Mixing
$C^1$-diffeomorphism $implies parallel (Dvarphi)^{-1} parallel$ is bounded
$begingroup$
Let $U,V$ be open subsets of $mathbb{R}^n$ and $varphi:U rightarrow V$ a $C^1$-diffeomorphism. We know that $varphi, D varphi$ and $(Dvarphi)^{-1}$ are defined on compact set $K$ with $U subset K$. This implies that $varphi$ is uniformly continuous on $K$ and also on $U$.
I would like to know how one can see that $parallel (Dvarphi)^{-1} parallel$ and $mid det(D varphi)mid$ are bounded by a constant $M$.
However what I know is that for $n=1$ a continuously differentiable function is Lipschitz continuous if its first derivative is bounded. This is easy to see from the mean value theorem.
I dont know if this is getting me somewhere.
For the $mid det(Dvarphi) mid$ part I guess its just a property of diffeomorphism but I dont know how to show this.
calculus diffeomorphism
asked Jan 7 at 15:40
ArjihadArjihad
390111
$endgroup$
add a comment |
$begingroup$
Let $U,V$ be open subsets of $mathbb{R}^n$ and $varphi:U rightarrow V$ a $C^1$-diffeomorphism. We know that $varphi, D varphi$ and $(Dvarphi)^{-1}$ are defined on compact set $K$ with $U subset K$. This implies that $varphi$ is uniformly continuous on $K$ and also on $U$.
I would like to know how one can see that $parallel (Dvarphi)^{-1} parallel$ and $mid det(D varphi)mid$ are bounded by a constant $M$.
However what I know is that for $n=1$ a continuously differentiable function is Lipschitz continuous if its first derivative is bounded. This is easy to see from the mean value theorem.
I dont know if this is getting me somewhere.
For the $mid det(Dvarphi) mid$ part I guess its just a property of diffeomorphism but I dont know how to show this.
calculus diffeomorphism
asked Jan 7 at 15:40
ArjihadArjihad
390111
$endgroup$
add a comment |
$begingroup$
Let $U,V$ be open subsets of $mathbb{R}^n$ and $varphi:U rightarrow V$ a $C^1$-diffeomorphism. We know that $varphi, D varphi$ and $(Dvarphi)^{-1}$ are defined on compact set $K$ with $U subset K$. This implies that $varphi$ is uniformly continuous on $K$ and also on $U$.
I would like to know how one can see that $parallel (Dvarphi)^{-1} parallel$ and $mid det(D varphi)mid$ are bounded by a constant $M$.
However what I know is that for $n=1$ a continuously differentiable function is Lipschitz continuous if its first derivative is bounded. This is easy to see from the mean value theorem.
I dont know if this is getting me somewhere.
For the $mid det(Dvarphi) mid$ part I guess its just a property of diffeomorphism but I dont know how to show this.
calculus diffeomorphism
asked Jan 7 at 15:40
ArjihadArjihad
390111
$endgroup$
Let $U,V$ be open subsets of $mathbb{R}^n$ and $varphi:U rightarrow V$ a $C^1$-diffeomorphism. We know that $varphi, D varphi$ and $(Dvarphi)^{-1}$ are defined on compact set $K$ with $U subset K$. This implies that $varphi$ is uniformly continuous on $K$ and also on $U$.
I would like to know how one can see that $parallel (Dvarphi)^{-1} parallel$ and $mid det(D varphi)mid$ are bounded by a constant $M$.
However what I know is that for $n=1$ a continuously differentiable function is Lipschitz continuous if its first derivative is bounded. This is easy to see from the mean value theorem.
I dont know if this is getting me somewhere.
For the $mid det(Dvarphi) mid$ part I guess its just a property of diffeomorphism but I dont know how to show this.
calculus diffeomorphism
calculus diffeomorphism
asked Jan 7 at 15:40
ArjihadArjihad
390111
asked Jan 7 at 15:40
ArjihadArjihad
390111
asked Jan 7 at 15:40
ArjihadArjihad
390111
asked Jan 7 at 15:40
ArjihadArjihad
390111
asked Jan 7 at 15:40
ArjihadArjihad
390111
390111
add a comment |
add a comment |
1 Answer
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$begingroup$
We say $varphi : U rightarrow V$ between $U,V$ open is a diffeomorphism if $varphi$ is $C^1$ and has a $C^1$ inverse $varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.
Indeed consider the map,
$$ varphi : (-pi/2,pi/2) rightarrow mathbb R, $$
which is the restriction of $tan$ to $(-pi/2,pi/2).$ This map is a diffeomorphism, but $varphi'(x) = sec^2(x)$ blows up as $x$ tends to $pmpi/2.$
Additionally your assertion about the existence of a compact subset $K supset U$ is not true in general. Indeed we see $varphi$ above does not even admit a continuous extension $[-pi/2,pi/2] rightarrow mathbb R.$
What we have is that $lVert Dvarphi rVert,$ $lVert Dvarphi^{-1}rVert,$ $|det Dvarphi|$ and $|det Dvarphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.
answered Jan 7 at 16:44
ktoiktoi
2,3961616
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add a comment |
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1 Answer
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$begingroup$
We say $varphi : U rightarrow V$ between $U,V$ open is a diffeomorphism if $varphi$ is $C^1$ and has a $C^1$ inverse $varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.
Indeed consider the map,
$$ varphi : (-pi/2,pi/2) rightarrow mathbb R, $$
which is the restriction of $tan$ to $(-pi/2,pi/2).$ This map is a diffeomorphism, but $varphi'(x) = sec^2(x)$ blows up as $x$ tends to $pmpi/2.$
Additionally your assertion about the existence of a compact subset $K supset U$ is not true in general. Indeed we see $varphi$ above does not even admit a continuous extension $[-pi/2,pi/2] rightarrow mathbb R.$
What we have is that $lVert Dvarphi rVert,$ $lVert Dvarphi^{-1}rVert,$ $|det Dvarphi|$ and $|det Dvarphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.
answered Jan 7 at 16:44
ktoiktoi
2,3961616
$endgroup$
add a comment |
$begingroup$
We say $varphi : U rightarrow V$ between $U,V$ open is a diffeomorphism if $varphi$ is $C^1$ and has a $C^1$ inverse $varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.
Indeed consider the map,
$$ varphi : (-pi/2,pi/2) rightarrow mathbb R, $$
which is the restriction of $tan$ to $(-pi/2,pi/2).$ This map is a diffeomorphism, but $varphi'(x) = sec^2(x)$ blows up as $x$ tends to $pmpi/2.$
Additionally your assertion about the existence of a compact subset $K supset U$ is not true in general. Indeed we see $varphi$ above does not even admit a continuous extension $[-pi/2,pi/2] rightarrow mathbb R.$
What we have is that $lVert Dvarphi rVert,$ $lVert Dvarphi^{-1}rVert,$ $|det Dvarphi|$ and $|det Dvarphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.
answered Jan 7 at 16:44
ktoiktoi
2,3961616
$endgroup$
add a comment |
$begingroup$
We say $varphi : U rightarrow V$ between $U,V$ open is a diffeomorphism if $varphi$ is $C^1$ and has a $C^1$ inverse $varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.
Indeed consider the map,
$$ varphi : (-pi/2,pi/2) rightarrow mathbb R, $$
which is the restriction of $tan$ to $(-pi/2,pi/2).$ This map is a diffeomorphism, but $varphi'(x) = sec^2(x)$ blows up as $x$ tends to $pmpi/2.$
Additionally your assertion about the existence of a compact subset $K supset U$ is not true in general. Indeed we see $varphi$ above does not even admit a continuous extension $[-pi/2,pi/2] rightarrow mathbb R.$
What we have is that $lVert Dvarphi rVert,$ $lVert Dvarphi^{-1}rVert,$ $|det Dvarphi|$ and $|det Dvarphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.
answered Jan 7 at 16:44
ktoiktoi
2,3961616
$endgroup$
We say $varphi : U rightarrow V$ between $U,V$ open is a diffeomorphism if $varphi$ is $C^1$ and has a $C^1$ inverse $varphi^{-1}.$ If we are only given this, then we may not be guaranteed boundedness of the first derivative globally.
Indeed consider the map,
$$ varphi : (-pi/2,pi/2) rightarrow mathbb R, $$
which is the restriction of $tan$ to $(-pi/2,pi/2).$ This map is a diffeomorphism, but $varphi'(x) = sec^2(x)$ blows up as $x$ tends to $pmpi/2.$
Additionally your assertion about the existence of a compact subset $K supset U$ is not true in general. Indeed we see $varphi$ above does not even admit a continuous extension $[-pi/2,pi/2] rightarrow mathbb R.$
What we have is that $lVert Dvarphi rVert,$ $lVert Dvarphi^{-1}rVert,$ $|det Dvarphi|$ and $|det Dvarphi^{-1}|$ are locally bounded. This follows because all these quantities are continuous on their respective domains, and hence their restrictions to compact subsets are bounded.
answered Jan 7 at 16:44
ktoiktoi
2,3961616
answered Jan 7 at 16:44
ktoiktoi
2,3961616
answered Jan 7 at 16:44
ktoiktoi
2,3961616
answered Jan 7 at 16:44
ktoiktoi
2,3961616
2,3961616
add a comment |
add a comment |
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