I tried to prove the Multivariable Mean Theorem (I hope this is its correct name). Please evaluate my approach for exercise 1 and lend me your help for the latter part of this exercise.
Use the chain rule to show the following. Let $A\subseteq X=\mathbb{R}^n$ be a convex set and $f:A\rightarrow\mathbb{R}$ be differentiable in every point with $a,b\in A$. Then there exists a $\xi\in A$ such that $f(b)-f(a)=f'(\xi)(b-a)$.
Does the same equality hold for $f:A\rightarrow\mathbb{R}^m$? Derive an appropiate Mean Value Inequality.
1.
Let $h:[0,1]\rightarrow A$:
$$h(x)=(a_1+(b_1-a_1)x,\ldots,a_n+(b_n-a_n)x).$$
Then $h$ is differentiable on $(0,1)$ and continuous on $[0,1]$, the same holds for $f \circ h$. Let $\alpha \in (0,1)$. Then we apply the Mean Value Theorem to $f\circ h$ to get
$$(f \circ h )'(\alpha)\underset{\text{chain rule}}=f'(h(\alpha))\cdot h'(\alpha)=(f \circ h )(1)-(f \circ h )(0)$$
which is equivalent to
$$f(\mathbf{b})-f(\mathbf{a})=f '(h(\alpha))(\mathbf{b}-\mathbf{a}).$$
If we now set $\xi=h(\alpha)$, we get the desired result.
2.
Now I understand that the equality doesn't hold and can provide a counterexample, say $f : \Bbb R \to \Bbb R^2$, $t\mapsto (\cos t, \sin t)$. Then $f(2\pi)-f(0) = (0,0)$. But I don't know how to derive Mean Value Inequality. Can anybody help out? Much appreciated.
$f(b)-f(a)=\int_0^{1} f'(tb+(1-t)a) \, dt$ so $||f(b)-f(a)|| \leq M ||b-a||$ where $M$ is the maximum of $||f'||$ in the line joining $a$ and $b$.