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

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 $f:U \to f(U)$ is globally invertible.

Choose an random but constant $x$ or $y$ in $U$. Rewrite $f(x) - f(y)$ and use 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.

I know that a function $f$ is globally invertible if $f$ is bijective.

This must imply that one has to prove that the function is injective and surjective. This requires the inverse function and my guess is that the implicit function theorem can prove what is asked for above, but I don't know how to apply the theorem in this case.

Can someone show how it's done?

Answer 1

I am afraid it is rather routine to prove that $f:U \to f(U)$ is (globally) invertible. We have,

• By definition, $f$ is surjective.
• Suppose $x,y\in\mathbb R^n$, $x\ne y$. Then $\lVert x - y\rVert \gt0$. Hence, $\lVert f(x) - f(y) \rVert \geq c \lVert x - y\rVert\gt 0$, which says $f(x)\ne f(y)$. $f$ is injective.

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An interesting question that might be in your mind is whether $f(\mathbb R^n)=\mathbb R^n$, assuming $f$ is defined over $\mathbb R^n$. If that is true, then we can just claim $f$ is bijective without restricting its codomain.

For simple examples, such as $f(x)=ax+b$ for some constant $a\ne0$ and $b$, we have $f(\mathbb R^n)=\mathbb R^n $.

It is, in fact, true that, under with the slightly stronger condition that $f$ be continuously differentiable, $$f(\mathbb R^n)=\mathbb R^n.$$

Proof: for the sake of contradiction, suppose $f(\mathbb R^n)\ne \mathbb R^n$.

Let $q$ be a point on the boundary of $f(\mathbb R^n)$, i.e, there is a sequence of points in $f(\mathbb R^n)$, say, $f(p_1), f(p_2), \cdots,$ whose limit is $q$. Since $f(p_1), f(p_2), \cdots,$ is a cauchy sequence in terms of $\lVert\cdot\rVert$, and $f$ extends the distance by at least a positive constant factor, so is the sequence $p_1, p_2, \cdots$. Let the limit of $p_1, p_2, \cdots$ be $p$. Since $f$ is continuous, $f(p)=q$.

Because $f$ extends distance by at least a positive constant factor, the derivative of $f$ at $p$ along any direction will be at least $c$ in magnitude, i.e, not zero. That means the differential of $f$ is a linear isomorphism at $p$. The inverse function theorem stipulates that $f$ must be a local diffeomorphism, which contradicts the fact that $f(p)=q$ is on the boundary of $f(\mathbb R^n)$. This proof is done.

In a nutshell, the proof shows that $f(\mathbb R^n)$ must be both closed and open in $\mathbb R^n$, hence it must be all of $\mathbb R^n$.

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