Graduate Smoothness of multivariable function

Click For Summary
The discussion focuses on proving the smoothness of the function \( f^* \), defined using a bump function \( h \) and a smooth function \( f \) on a specified domain. It is established that \( f^* \) is smooth within the region \( B_{2\epsilon}^m(0) \) since it is the product of two smooth functions, and it is also smooth outside the closure of \( B_\epsilon^m(0) \) where it equals zero. The agreement of definitions in the overlapping region \( B_{2\epsilon}^m(0) - \overline{B_\epsilon}^m(0) \) further supports the conclusion. The reasoning presented confirms that \( f^* \) is indeed smooth across all of \( \mathbb{R^m} \). The argument is logically sound and effectively demonstrates the desired property.
JYM
Messages
14
Reaction score
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.
 
Physics news on Phys.org
sounds exactly right, but hard to read.
 
Thanks.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K