Theorem that I have to prove: Let $M$ be a smooth manifold, $f\colon M\to\mathbb{R}^n$ be a continuous map and let $S\subseteq M$ be a closed subset of $M$ such that the restriction $f_{|S}\colon S\to\mathbb R^n$ is smooth (this means that for all $p\in S$ there exists a neighbourhood $U_p$ of $p$ and a smooth map $g_p\colon U_p\to\mathbb{R}^n$ such that $g_{p|U_p\cap S}=f_{|U_p\cap S}$). Let $\epsilon\colon M\to (0,+\infty)$ be a continuous map. Then, there exists a smooth map $g\colon M\to\mathbb{R}^n$ such that
- $g(x)=f(x)\quad\forall x\in S$
- $\lVert {g(x)-f(x)} \rVert<\epsilon(x)\quad\forall x\in S$
In order to prove it I have to prove the fact that:
Fact 1: $\forall p\in M\quad\exists U_p$ neoighbourhood of $p$ , $g_p\colon U_p\to\mathbb R^n$ such that $g_p(x)=f(x)\qquad\forall x\in U_p\cap S\\ ||{g_p(x)-f(x)}||<\epsilon(x)\qquad\forall x\in U_p$.
My task is to use the following fact $0$ that is a consequence of theorem of partition of unity.
Fact 0: if $f\colon S\to\mathbb{R}^n$ is a smooth map, where $S\subseteq M$ is closed, then there exists a smooth map $F\colon M\to\mathbb{R}^n$ such that $F_{|S}=f$. Can anyone help me by proving FACT 1, please?
Given $F_S$ and $p$ by fact 0, it suffices to find some neighbourhood $U_p$ of $p$ where $|F_S(x)-f(x)|<\epsilon(x)$ for all $x \in U_p$, then you can just take $g_p:=F_S|_{U_p}$. There is a property of the function $h(x):=F_S(x) - f(x)$ which means you can always choose such a neighbourhood - what do you know about the function $f$, that is true on all of $M$?
The continuity of $\epsilon$ means that $\epsilon$ does not approach 0 on a sequence of points approaching $p$. Think about why this wouldn't work at $p=(0,0)$ if $M=\mathbb{R}^2$, and $\epsilon(x):=\frac{1}{n}$ at $(\frac{1}{n},0)$, $n \in \mathbb{Z}_+$ and $\epsilon(x):=1$ otherwise.