Prove that function is bi-lipshitz given it'a differential is zero Let $f$ be a function $f:R^n -> R^n$ which is differentially continuous. Moreover, it is given that in the origin: $D_f(0)=Id$, which is the identity function. 
I need to prove that around the origin, there exist $A,B>0$ that satisfy:
$$A||x-y|| \le ||f(x)-f(y)|| \le B||x-y||$$
I tried to maybe somehow use the Implicit Function Theorem, but no luck yet.
If someone could give me a hint or some direction, it will be very appreciated.
 A: By the Inverse Function Theorem, there exists two open sets $U$ and $V$ such that $0\in U$, $f(U)=V$ and the restriction $f:U \to V$ is a diffeomorphism. Let $\varepsilon$ be a positive real number such that the closed ball $\overline{B_\varepsilon(0)}$ is included in $U$. Recall that the norm of $||\cdot||$ of $\mathbb{R}^n$ induces a norm on the vector space $\mathbb{R}^{n\times n}$ of real matrices of size $n\times n$ defined by
$$
||A||=\sup\{||Ax||:||x||=1\},
$$
for every $A$ in $\mathbb{R}^{n\times n}$. The main property of this induced norm is the inequality $||Ax||\leq ||A||.||x||$, which holds for every matrix $A$ and every vector $x$. Now, the functions $\Phi:\xi\in\overline{B_\varepsilon(0)}\mapsto Df(\xi)\in \mathbb{R}^{n\times n}$ and $\Psi:\xi\in \overline{B_\varepsilon(0)}\mapsto Df^{-1}(f(\xi))\in \mathbb{R}^{n\times n}$ are continuous and they are defined on a compact space. So there exists positive real numbers $A$ and $C$ such that
$$
||\Phi(\xi)||=||Df(\xi)||\leq A \quad\text{ and }\quad ||\Psi(\xi)||=||Df^{-1}(f(\xi))||\leq C,
$$
for every $\xi$ in $\overline{B_\varepsilon(0)}$. Now, let $x$ and $y$ be different points in $B_\varepsilon(0)$. By the Mean Value Theorem for Several Variables (if you do not know this version let me know and I will add the details), there exists $\xi$ in the  segment between $x$ and $y$ such that
$$
f(x)-f(y)=Df(\xi).(x-y)
$$
And we know that 
$$
||Df(\xi).(x-y)||\leq ||Df(\xi)||.||x-y||\leq A||x-y||
$$
and
$$
||x-y||=||Df^{-1}(f(\xi)).Df(\xi).(x-y)||\leq ||Df^{-1}(f(\xi))||.||Df(\xi).(x-y)||\leq C||Df(\xi).(x-y)||,
$$
so $\frac{1}{C}||x-y||\leq ||Df(\xi).(x-y)||$. If we set $B=\frac{1}{C}$, it follows that
$$
B||x-y||\leq ||f(x)-f(y)|| \leq A||x-y||,
$$
for every $x$ and $y$ in $B_\epsilon(0)$.
Note that we only need $Df(0)$ to be invertible.
