0
$\begingroup$

Suppose there exists a smooth function $f$ from $U \subset R^n \rightarrow R$ (smooth at every point in $U$). Then does that mean there exists a smooth function in any open set containing $U$ to $R$

I see an answer here which provided a counter example. Now as a follow up I’m interested to know is it true that there exist any open set containing $U$ such that f is smooth in that set?

Definition of “smooth at a point”

For any set $A \subset R^n$, and $a \in A$, we say that a function $ f : A \rightarrow R$ is smooth at a if there is a smooth function $g: U \rightarrow R$ defined in a neighbourhood of a such that $g=f$ on $U \cap S$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Let $n=1$ and consider the set $U=\mathbb{R}\setminus \{0\}$. Let $f$ be the function that is $0$ on $x<0$ and $1$ on $x>0$. Then $f$ is smooth, but does not have a smooth extension.

$\endgroup$
1
  • $\begingroup$ I see. As a follow up is there any open set where f has a smooth extension? I have updated the question with more details on this. $\endgroup$
    – dev
    Oct 26, 2021 at 17:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .