# differentiable isometry

• Feb 1st 2010, 06:58 PM
miguemate
differentiable isometry
Hello everybody, I am new in the forum, nice to meet you, please help me with this problem.

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

Thanks (Rofl)(Rofl)(Rofl)
• Feb 1st 2010, 06:59 PM
Drexel28
Quote:

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

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

Thanks (Rofl)(Rofl)(Rofl)

What have you tried?
• Feb 1st 2010, 07:13 PM
miguemate

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

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

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

Hugs
• Feb 1st 2010, 07:14 PM
miguemate

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

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

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

Hugs
• Feb 9th 2010, 09:28 PM
miguemate
any idea?