Smooth map on a “non-open” subset

Let $A\subset\mathbb{R}^n$ be a subset (not necessarily open) and $f:A\to\mathbb{R}$ be a map. As far as I have read, there are two definitions of the smoothness of $f$.

[Definition 1] The map $f$ is smooth if and only if there exists an open set $U\subset\mathbb{R}^n$ and a smooth map $F:U\to\mathbb{R}$ such that $A\subset U$ and $f=F$ on $A\cap U$.

[Definition 2] The map $f$ is smooth if and only if for every $x\in\mathbb{R}^n$ there exists an open neighborhood $U$ of $x$ and a smooth map $F:U\to\mathbb{R}$ such that $f=F$ on $A\cap U$.

Maybe Definition 1 is more usual than Definition 2. Definition 2 is seen in Lee's "Introduction to smooth manifolds."

Now, I have a question. Are these two definitions equivalent?

The implication [Definition 1]$\Rightarrow$[Definition 2] is clear. But the converse does not seem to be obvious. Is there any counterexample?

1 Answer

Yes, they are equivalent. Suppose $f : A \to \mathbb R$ satisfies definition 2, and for each $x \in A$ denote by $F_x : U_x \to \mathbb R$ the smooth extension of $f$ defined on a neighbourhood of $x$. The family $\{U_x\}_{x \in A}$ is an open cover of the manifold $U = \bigcup_{x \in A} U_x;$ so there is a corresponding locally finite partition of unity $\{\phi_x\}_{x \in A}$ such that $\sum_x \phi_x = 1$ and $\phi_x$ is supported in $U_x.$

Defining $F : U \to \mathbb R$ by $F = \sum_x \phi_x F_x$ and remembering that for $x,y\in A$ we have $F_x(y) = f(y),$ we see that $y \in A$ implies $$F(y) = \sum_x \phi_x(y) f(y) =\left(\sum_x \phi_x(y)\right) f(y) = f(y)$$ as desired. Since the partition of unity is locally finite, $F$ is locally a finite sum of smooth functions and thus is smooth.

• Thank you for your answer. It is easy to understand. I didn't hit on the partition of unity. – user356126 Dec 20 '17 at 7:58