# counterexample for direct generalization of the one-dimensional Mean Value Theorem

It is written in our textbook that the Mean Value Theorem is just correct in one-dimensional spaces and its direct generalization is not true in higher dimensions. Instead it has used an inequality for generalization of Mean Value Theorem.

I want to find a counterexample for direct generalization of the theorem. say a fuction $f: U \rightarrow \mathbb{R}^n$ where $U$ is an open set and $[p,q] \subset U \subset \mathbb{R}^m$ which there is no point $\theta \in [p,q]$ such that $$f(q)-f(p)=(Df)_{\theta}(q-p)$$. By $(Df)_\theta$ I mean derivative of $f$ in $\theta$.

Any idea for finding this counterexample would be appreciated.

• You could start by looking at a nonconvex region so that $p$ and $q$ are not connected by a segment that stays in the region. – Umberto P. Mar 28 '17 at 11:51
• The basic reason that it can fail is that, although you can find a point $\theta_i$ for each component individually (by using one-dimensional MVT), you can’t guarantee that all of these points are the same. – amd Mar 28 '17 at 19:17

Take $f(x) = (\cos x, \sin x)$ on $[0,2\pi]$. Then, take $q=x>0$, $p=0$ and suppose
$$(\cos x-1, \sin x) =f(x)-f(0) = f'(\xi)x = (-\sin \xi , cos \xi)x$$
for some $\xi \in (0,x)$. This is impossible, since the magnitude of the right hand side is $x$, while the magnitude of the left hand side is
$$\sqrt{(\cos x-1)^2+\sin^2 x} = \sqrt{2- 2 \cos x} = 2 \sin \frac{x}{2} < x$$