Mixing
Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$ where $f(x) = Vert x Vert^{2}x$
$begingroup$
Let $f: mathbb{R}^{n}to mathbb{R}^{n}$ defined by $Vert x Vert^{2} x$.
(a) Show that $f$ is of class $C^{infty}$.
(b) Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$.
(c) Prove that $g^{-1}$ (defined on $g(B_{1}(0))$) is not differentiable in $0$.
My attempt.
(a) Write $f = phicdot psi$ where $phi(x) = Vert x Vert^2$ and $psi = x$. But $D_{x}phi(h) = 2langle x : h rangle$, that is,$D_{x}phi(h) in C^{infty}$ and obviously, $psi(x) in C^{infty}$. Therefore, $f in C^{infty}$
(b) Take $x, y in B_{1}(0)setminus{0}$ such that $g(x) = g(y)$, that is, $Vert x Vert^{2}x = Vert y Vert^{2}y$. But $Vert x^{2} Vert = k_{1} leq 1$ and $Vert y Vert^{2} = k_{2} leq 1$, then $displaystyle x_{i} = frac{k_{2}}{k_{1}}y_{i}$. Is there a way to show that $displaystyle frac{k_{2}}{k_{1}} = 1$ or is there a better way to show injectivity?
Im not sure about the range of $g$. We have $Vert x Vert leq 1$ so, $Vert Vert x Vert^{2} x Vert = Vert x Vert^{3} leq 1$, then $g(B_{1}(0)) subset B_{1}(0)$?
I have not tried it since Im not sure about the range of $g$. But I think I should try to show that some of the partial derivatives are not continuous at $0$.
real-analysis derivatives
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb{R}^{n}to mathbb{R}^{n}$ defined by $Vert x Vert^{2} x$.
(a) Show that $f$ is of class $C^{infty}$.
(b) Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$.
(c) Prove that $g^{-1}$ (defined on $g(B_{1}(0))$) is not differentiable in $0$.
My attempt.
(a) Write $f = phicdot psi$ where $phi(x) = Vert x Vert^2$ and $psi = x$. But $D_{x}phi(h) = 2langle x : h rangle$, that is,$D_{x}phi(h) in C^{infty}$ and obviously, $psi(x) in C^{infty}$. Therefore, $f in C^{infty}$
(b) Take $x, y in B_{1}(0)setminus{0}$ such that $g(x) = g(y)$, that is, $Vert x Vert^{2}x = Vert y Vert^{2}y$. But $Vert x^{2} Vert = k_{1} leq 1$ and $Vert y Vert^{2} = k_{2} leq 1$, then $displaystyle x_{i} = frac{k_{2}}{k_{1}}y_{i}$. Is there a way to show that $displaystyle frac{k_{2}}{k_{1}} = 1$ or is there a better way to show injectivity?
Im not sure about the range of $g$. We have $Vert x Vert leq 1$ so, $Vert Vert x Vert^{2} x Vert = Vert x Vert^{3} leq 1$, then $g(B_{1}(0)) subset B_{1}(0)$?
I have not tried it since Im not sure about the range of $g$. But I think I should try to show that some of the partial derivatives are not continuous at $0$.
real-analysis derivatives
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39
add a comment |
$begingroup$
Let $f: mathbb{R}^{n}to mathbb{R}^{n}$ defined by $Vert x Vert^{2} x$.
(a) Show that $f$ is of class $C^{infty}$.
(b) Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$.
(c) Prove that $g^{-1}$ (defined on $g(B_{1}(0))$) is not differentiable in $0$.
My attempt.
(a) Write $f = phicdot psi$ where $phi(x) = Vert x Vert^2$ and $psi = x$. But $D_{x}phi(h) = 2langle x : h rangle$, that is,$D_{x}phi(h) in C^{infty}$ and obviously, $psi(x) in C^{infty}$. Therefore, $f in C^{infty}$
(b) Take $x, y in B_{1}(0)setminus{0}$ such that $g(x) = g(y)$, that is, $Vert x Vert^{2}x = Vert y Vert^{2}y$. But $Vert x^{2} Vert = k_{1} leq 1$ and $Vert y Vert^{2} = k_{2} leq 1$, then $displaystyle x_{i} = frac{k_{2}}{k_{1}}y_{i}$. Is there a way to show that $displaystyle frac{k_{2}}{k_{1}} = 1$ or is there a better way to show injectivity?
Im not sure about the range of $g$. We have $Vert x Vert leq 1$ so, $Vert Vert x Vert^{2} x Vert = Vert x Vert^{3} leq 1$, then $g(B_{1}(0)) subset B_{1}(0)$?
I have not tried it since Im not sure about the range of $g$. But I think I should try to show that some of the partial derivatives are not continuous at $0$.
real-analysis derivatives
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
$endgroup$
Let $f: mathbb{R}^{n}to mathbb{R}^{n}$ defined by $Vert x Vert^{2} x$.
(a) Show that $f$ is of class $C^{infty}$.
(b) Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$.
(c) Prove that $g^{-1}$ (defined on $g(B_{1}(0))$) is not differentiable in $0$.
My attempt.
(a) Write $f = phicdot psi$ where $phi(x) = Vert x Vert^2$ and $psi = x$. But $D_{x}phi(h) = 2langle x : h rangle$, that is,$D_{x}phi(h) in C^{infty}$ and obviously, $psi(x) in C^{infty}$. Therefore, $f in C^{infty}$
(b) Take $x, y in B_{1}(0)setminus{0}$ such that $g(x) = g(y)$, that is, $Vert x Vert^{2}x = Vert y Vert^{2}y$. But $Vert x^{2} Vert = k_{1} leq 1$ and $Vert y Vert^{2} = k_{2} leq 1$, then $displaystyle x_{i} = frac{k_{2}}{k_{1}}y_{i}$. Is there a way to show that $displaystyle frac{k_{2}}{k_{1}} = 1$ or is there a better way to show injectivity?
Im not sure about the range of $g$. We have $Vert x Vert leq 1$ so, $Vert Vert x Vert^{2} x Vert = Vert x Vert^{3} leq 1$, then $g(B_{1}(0)) subset B_{1}(0)$?
I have not tried it since Im not sure about the range of $g$. But I think I should try to show that some of the partial derivatives are not continuous at $0$.
real-analysis derivatives
real-analysis derivatives
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
asked Jan 24 at 3:34
Lucas CorrêaLucas Corrêa
1,5751421
1,5751421
$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39
add a comment |
$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39
$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39
$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The range is all of $mathbb R^{n}$. Clearly, $0$ is in the range. If $y neq 0$ let $x=|y|^{-2/3} y$. Then $|x||^{2}x=y$. Also, if $y in B_1(0) setminus {0}$ then $x in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $|x|^{2}x=|y|^{2}y$. Take norm on both sides. We get $|x|^{3}=|y|^{3}$ which implies $|x|=|y|$. If $x=0$ then so is $y$. Otherise, we can divide by $|x|^{3}$ to get $x=y$.
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
add a comment |
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1 Answer
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$begingroup$
The range is all of $mathbb R^{n}$. Clearly, $0$ is in the range. If $y neq 0$ let $x=|y|^{-2/3} y$. Then $|x||^{2}x=y$. Also, if $y in B_1(0) setminus {0}$ then $x in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $|x|^{2}x=|y|^{2}y$. Take norm on both sides. We get $|x|^{3}=|y|^{3}$ which implies $|x|=|y|$. If $x=0$ then so is $y$. Otherise, we can divide by $|x|^{3}$ to get $x=y$.
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
add a comment |
$begingroup$
The range is all of $mathbb R^{n}$. Clearly, $0$ is in the range. If $y neq 0$ let $x=|y|^{-2/3} y$. Then $|x||^{2}x=y$. Also, if $y in B_1(0) setminus {0}$ then $x in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $|x|^{2}x=|y|^{2}y$. Take norm on both sides. We get $|x|^{3}=|y|^{3}$ which implies $|x|=|y|$. If $x=0$ then so is $y$. Otherise, we can divide by $|x|^{3}$ to get $x=y$.
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
add a comment |
$begingroup$
The range is all of $mathbb R^{n}$. Clearly, $0$ is in the range. If $y neq 0$ let $x=|y|^{-2/3} y$. Then $|x||^{2}x=y$. Also, if $y in B_1(0) setminus {0}$ then $x in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $|x|^{2}x=|y|^{2}y$. Take norm on both sides. We get $|x|^{3}=|y|^{3}$ which implies $|x|=|y|$. If $x=0$ then so is $y$. Otherise, we can divide by $|x|^{3}$ to get $x=y$.
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
The range is all of $mathbb R^{n}$. Clearly, $0$ is in the range. If $y neq 0$ let $x=|y|^{-2/3} y$. Then $|x||^{2}x=y$. Also, if $y in B_1(0) setminus {0}$ then $x in B_1(0)$ (and $g(x)=y$) so $g$ maps $B_1(0)$ onto itself. Injectivity of $f$ and $g$: let $|x|^{2}x=|y|^{2}y$. Take norm on both sides. We get $|x|^{3}=|y|^{3}$ which implies $|x|=|y|$. If $x=0$ then so is $y$. Otherise, we can divide by $|x|^{3}$ to get $x=y$.
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
edited Jan 24 at 23:19
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 24 at 8:23
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
67.5k53067
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
add a comment |
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
$begingroup$
Thank you, Kavi! Do you have any hints for injectivity?
$endgroup$
– Lucas Corrêa
Jan 24 at 17:38
1
1
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
$begingroup$
@LucasCorrêa I have now included a proof of injectivity.
$endgroup$
– Kavi Rama Murthy
Jan 24 at 23:20
add a comment |
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$begingroup$
Pretty much. Looks like mapping the unit vectors the give the direction of the vector with a cubic resizing factor based on its norm. Also your argument about the ball centered at 0 is correct.
$endgroup$
– Lucas Henrique
Jan 24 at 3:39