Prove $\det f'(x) \neq 0$ with $\forall x,y \in \mathbb{R^n}: x,y \in U => \left\lVert f(x) - f(y) \right\rVert \geq c \left\lVert x - y \right\rVert$

Question

Let $f:U \subset \mathbb{R^n} \to \mathbb{R}^n$ be totally differentiable and there exists a constant $c > 0$, so that

$$\forall x,y \in \mathbb{R^n}: x,y \in U => \left\lVert f(x) - f(y) \right\rVert \geq c \left\lVert x - y \right\rVert$$

Prove that $\det f'(x) \neq 0$ for all $x \in U$ and that $f:U \to f(U)$ is globally invertible.

I thought of choosing arbitrary $x$ or $y$, but constant in $U$. What I had problems with was trying to rewrite $f(x) - f(y)$ and using the fact that a linear function of $\mathbb{R^n}$ to $\mathbb{R}^n$ is injective iff only the zero vector maps to the zero vector.

Could someone help me with this?

It was a joke, words such as totally, hella have some association with Southern California. — copper.hat, Apr 07 '20 at 22:26

score 2 · Accepted Answer · answered Apr 07 '20 at 22:33

Partial proof:

If $f$ is differentiable at $x$ then we can find $\delta>0$ such that $\|f(x+h)-f(x)-Df(x)h\| \le {1 \over 2}x \|h\|$ for $\|h\| < \delta$.

Then $c \|h\| \le \|f(x+h)-f(x) \| \le {1 \over 2}c \|h\| + \|Df(x)h\|$ and so $\|Df(x)h\| \ge {1 \over 2}c \|h\|$. It follows that $Df(x)$ is invertible and hence the determinant is non zero.

Showing that $f$ is globally invertible is a lot more work.

Prove $\det f'(x) \neq 0$ with $\forall x,y \in \mathbb{R^n}: x,y \in U => \left\lVert f(x) - f(y) \right\rVert \geq c \left\lVert x - y \right\rVert$

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