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How to prove that $f(x)=fracxsqrtx$ is smooth?

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1

$begingroup$

I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?

Thanks!

differential-geometry smooth-manifolds

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edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

$endgroup$

add a comment | 

1

$begingroup$

I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?

Thanks!

differential-geometry smooth-manifolds

share|cite|improve this question

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

$endgroup$

add a comment | 

$begingroup$

I have a function $f(x)=fracxsqrtx$ on the open ball $B_1(0)$ in $mathbbR^n$. I have to show that this function is a diffeomorphism between $mathbbR^n$ and $B_1(0)$. I have already proved that the function is bijective with inverse $fracxsqrtx$ and it is continuous. How to prove that the function is smooth?

Thanks!

differential-geometry smooth-manifolds

share|cite|improve this question

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

$endgroup$

share|cite|improve this question

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

share|cite|improve this question

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

edited May 7 at 13:33

Asaf Karagila♦

310k33443776

asked May 7 at 8:56

sakasaka

504

asked May 7 at 8:56

sakasaka

504

2 Answers
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4

$begingroup$

You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.

share|cite|improve this answer

answered May 7 at 9:06

Loring TuLoring Tu

1912

$endgroup$

add a comment | 

2

$begingroup$

Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.

share|cite|improve this answer

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

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add a comment | 

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4

$begingroup$

You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.

share|cite|improve this answer

answered May 7 at 9:06

Loring TuLoring Tu

1912

$endgroup$

add a comment | 

4

$begingroup$

You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.

share|cite|improve this answer

answered May 7 at 9:06

Loring TuLoring Tu

1912

$endgroup$

add a comment | 

$begingroup$

You can apply the theorem that the sum, product, quotient (whenever defined), square root (of a positive smooth function), and composition of smooth functions are smooth. In your examples, all the component functions $x$, $|x|^2$, $1 pm |x|^2$, and so on are all smooth.

share|cite|improve this answer

answered May 7 at 9:06

Loring TuLoring Tu

1912

$endgroup$

share|cite|improve this answer

answered May 7 at 9:06

Loring TuLoring Tu

1912

answered May 7 at 9:06

Loring TuLoring Tu

1912

answered May 7 at 9:06

Loring TuLoring Tu

1912

2

$begingroup$

Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.

share|cite|improve this answer

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

$endgroup$

add a comment | 

2

$begingroup$

Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.

share|cite|improve this answer

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

$endgroup$

add a comment | 

$begingroup$

Just use the fact that compositions of smooth functions is smooth and ratio of two smooth functions is smooth as long as the denominator never vanishes.

share|cite|improve this answer

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

$endgroup$

share|cite|improve this answer

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673

answered May 7 at 9:04

Kavi Rama MurthyKavi Rama Murthy

82k53673