“Counterexample” for the Inverse function theoremApplication 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?
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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?
$begingroup$
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
asked 2 hours ago
PedroPedro
661212
$endgroup$
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
asked 2 hours ago
PedroPedro
661212
$endgroup$
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
asked 2 hours ago
PedroPedro
661212
$endgroup$
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
real-analysis examples-counterexamples inverse-function-theorem
asked 2 hours ago
PedroPedro
661212
asked 2 hours ago
PedroPedro
661212
asked 2 hours ago
PedroPedro
661212
asked 2 hours ago
PedroPedro
661212
asked 2 hours ago
PedroPedro
661212
661212
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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.
answered 2 hours ago
MindlackMindlack
5,100312
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1 Answer
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$begingroup$
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.
answered 2 hours ago
MindlackMindlack
5,100312
$endgroup$
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.
answered 2 hours ago
MindlackMindlack
5,100312
$endgroup$
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.
answered 2 hours ago
MindlackMindlack
5,100312
$endgroup$
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.
answered 2 hours ago
MindlackMindlack
5,100312
answered 2 hours ago
MindlackMindlack
5,100312
answered 2 hours ago
MindlackMindlack
5,100312
answered 2 hours ago
MindlackMindlack
5,100312
5,100312
add a comment |
add a comment |
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