Inverse Function Theorem proof: f is injective
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I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:
Let $f:mathbf{X}to mathbf{Y}$ be a $mathcal{C}^k$ function between Banach spaces, and let $x^*inmathbf{X}$ and $y^*:=f(x^*)$ be such that the Fréchet derivative $mathrm{d},f(x^*):mathbf{X}tomathbf{Y}$ is bounded with bounded inverse.
Consider $F:mathbf{X}timesmathbf{Y}tomathbf{Y}$ given by $F(x,y):=f(x)-y$. Then the partial Fréchet derivative $partial_xF(x^*,y^*)equiv mathrm{d},f(x^*)$ is bounded with a bounded inverse, so $F$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $Usubseteqmathbf{X}$ and $Vsubseteqmathbf{Y}$ containing $x^*$ and $y^*$, respectively, and a $mathcal{C}^k$ function $g:V to U$ such that $F(g(y), y) = 0$ for all $yin V$, i.e. $f(g(y)) = y$.
The above shows that $g$ is a right inverse of $f$, but I want to be able to conclude, further, that $g$ is a left inverse of $f$ on $U$, i.e. $g(f(x)) = x$ for all $x in U$. I know this to be equivalent to showing that the restriction $f|_U$ is injective (possibly by choosing a smaller neighbourhood of $x^*$ in $U$), but I am not sure how to do this.
I would greatly appreciate any hints!
functional-analysis multivariable-calculus implicit-function-theorem frechet-derivative inverse-function-theorem
asked May 16 '17 at 10:38
ryan221bryan221b
9510
$endgroup$
add a comment |
$begingroup$
I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:
Let $f:mathbf{X}to mathbf{Y}$ be a $mathcal{C}^k$ function between Banach spaces, and let $x^*inmathbf{X}$ and $y^*:=f(x^*)$ be such that the Fréchet derivative $mathrm{d},f(x^*):mathbf{X}tomathbf{Y}$ is bounded with bounded inverse.
Consider $F:mathbf{X}timesmathbf{Y}tomathbf{Y}$ given by $F(x,y):=f(x)-y$. Then the partial Fréchet derivative $partial_xF(x^*,y^*)equiv mathrm{d},f(x^*)$ is bounded with a bounded inverse, so $F$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $Usubseteqmathbf{X}$ and $Vsubseteqmathbf{Y}$ containing $x^*$ and $y^*$, respectively, and a $mathcal{C}^k$ function $g:V to U$ such that $F(g(y), y) = 0$ for all $yin V$, i.e. $f(g(y)) = y$.
The above shows that $g$ is a right inverse of $f$, but I want to be able to conclude, further, that $g$ is a left inverse of $f$ on $U$, i.e. $g(f(x)) = x$ for all $x in U$. I know this to be equivalent to showing that the restriction $f|_U$ is injective (possibly by choosing a smaller neighbourhood of $x^*$ in $U$), but I am not sure how to do this.
I would greatly appreciate any hints!
functional-analysis multivariable-calculus implicit-function-theorem frechet-derivative inverse-function-theorem
asked May 16 '17 at 10:38
ryan221bryan221b
9510
$endgroup$
$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15
add a comment |
$begingroup$
I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:
Let $f:mathbf{X}to mathbf{Y}$ be a $mathcal{C}^k$ function between Banach spaces, and let $x^*inmathbf{X}$ and $y^*:=f(x^*)$ be such that the Fréchet derivative $mathrm{d},f(x^*):mathbf{X}tomathbf{Y}$ is bounded with bounded inverse.
Consider $F:mathbf{X}timesmathbf{Y}tomathbf{Y}$ given by $F(x,y):=f(x)-y$. Then the partial Fréchet derivative $partial_xF(x^*,y^*)equiv mathrm{d},f(x^*)$ is bounded with a bounded inverse, so $F$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $Usubseteqmathbf{X}$ and $Vsubseteqmathbf{Y}$ containing $x^*$ and $y^*$, respectively, and a $mathcal{C}^k$ function $g:V to U$ such that $F(g(y), y) = 0$ for all $yin V$, i.e. $f(g(y)) = y$.
The above shows that $g$ is a right inverse of $f$, but I want to be able to conclude, further, that $g$ is a left inverse of $f$ on $U$, i.e. $g(f(x)) = x$ for all $x in U$. I know this to be equivalent to showing that the restriction $f|_U$ is injective (possibly by choosing a smaller neighbourhood of $x^*$ in $U$), but I am not sure how to do this.
I would greatly appreciate any hints!
functional-analysis multivariable-calculus implicit-function-theorem frechet-derivative inverse-function-theorem
asked May 16 '17 at 10:38
ryan221bryan221b
9510
$endgroup$
I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:
Let $f:mathbf{X}to mathbf{Y}$ be a $mathcal{C}^k$ function between Banach spaces, and let $x^*inmathbf{X}$ and $y^*:=f(x^*)$ be such that the Fréchet derivative $mathrm{d},f(x^*):mathbf{X}tomathbf{Y}$ is bounded with bounded inverse.
Consider $F:mathbf{X}timesmathbf{Y}tomathbf{Y}$ given by $F(x,y):=f(x)-y$. Then the partial Fréchet derivative $partial_xF(x^*,y^*)equiv mathrm{d},f(x^*)$ is bounded with a bounded inverse, so $F$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $Usubseteqmathbf{X}$ and $Vsubseteqmathbf{Y}$ containing $x^*$ and $y^*$, respectively, and a $mathcal{C}^k$ function $g:V to U$ such that $F(g(y), y) = 0$ for all $yin V$, i.e. $f(g(y)) = y$.
The above shows that $g$ is a right inverse of $f$, but I want to be able to conclude, further, that $g$ is a left inverse of $f$ on $U$, i.e. $g(f(x)) = x$ for all $x in U$. I know this to be equivalent to showing that the restriction $f|_U$ is injective (possibly by choosing a smaller neighbourhood of $x^*$ in $U$), but I am not sure how to do this.
I would greatly appreciate any hints!
functional-analysis multivariable-calculus implicit-function-theorem frechet-derivative inverse-function-theorem
functional-analysis multivariable-calculus implicit-function-theorem frechet-derivative inverse-function-theorem
asked May 16 '17 at 10:38
ryan221bryan221b
9510
asked May 16 '17 at 10:38
ryan221bryan221b
9510
edited Dec 3 '18 at 17:59
ryan221b
asked May 16 '17 at 10:38
ryan221bryan221b
9510
asked May 16 '17 at 10:38
ryan221bryan221b
9510
asked May 16 '17 at 10:38
ryan221bryan221b
9510
9510
$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15
add a comment |
$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15
$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15
add a comment |
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$begingroup$
Doesn't the classical formulation of the implicit function theorem say $f:Xoplus Y to Z$ etc
$endgroup$
– JJR
May 16 '17 at 11:11
$begingroup$
@JJR Are you referring to my writing '$X times Y$' instead of '$X oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W subseteq X times Y$ (I have taken $W = X times Y$ for simplicity).
$endgroup$
– ryan221b
May 16 '17 at 13:10
$begingroup$
Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around
$endgroup$
– JJR
May 16 '17 at 13:15