for some $x \in \partial S_\phi$, $\phi(x) = a \not = 0$, then, choose $0<\epsilon < a$, so let $0< \delta $ be arbitrary, and consider $y \in Ext(S_\phi)$ and $0 < |x-y| < \delta$. it is clear that $|\phi (x) - \phi (y) |= |a-0| > \epsilon$, hence a contradiction.QED
Now, lets prove the part a-).
We first show the continuity of $h$, and then use induction on $r$.
Since $S_\phi$ is compact and $g$ is cont. on $S_\phi \subset U$, $g$ is bounded, but as we have shown in the lemma, $\phi (x) = 0$ for $x\in \partial S_\phi$, hence $\phi(x) g(x) = 0$ for $x\in \partial S_\phi$, hence $h$ is cont. on $\mathbb{R}^n$.
Now, by induction hypothesis, let the result hold for some $n < r$, then
$D^{n+1} h(x) = \begin{cases} D^{n+1} (\phi * g) & x \in S_\phi \\ 0 & otherwise \end{cases} $
It is clear that $D^{n+1} h(x)$ is well-defined and exists everywhere. Moreover, $S_{D^{n+1} \phi * g} \subset S_\phi$, by the same method before, we can show that this function is again cont. hence $D^{n+1} h(x)$ is cont. . Hence by induction, we have desired result. QED
We now prove the part b-).
Lets define $$\mathscr{A} = \{ U_x \subset \mathbb{R}^n | U_x \text{is an open nbd of } x\in S \}.$$
and $$A = \bigcup_{U_x \in \mathscr{A}} U_x.$$
It is clear that the sets $U_x$s form a cover for $S$. By a previous theorem on the book, we can find a partition of unit $\phi_i $on $A$ dominated by $\mathscr{A}$, s.t each $U_x$ intersect with finitely many $Support \phi_i = S_{\phi_i}$.
Hence, we define $h$ as
$ h(x) = \begin{cases} (\phi = \sum_i \phi_i)(x) \cdot g(x) & x \in U_x \\ 0 & otherwise \end{cases} ,$ where $g$ comes from the hypothesis that $f$ is differentiable at $x$. By the construction of $\phi_i$s, (it is given in the theorem that is mentioned earlier), each $S_{\phi_i}$ is contained in one of $U_y$s, so for a fix $x \in S$, there is only finitely many $\phi_i (x) \not = 0$ , but if we can show that $Support of [\phi = \sum_i \phi_i)(x)]$ is contained in $U_x$, we could use the previous part to conclude the theorem, however, it does not have to be the case, so what can we do at this point ?