$\|D_{f}(x) v\|=\|v\|$ $\implies$ $f$ is an isometry

Question

Let $f\colon \mathbb{R}^m \to \mathbb{R}^m$ be a $C^2$ map such that $\|D_f(x)v\|=\|v\|$ for all $v\in\mathbb{R}^m$, where $D_f(x)$ is the derivative of $f$ at $x$. Then I am asked to prove that $\|f(x)-f(y)\|=\|x-y\|$ for all $x,y\in\mathbb{R}^m$. There's a hint that says to use the Schwarz theorem.

I just can prove that $\|f(x)-f(y)\| \le \|x-y\|$, using the image of $\alpha \colon [0,1] \to \mathbb{R}^m$, $\alpha(t)=x+(y-x)t$.

Consider $\gamma\colon [0,1] \to \mathbb{R}^m$, $\gamma(t)=f(\alpha(t))$, we have that $\gamma '(t)=D_{f}(x +(y-x)t).(y-x)$. Since $\gamma$ connect $f(x)$ and $f(y)$ and it is $C^{1}$ we have:

$\|f(x)-f(y)\| \leq L(\gamma) = \displaystyle\int_{[0,1]} \|\gamma '(t)\|dt=\displaystyle\int_{[0,1]}\|D_{f}(x +(y-x)t).(y-x)\|dt$. By assumtion this is equals to $\displaystyle\int_{[0,1]} \|x-y\| \leq \|\displaystyle\int_{[0,1]} x-y\space dt\|= \|x-y\|$.

I have no idea how use schwarz theorem in this case.

Any help is welcome.

Answer 1

$f$ is an isometry if its second partials vanish. Here's a quick explanation why: If the second partials vanish, then $D_f(\cdot)$ is constant, so that $$ \begin{eqnarray} \forall(v\in\mathbb{R}^m) \: f(v) & = & f(0) + \int_0^1 \frac{d}{dt} f(tv)\: dt \\ & =& f(0) + \int_0^1 D_f(x) \cdot v \:dt \\ & = & f(0) + \int_0^1 D_f(0) \cdot v \:dt \\ & = & f(0) + D_f(0)\cdot v. \end{eqnarray} $$ Then $\forall(v,w \in \mathbb{R}^m) \: ||f(v)-f(w)|| = ||D_f(0)v - D_f(0)w|| = ||D_f(0) (v-w)|| = ||v-w||$, as desired. The more interesting part is showing that the second partials all vanish:

Some notation: Let $f^i$ be component $i$ of $f$ (so $f=(f^1,f^2,\dots,f^m)^T$); let $f^i_j := \partial_j f^i$ be the partial of $f^i$ in direction $e_j$; and let $f_j := \partial_j f = (f^1_j,f^2_j,\dots,f^m_j)^T$.

$||D_f(x) v|| = ||v||$ for all $v$ means that $||D_f(x)||$ is an orthogonal matrix. So $D_f(x)^T D_f(x) =D_f(x) D_f(x)^T = \mathcal{I}_{m\times m}$, or in components $\sum_j f^i_j f^k_j = \delta^{ik}$ and $\sum_if^i_j f^i_k = \delta_{jk} = f_j \cdot f_k$. Apply a partial derivative to the second of these relations to get $0 = \partial_l (f_j \cdot f_k) = f_{jl} \cdot f_k + f_j \cdot f_{kl}$, so that $f_{jl} \cdot f_k = -f_j \cdot f_{kl}$, which holds for all indices $j,k,l$. Now we have $$ \begin{eqnarray} f_{jl} \cdot f_k & = & -f_j \cdot f_{kl} \\ & = & - f_{j} \cdot f_{lk} \\ & = & f_{jk} \cdot f_{l} \\ & = & f_{kj} \cdot f_{l} \\ & = & -f_{k} \cdot f_{lj} \\ & = & -f_k \cdot f_{jl} \\ \end{eqnarray} $$ so that $f_{jl}\cdot f_k = 0$. Note that I have swapped the order of differentiation (i.e. used Schwartz' theorem) numerous times above. Now since $D_f(x)$ is an orthogonal matrix, the columns $f_i$ form an orthonormal basis for $\mathbb{R}^m$. Since $f_{jk}$ is orthogonal to all $f_i$, it must therefore be the zero vector. This shows that all second partials of $f$ vanish.