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“Counterexample” for the Inverse function theorem

Application of the Inverse Function TheoremQuestion regarding the Kolmogorov-Riesz theorem on relatively compact subsets of $L^p(Omega)$.Looking for a special kind of injective functionInverse Function Theorem and InjectivityFind all $(x,y,z) in mathbbR^3$ where $f(x,y,z)=(xy,xz,yz)$ is locally invertibleInverse Function Theorem and global inversesInverse function theorem local injectivity proofFunction satisfying $(Df(x)h,h) geq alpha(h,h), forall x,h in mathbbR^n$ has an inverse around every point?Jordan Regions and the Inverse Function TheoremHow is this not a proof of the Jacobian conjecture in the complex case?

10

1

$begingroup$

In my class we stated the theorem as follows:

Let $OmegasubseteqmathbbR^n$ be an open set and $f:OmegatomathbbR^n$ a $mathscrC^1(Omega)$ function. If $|J_f(a)|ne0$ for some $ainOmega$ then there exists $delta>0$ such that $g:=fvert_B(a,delta)$ is injective and ...

This only is a sufficient condition, so is there any function whose jacobian has determinant $0$ at every point but still is injective? If the determinant only vanished on one single point something similar to $f(x)=x^3$ at $x=0$ in $mathbbR$ would do the trick, but if $|f'(x)|=0$ for every $xinOmegasubseteqmathbbR$ then $f$ is constant and not injective. Does the same hold in $mathbbR^n$?

Thanks

real-analysis examples-counterexamples inverse-function-theorem

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asked May 15 at 21:59

PedroPedro

688213

$endgroup$

add a comment | 

10

1

$begingroup$

In my class we stated the theorem as follows:

Let $OmegasubseteqmathbbR^n$ be an open set and $f:OmegatomathbbR^n$ a $mathscrC^1(Omega)$ function. If $|J_f(a)|ne0$ for some $ainOmega$ then there exists $delta>0$ such that $g:=fvert_B(a,delta)$ is injective and ...

This only is a sufficient condition, so is there any function whose jacobian has determinant $0$ at every point but still is injective? If the determinant only vanished on one single point something similar to $f(x)=x^3$ at $x=0$ in $mathbbR$ would do the trick, but if $|f'(x)|=0$ for every $xinOmegasubseteqmathbbR$ then $f$ is constant and not injective. Does the same hold in $mathbbR^n$?

Thanks

real-analysis examples-counterexamples inverse-function-theorem

share|cite|improve this question

asked May 15 at 21:59

PedroPedro

688213

$endgroup$

add a comment | 

$begingroup$

In my class we stated the theorem as follows:

Let $OmegasubseteqmathbbR^n$ be an open set and $f:OmegatomathbbR^n$ a $mathscrC^1(Omega)$ function. If $|J_f(a)|ne0$ for some $ainOmega$ then there exists $delta>0$ such that $g:=fvert_B(a,delta)$ is injective and ...

This only is a sufficient condition, so is there any function whose jacobian has determinant $0$ at every point but still is injective? If the determinant only vanished on one single point something similar to $f(x)=x^3$ at $x=0$ in $mathbbR$ would do the trick, but if $|f'(x)|=0$ for every $xinOmegasubseteqmathbbR$ then $f$ is constant and not injective. Does the same hold in $mathbbR^n$?

Thanks

real-analysis examples-counterexamples inverse-function-theorem

share|cite|improve this question

asked May 15 at 21:59

PedroPedro

688213

$endgroup$

share|cite|improve this question

asked May 15 at 21:59

PedroPedro

688213

share|cite|improve this question

asked May 15 at 21:59

PedroPedro

688213

asked May 15 at 21:59

PedroPedro

688213

asked May 15 at 21:59

PedroPedro

688213

1 Answer
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10

$begingroup$

Actually, this is not possible in $mathbbR^n$ either.

Indeed, if you have any $mathscrC^1$ injective function $f: Omega rightarrow mathbbR^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).

From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $mathbbR^n$, thus has empty interior, thus the set of critical points has no interior as well.

share|cite|improve this answer

answered May 15 at 22:16

MindlackMindlack

5,290413

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10

$begingroup$

Actually, this is not possible in $mathbbR^n$ either.

Indeed, if you have any $mathscrC^1$ injective function $f: Omega rightarrow mathbbR^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).

From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $mathbbR^n$, thus has empty interior, thus the set of critical points has no interior as well.

share|cite|improve this answer

answered May 15 at 22:16

MindlackMindlack

5,290413

$endgroup$

add a comment | 

10

$begingroup$

Actually, this is not possible in $mathbbR^n$ either.

Indeed, if you have any $mathscrC^1$ injective function $f: Omega rightarrow mathbbR^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).

From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $mathbbR^n$, thus has empty interior, thus the set of critical points has no interior as well.

share|cite|improve this answer

answered May 15 at 22:16

MindlackMindlack

5,290413

$endgroup$

add a comment | 

$begingroup$

Actually, this is not possible in $mathbbR^n$ either.

Indeed, if you have any $mathscrC^1$ injective function $f: Omega rightarrow mathbbR^n$, then $f$ is open and a homeomorphism on its image (invariance of domain : https://en.m.wikipedia.org/wiki/Invariance_of_domain ).

From Sard’s theorem (https://en.m.wikipedia.org/wiki/Sard%27s_theorem ), the set of critical values has null measure in $mathbbR^n$, thus has empty interior, thus the set of critical points has no interior as well.

share|cite|improve this answer

answered May 15 at 22:16

MindlackMindlack

5,290413

$endgroup$

share|cite|improve this answer

answered May 15 at 22:16

MindlackMindlack

5,290413

answered May 15 at 22:16

MindlackMindlack

5,290413

answered May 15 at 22:16

MindlackMindlack

5,290413