Prove that $|f(x)-f(y)| \geq k|x-y|$

Question

Let $f:R^m \rightarrow R^m$ be $C^1$. And there exists $k>0$ s.t. $\forall x,h \in R^m$, $h^{T}f'(x)h \geq k|h|^2$. Prove that $\forall x,y\in R^m$, $|f(x)-f(y)| \geq k|x-y|$.
Well, if $m=1$ then $f'(x) \in R^1$, hence $f'(x) \geq k$. And just use the mean value theorem to obtain the inequality.
In general, we can deduce that $\forall x\in R^m,f'(x) \neq0$. Hence we can use the inverse function theorem to find $g=f^{-1}$. And, $\forall x,y \in R^m, \exists \xi \in(x,y)$ s.t. $$ |g(x)-g(y)| \leq \Vert g'(\xi) \Vert \cdot |x-y|, \\ |g(f(x))-g(f(y))| \leq \Vert g'(\xi) \Vert \cdot |f(x)-f(y)|, \\ |x-y| \leq \Vert g'(\xi)\Vert \cdot |f(x)-f(y)|, \\ |f(x)-f(y)| \geq \frac{1}{\Vert g'(\xi)\Vert} |x-y|. $$ So if we can prove that $1/ \Vert g'(\xi)\Vert \geq k$ then we are done. But after some trials it seems hopeless. Please help me.
EDIT: I just realize that we cannot use the inverse function theorem immediately to get $g=f^{-1}$.

Answer 1

I'll assume that $| \cdot |$ refers to the Euclidean norm $\Vert x \Vert_2 = \sqrt{ x^T x}$ on $\Bbb R^m$.

One can reduce the problem to the one dimensional case: For fixed $x, y \in \Bbb R^m$ define the function $h: [0, 1] \to \Bbb R$ as $$ h(t) = (y-x)^T f(x + (y-x)t) \, . $$ Then $$ h'(t) = (y-x)^T f'(x + (y-x)t) (y-x) \ge k \Vert y-x \Vert^2_2 $$ so that the mean value theorem gives $$ |h(1) - h(0)| \ge k \Vert y-x \Vert^2_2 \, . $$ On the other hand we have $$ |h(1) - h(0)| = |(y-x)^T (f(y) - f(x))| \le \Vert y-x \Vert_2 \Vert f(y) - f(x) \Vert_2 $$ from the Cauchy Schwarz inequality. Combining these estimates gives the desired conclusion.