Smoothness of multivariable function

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SUMMARY

The discussion centers on the smoothness of the function $f^*$ defined as $f^*(x)=h(x)f(x)$ for $x\in B_{2\epsilon}^m(0)$ and $f^*(x)=0$ for $x\in \mathbb{R^m}-B_\epsilon^m(0)$. The reasoning presented confirms that $f^*$ is smooth on $\mathbb{R^m}$ by demonstrating its smoothness in the regions $B_{2\epsilon}^m(0)$ and $\mathbb{R^m}-\overline{B_\epsilon^m(0)}$, as well as the agreement of definitions in the intersection $B_{2\epsilon}^m(0)-\overline{B_\epsilon^m(0)}$. Therefore, the conclusion that $f^*$ is smooth on the entirety of $\mathbb{R^m}$ is established as correct.

PREREQUISITES
  • Understanding of bump functions in mathematical analysis
  • Knowledge of smooth functions and their properties
  • Familiarity with the notation and concepts of open and closed sets in $\mathbb{R^m}$
  • Basic comprehension of the concept of function extension
NEXT STEPS
  • Study the properties of bump functions in detail
  • Explore the implications of smoothness in multivariable calculus
  • Learn about the concept of function extension and its applications
  • Investigate the role of partitions of unity in analysis
USEFUL FOR

Mathematicians, students of advanced calculus, and anyone interested in the properties of smooth functions and their applications in multivariable analysis.

JYM
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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.
 
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sounds exactly right, but hard to read.
 
Thanks.
 

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