- #1
JYM
- 14
- 0
Let $h$ be a bump function that is $0$ outside $B_\epsilon^m(0)$ and posetive on its interior.
Let $f$ be smooth function on $B_{2\epsilon}^m(0)$.
Define $f^*(x)=h(x)f(x)$ if $x\in B_{2\epsilon}^m(0)$ and $=0$ if $x\in \mathbb{R^m}-B_\epsilon^m(0)$.
I want to show that $f^*$ is smooth on $\mathbb{R^m}$.
I proceed as follows.
Clearly $f^*$ is smooth on $x\in B_{2\epsilon}^m(0)$ as it is equal to $hf$ there, and also smooth on $\mathbb{R^m}-\overline{B_\epsilon^m(0)}$ as it is zero there.
Moreover, on the open intersection $B_{2\epsilon}^m(0)-\overline{B_\epsilon^m(0)}$ both definitions agree.
Thus $f^*$ is smooth on $\mathbb{R^m}$. Is my reasoning correct? Thanks.
Let $f$ be smooth function on $B_{2\epsilon}^m(0)$.
Define $f^*(x)=h(x)f(x)$ if $x\in B_{2\epsilon}^m(0)$ and $=0$ if $x\in \mathbb{R^m}-B_\epsilon^m(0)$.
I want to show that $f^*$ is smooth on $\mathbb{R^m}$.
I proceed as follows.
Clearly $f^*$ is smooth on $x\in B_{2\epsilon}^m(0)$ as it is equal to $hf$ there, and also smooth on $\mathbb{R^m}-\overline{B_\epsilon^m(0)}$ as it is zero there.
Moreover, on the open intersection $B_{2\epsilon}^m(0)-\overline{B_\epsilon^m(0)}$ both definitions agree.
Thus $f^*$ is smooth on $\mathbb{R^m}$. Is my reasoning correct? Thanks.