If $f$ is $C^1$ s. t. $\Vert f(x) - f(y) \Vert \geq k \Vert x - y \Vert$, then $f$ is a diffeomorphism of $\mathbb{R}^{n}$

Question

Let $f:\mathbb{R}^{n} \to \mathbb{R}^{n}$ be a function of class $C^1$ and suppose that there is $k>0$ such that $$\Vert f(x) - f(y) \Vert \geq k \Vert x - y \Vert$$ for any $x,y \in \mathbb{R}^{n}$.

(a) Prove that $f$ is injective and $f(\mathbb{R}^{n})$ is closed.

(b) Prove that $f'$ is invertible.

(c) Prove that $f(\mathbb{R}^{n})$ is open. Conclude that $f$ is a $C^1$ diffeomorphism of $\mathbb{R}^{n}$.

My attempt.

(a) If $f(x) = f(y)$, then $\Vert x - y \Vert = 0$, that is, $x=y$. Take $(f(x_{n}))$ such that $f(x_{n}) \to p$. By hypothesis, $(x_{n})$ is Cauchy and by continuity, $f(x_{n}) \to f(q) = p$ where $x_{n} \to q$. Thus, $f(\mathbb{R}^{n})$ is closed.

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(b) I dont know how to prove.

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(c) I proved in a previous question that

if $f: U \to \mathbb{R}^{n}$ with $U \subset \mathbb{R}^{n}$ is $C^1$ with $\det Df \neq 0$, then $f$ is an open map.

Suppose (b), $f$ is an open map and, therefore, $f(\mathbb{R}^{n})$ is open. So, $f(\mathbb{R}^{n})$ is non-empty clopen set, that is, $f(\mathbb{R}^{n}) = \mathbb{R}^{n}$. By Inverse Function Theorem, $f$ is a local diffeomorphism. Note that $f: \mathbb{R}^{n} \to \mathbb{R}^{n}$ is bijective and, since $f$ is a local diffeomorphism, $f^{-1}$ is differentiable for each $x$. Then $f$ is a $C^1$ diffeormorphism of $\mathbb{R}^{n}$.

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Is (a) and (c) corrects? I need help with item (b).

Answer 1

Let $y\in \mathbb{R}^{n}\setminus\{0 \}$, $t\in\mathbb{R}$. Then by definition of $Df(x_0)$ $$\lim_{t\rightarrow0}\frac{\Vert{f(x_0+ty)-f(x_0)-Df(x_0)\cdot ty}\Vert}{\Vert ty\Vert}=0$$ Then:

$$\frac{\Vert{f(x_0+ty)-f(x_0)-Df\cdot ty}\Vert}{\Vert ty\Vert}\geq\frac{\Vert{f(x_0+ty)-f(x_0)\Vert-\Vert Df(x_0)\cdot ty}\Vert}{\Vert ty\Vert}\geq \frac{k\Vert ty \Vert-\Vert Df(x_0)\cdot ty\Vert}{\Vert ty \Vert}=k-\frac{\Vert Df(x_0) \cdot y \Vert}{\Vert y \Vert}$$

This leads to: $$0=\lim_{t\rightarrow0}\frac{\Vert{f(x_0+ty)-f(x_0)-Df(x_0)\cdot ty}\Vert}{\Vert ty\Vert}\geq\lim_{t\rightarrow0}k-\frac{\Vert Df(x_0) \cdot y \Vert}{\Vert y \Vert}=k-\frac{\Vert Df(x_0) \cdot y \Vert}{\Vert y \Vert}$$ $$\Rightarrow \Vert Df(x_0) \cdot y \Vert\geq k \Vert y \Vert >0$$

So $Df(x_0)$ is invertible.

(b) and (c) are correct.