Show that $g=fmid_{B_{1}(0)}$ is injective and determines the range of $g$ where $f(x) = Vert x Vert^{2}x$












0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$



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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$



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







share|cite|improve this question









$endgroup$





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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 24 at 3:34









Lucas CorrêaLucas Corrêa

1,6231321




1,6231321












  • $begingroup$
    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




$begingroup$
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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $x=y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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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 $x=y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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 $x=y$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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 $x=y$.






share|cite|improve this answer











$endgroup$



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 $x=y$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 24 at 23:19

























answered Jan 24 at 8:23









Kavi Rama MurthyKavi Rama Murthy

65.1k42766




65.1k42766












  • $begingroup$
    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






  • 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


















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