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Implicit function theorem implies inverse function theorem proof
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I believe this will be a very long answer if anyone tries to write the full proof or anything so I'll specify which specific parts I am having trouble with to save people's time.
I want prove three things:
First we know, $f: U to V$ where $U subset mathbb{R^n}$ and $V subset mathbb{R^n}$. And let $a in U$, s.t. $Df(a)$ is invertible.
(1) $f$ is one-to-one and onto (hence invertible)
(2) The inverse function $f^{-1}$ is of class $C^1$
(3) If $x in mathbb{R^n}$ and $y=f(x)inmathbb{R^n}$ then $D(f^{-1})(y)=(Df(x))^{-1}$.
Proof: So I know that the (2) follows directly from one of the conclusions in the implicit function theorem.
I am needing help in proving that $f$ is invertible. I know from implicit function theorem that there is a function $g$ such that $g(f(a))=a$. My idea was to prove that $f(g(b))=b$? But I'm not sure if this will work and if this approach can work not sure how to make it work. If this works we know that $g=f^{-1}$.
Also, for (3) what I got to was $Dg(y)=Dg(f(x))Df(x)$ by chain rule. And since we know that $g=f^{-1}$ we can simplify the above equation to: $Dg(y)=DxDf(x)$ but not sure how to work with the $DxDf(x)$ to $(Df(x))^{-1}$.
Explicit hints will be greatly appreciate, thanks in advance.
multivariable-calculus implicit-function-theorem inverse-function-theorem
asked Jan 26 at 0:46
javacoderjavacoder
848
$endgroup$
add a comment |
$begingroup$
I believe this will be a very long answer if anyone tries to write the full proof or anything so I'll specify which specific parts I am having trouble with to save people's time.
I want prove three things:
First we know, $f: U to V$ where $U subset mathbb{R^n}$ and $V subset mathbb{R^n}$. And let $a in U$, s.t. $Df(a)$ is invertible.
(1) $f$ is one-to-one and onto (hence invertible)
(2) The inverse function $f^{-1}$ is of class $C^1$
(3) If $x in mathbb{R^n}$ and $y=f(x)inmathbb{R^n}$ then $D(f^{-1})(y)=(Df(x))^{-1}$.
Proof: So I know that the (2) follows directly from one of the conclusions in the implicit function theorem.
I am needing help in proving that $f$ is invertible. I know from implicit function theorem that there is a function $g$ such that $g(f(a))=a$. My idea was to prove that $f(g(b))=b$? But I'm not sure if this will work and if this approach can work not sure how to make it work. If this works we know that $g=f^{-1}$.
Also, for (3) what I got to was $Dg(y)=Dg(f(x))Df(x)$ by chain rule. And since we know that $g=f^{-1}$ we can simplify the above equation to: $Dg(y)=DxDf(x)$ but not sure how to work with the $DxDf(x)$ to $(Df(x))^{-1}$.
Explicit hints will be greatly appreciate, thanks in advance.
multivariable-calculus implicit-function-theorem inverse-function-theorem
asked Jan 26 at 0:46
javacoderjavacoder
848
$endgroup$
$begingroup$
Anyone willing to help?
$endgroup$
– javacoder
Jan 26 at 20:26
add a comment |
$begingroup$
I believe this will be a very long answer if anyone tries to write the full proof or anything so I'll specify which specific parts I am having trouble with to save people's time.
I want prove three things:
First we know, $f: U to V$ where $U subset mathbb{R^n}$ and $V subset mathbb{R^n}$. And let $a in U$, s.t. $Df(a)$ is invertible.
(1) $f$ is one-to-one and onto (hence invertible)
(2) The inverse function $f^{-1}$ is of class $C^1$
(3) If $x in mathbb{R^n}$ and $y=f(x)inmathbb{R^n}$ then $D(f^{-1})(y)=(Df(x))^{-1}$.
Proof: So I know that the (2) follows directly from one of the conclusions in the implicit function theorem.
I am needing help in proving that $f$ is invertible. I know from implicit function theorem that there is a function $g$ such that $g(f(a))=a$. My idea was to prove that $f(g(b))=b$? But I'm not sure if this will work and if this approach can work not sure how to make it work. If this works we know that $g=f^{-1}$.
Also, for (3) what I got to was $Dg(y)=Dg(f(x))Df(x)$ by chain rule. And since we know that $g=f^{-1}$ we can simplify the above equation to: $Dg(y)=DxDf(x)$ but not sure how to work with the $DxDf(x)$ to $(Df(x))^{-1}$.
Explicit hints will be greatly appreciate, thanks in advance.
multivariable-calculus implicit-function-theorem inverse-function-theorem
asked Jan 26 at 0:46
javacoderjavacoder
848
$endgroup$
I believe this will be a very long answer if anyone tries to write the full proof or anything so I'll specify which specific parts I am having trouble with to save people's time.
I want prove three things:
First we know, $f: U to V$ where $U subset mathbb{R^n}$ and $V subset mathbb{R^n}$. And let $a in U$, s.t. $Df(a)$ is invertible.
(1) $f$ is one-to-one and onto (hence invertible)
(2) The inverse function $f^{-1}$ is of class $C^1$
(3) If $x in mathbb{R^n}$ and $y=f(x)inmathbb{R^n}$ then $D(f^{-1})(y)=(Df(x))^{-1}$.
Proof: So I know that the (2) follows directly from one of the conclusions in the implicit function theorem.
I am needing help in proving that $f$ is invertible. I know from implicit function theorem that there is a function $g$ such that $g(f(a))=a$. My idea was to prove that $f(g(b))=b$? But I'm not sure if this will work and if this approach can work not sure how to make it work. If this works we know that $g=f^{-1}$.
Also, for (3) what I got to was $Dg(y)=Dg(f(x))Df(x)$ by chain rule. And since we know that $g=f^{-1}$ we can simplify the above equation to: $Dg(y)=DxDf(x)$ but not sure how to work with the $DxDf(x)$ to $(Df(x))^{-1}$.
Explicit hints will be greatly appreciate, thanks in advance.
multivariable-calculus implicit-function-theorem inverse-function-theorem
multivariable-calculus implicit-function-theorem inverse-function-theorem
asked Jan 26 at 0:46
javacoderjavacoder
848
asked Jan 26 at 0:46
javacoderjavacoder
848
asked Jan 26 at 0:46
javacoderjavacoder
848
asked Jan 26 at 0:46
javacoderjavacoder
848
asked Jan 26 at 0:46
javacoderjavacoder
848
848
$begingroup$
Anyone willing to help?
$endgroup$
– javacoder
Jan 26 at 20:26
add a comment |
$begingroup$
Anyone willing to help?
$endgroup$
– javacoder
Jan 26 at 20:26
$begingroup$
Anyone willing to help?
$endgroup$
– javacoder
Jan 26 at 20:26
$begingroup$
Anyone willing to help?
$endgroup$
– javacoder
Jan 26 at 20:26
add a comment |
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$begingroup$
Anyone willing to help?
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– javacoder
Jan 26 at 20:26