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Let $f: U \rightarrow \mathbb{R}^n$ be a differentiable function on an open subset $U$ of $\mathbb{R}^n$. Assume that there exists $c>0$ such that $|f(x)-f(y)| \geq c |x-y|$ for all $x,y \in U$. Then I want to show that $\det \mathbf{J}_f(x)$ is nonzero for all $x \in U$.

I think I need to apply the Inverse function theorem but I cannot see how.

Karatuğ Ozan Bircan
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    Note that if $\det J_f(x)$ is zero, then there must be a nonzero vector $v$ so that $J_f(x) v = 0$. Now if $y = x + tv$...... –  Nov 29 '14 at 04:00

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Suppose there is some $x\in U$ such that $\det\mathbf{J}_f(x)=0$. Then there must be some $v\in \mathbb{R}^n$ such that $\mathbf{J}_f(x)\cdot v=0$. Let $y_t=x+tv$. Since $U$ is open, there exists $r>0$ such that $B_r(x)\subset U$. Then $|x-y_t|=t|v|$. If we make $t$ small enough, say some $t_0$, then $t_0|v|<r$. But since $B_r(x)$ is convex, for all of the $t\leq t_0$, $y_t\in B_r(x)\subset U$.

Then $|f(x)-f(y_t)-\mathbf{J}_f(x)\cdot tv|\geq ||f(x)-f(y_t)|-|t\mathbf{J}_f(x)\cdot v|=|f(x)-f(y_t)|\geq c|x-x-tv|=c|tv|$. Now divide to get $\dfrac{|f(x)-f(y_t)-\mathbf{J}_f(x)\cdot tv|}{t|v|}\geq c>0$. Taking the limit as $t\to 0$, since $f$ is differentiable, the left hand side of this equation is $0$, so we get $0\geq c>0$. Contradiction. So $x$ doesn't exist, and $\mathbf{J}_f(x)\neq 0$ for all $x\in U$.

Moya
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