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Thread: differentiable isometry

  1. #1
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    differentiable isometry

    Hello everybody, I am new in the forum, nice to meet you, please help me with this problem.

    Let $\displaystyle f\in{C^1(\mathbb{R}^m,\mathbb{R}^m)}$ such that: $\displaystyle \left\|{f'(x)(v)}\right\|= \left\|{v}\right\|$ for all $\displaystyle x,v\in{\mathbb{R}^m}$, i.e. $\displaystyle f'(x)$ is an isometry, for all $\displaystyle x\in{\mathbb{R}^m}$. Show that $\displaystyle \left\|{f(x)-f(y)}\right\|= \left\|{x-y}\right\|$, for all $\displaystyle x,y\in{\mathbb{R}^m}$. Conclude that there is $\displaystyle a\in{}\mathbb{R}^m$ and a $\displaystyle T\in{L(\mathbb{R}^m)}$ such that $\displaystyle f(x)=T(x)+a$.

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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by miguemate View Post
    Hello everybody, I am new in the forum, nice to meet you, please help me with this problem.

    Let $\displaystyle f\in{C^1(\mathbb{R}^m,\mathbb{R}^m)}$ such that: $\displaystyle \left\|{f'(x)(v)}\right\|= \left\|{v}\right\|$ for all $\displaystyle x,v\in{\mathbb{R}^m}$, i.e. $\displaystyle f'(x)$ is an isometry, for all $\displaystyle x\in{\mathbb{R}^m}$. Show that $\displaystyle \left\|{f(x)-f(y)}\right\|= \left\|{x-y}\right\|$, for all $\displaystyle x,y\in{\mathbb{R}^m}$. Conclude that there is $\displaystyle a\in{}\mathbb{R}^m$ and a $\displaystyle T\in{L(\mathbb{R}^m)}$ such that $\displaystyle f(x)=T(x)+a$.

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  3. #3
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    Hello Drexel, thanks for your answer, here is my work

    Using Cauchy’s inequality
    $\displaystyle \left\|{f'(x)(v)}\right\|\cdot{ \left\|{v}\right\|}\geq{\left<{f'(x)(v),v}\right>}$.
    By hypotesis, $\displaystyle \left\|{f'(x)(v)}\right\|= \left\|{v}\right\|$, then
    $\displaystyle \left\|{v}\right\|^2\geq{\left<{f'(x)(v),v}\right> }$ (1)
    Taking $\displaystyle h:[0,1]\rightarrow{R}$ by
    $\displaystyle h(t)=\left<{f((1-t)x+ty)-f(x),y-x}\right>$,
    I get $\displaystyle h'(t)=\left<{f'((1-t)x+ty)(y-x),y-x}\right>$.

    By the mean value theorem, there is a c in <0,1> such that
    $\displaystyle h(1)-h(0)=h'(c)$
    $\displaystyle \left<{f(y)-f(x),y-x}\right>=\left<{f'((1-c)x+cy)(y-x),(y-x)}\right>\leq{ \left\|{y-x}\right\|}^2$
    This latter using (1)
    Hence
    $\displaystyle \left\|{y-x}\right\|^2\geq{\left<{f(y)-f(x),y-x}\right>}$.

    That's all, but i do not what else to do

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  4. #4
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    Hello Drexwl, thanks for your answer, here is my work

    Using Cauchy’s inequality
    $\displaystyle \left\|{f'(x)(v)}\right\|\cdot{ \left\|{v}\right\|}\geq{\left<{f'(x)(v),v}\right>}$.
    By hypotesis, $\displaystyle \left\|{f'(x)(v)}\right\|= \left\|{v}\right\|$, then
    $\displaystyle \left\|{v}\right\|^2\geq{\left<{f'(x)(v),v}\right> }$ (1)
    Taking $\displaystyle h:[0,1]\rightarrow{R}$ by
    $\displaystyle h(t)=\left<{f((1-t)x+ty)-f(x),y-x}\right>$,
    I get $\displaystyle h'(t)=\left<{f'((1-t)x+ty)(y-x),y-x}\right>$.

    By the mean value theorem, there is a c in <0,1> such that
    $\displaystyle h(1)-h(0)=h'(c)$
    $\displaystyle \left<{f(y)-f(x),y-x}\right>=\left<{f'((1-c)x+cy)(y-x),(y-x)}\right>\leq{ \left\|{y-x}\right\|}^2$
    This latter using (1)
    Hence
    $\displaystyle \left\|{y-x}\right\|^2\geq{\left<{f(y)-f(x),y-x}\right>}$.

    That's all, but i do not what else to do

    Hugs
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  5. #5
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    any idea?
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