Error in Statement of Inverse Function Theorem?
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$boldsymbol{6.8.2.}$ Theorem (Inverse Function Theorem). Let $G$ be an open subset of $mathbb{R}^p$ and assume $f:Gtomathbb{R}^p$ is a continuously differentiable function. If $ain G$ with $b=f(a)$ and $Df(a)$ is an invertible linear transformation, then there is an open neighborhood $U$ of $a$ such that $Omega=f(U)$ is open in $mathbb{R}^p$ and $f$ is a bijection of $U$ onto $Omega$. If $g:Omegato U$ is the inverse of $f:UtoOmega$, then $g$ is continuously differentiable and for every $zeta$ in $Omega$, $$[Dg](zeta)=[Df(g(zeta))]^{-1}.$$
In the bottom line, isn't $f(g(zeta))$ just equal to the identity map from $Omega$ to $U$ and therefore the derivative of it would just be the identity matrix, right? Am I misunderstanding something? Should the last line instead be
$$[Dg](zeta)=[Df(zeta)]^{-1} ~?$$
real-analysis multivariable-calculus inverse-function-theorem
asked Nov 18 at 18:54
user193319
2,2862922
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0
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$boldsymbol{6.8.2.}$ Theorem (Inverse Function Theorem). Let $G$ be an open subset of $mathbb{R}^p$ and assume $f:Gtomathbb{R}^p$ is a continuously differentiable function. If $ain G$ with $b=f(a)$ and $Df(a)$ is an invertible linear transformation, then there is an open neighborhood $U$ of $a$ such that $Omega=f(U)$ is open in $mathbb{R}^p$ and $f$ is a bijection of $U$ onto $Omega$. If $g:Omegato U$ is the inverse of $f:UtoOmega$, then $g$ is continuously differentiable and for every $zeta$ in $Omega$, $$[Dg](zeta)=[Df(g(zeta))]^{-1}.$$
In the bottom line, isn't $f(g(zeta))$ just equal to the identity map from $Omega$ to $U$ and therefore the derivative of it would just be the identity matrix, right? Am I misunderstanding something? Should the last line instead be
$$[Dg](zeta)=[Df(zeta)]^{-1} ~?$$
real-analysis multivariable-calculus inverse-function-theorem
asked Nov 18 at 18:54
user193319
2,2862922
I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$boldsymbol{6.8.2.}$ Theorem (Inverse Function Theorem). Let $G$ be an open subset of $mathbb{R}^p$ and assume $f:Gtomathbb{R}^p$ is a continuously differentiable function. If $ain G$ with $b=f(a)$ and $Df(a)$ is an invertible linear transformation, then there is an open neighborhood $U$ of $a$ such that $Omega=f(U)$ is open in $mathbb{R}^p$ and $f$ is a bijection of $U$ onto $Omega$. If $g:Omegato U$ is the inverse of $f:UtoOmega$, then $g$ is continuously differentiable and for every $zeta$ in $Omega$, $$[Dg](zeta)=[Df(g(zeta))]^{-1}.$$
In the bottom line, isn't $f(g(zeta))$ just equal to the identity map from $Omega$ to $U$ and therefore the derivative of it would just be the identity matrix, right? Am I misunderstanding something? Should the last line instead be
$$[Dg](zeta)=[Df(zeta)]^{-1} ~?$$
real-analysis multivariable-calculus inverse-function-theorem
asked Nov 18 at 18:54
user193319
2,2862922
$boldsymbol{6.8.2.}$ Theorem (Inverse Function Theorem). Let $G$ be an open subset of $mathbb{R}^p$ and assume $f:Gtomathbb{R}^p$ is a continuously differentiable function. If $ain G$ with $b=f(a)$ and $Df(a)$ is an invertible linear transformation, then there is an open neighborhood $U$ of $a$ such that $Omega=f(U)$ is open in $mathbb{R}^p$ and $f$ is a bijection of $U$ onto $Omega$. If $g:Omegato U$ is the inverse of $f:UtoOmega$, then $g$ is continuously differentiable and for every $zeta$ in $Omega$, $$[Dg](zeta)=[Df(g(zeta))]^{-1}.$$
In the bottom line, isn't $f(g(zeta))$ just equal to the identity map from $Omega$ to $U$ and therefore the derivative of it would just be the identity matrix, right? Am I misunderstanding something? Should the last line instead be
$$[Dg](zeta)=[Df(zeta)]^{-1} ~?$$
real-analysis multivariable-calculus inverse-function-theorem
real-analysis multivariable-calculus inverse-function-theorem
asked Nov 18 at 18:54
user193319
2,2862922
asked Nov 18 at 18:54
user193319
2,2862922
edited Nov 18 at 19:56
asked Nov 18 at 18:54
user193319
2,2862922
asked Nov 18 at 18:54
user193319
2,2862922
asked Nov 18 at 18:54
user193319
2,2862922
2,2862922
I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20
add a comment |
I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20
I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20
I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20
add a comment |
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I guess the operator $D$ must bind more tightly than parentheses, so that $Df(ldots)$ is parsed as $(Df)(ldots)$. (I'll delete this comment when the question is answered by someone who actually knows the convention!)
– Calum Gilhooley
Nov 18 at 21:20