Mean value theorem for functions from $\mathbb R^n \to \mathbb R^n$

Question

I know of the mean value inequality where if $f$ is $C^1$ and norm of $D f$ is less than $M$ at each point then $|f(x)-f(y)| \le M |x-y|$ but this is not quite as useful as the mean value theorem. I am wondering if there is an analogous statement to that in $\mathbb R$. If not is there a counterexample? In particular I find myself needing to prove that $|f(x)-f(y)| \ge M |x-y|$ and I know $D f$ is greater than $M$ at every point. Is there a mean value inequality in this direction too? I have tried modifying the proof of the original but it uses the Cauchy Schwarz which doesn't work in the other direction.

Answer 1

No, there is not. Take $f\colon\mathbb{R}\longrightarrow\mathbb{R}^2$ defined by $f(x)=\bigl(\cos(x),\sin(x)\bigr)$. Then $(\forall x\in\mathbb{R}):\|f'(x)\|=1$, but $f(0)=f(2\pi)=(1,0)$ and therefore $f(2\pi)-f(0)=0$.

Answer 2

The "real form" of the Mean Value Theorem doesn't hold in general for $n \ge 2$.

Counterexample: $f : \theta \mapsto (\cos \theta, \sin \theta)$. You have $f(0)=f(2 \pi)$ however there is no $\theta \in \mathbb R$ with $f^\prime(\theta) = (0,0)$.

Regarding your second question, if you have $|f(x)-f(y)| \ge M |x-y|$ for all $x, y$, $f$ in one-to-one. Hence you have a bijection from $E$ onto $f(E)$ and following inequality holds: $\vert f^{-1}(x)-f^{-1}(y)\vert \le \frac{1}{M} \vert x - y \vert$.