TITLE: Convex set of derivatives implies mean value theorem
QUESTION [9 upvotes]: Let  U$ \subset$ $R^{^{n}}\ $be open, $f:U\rightarrow R^{m}$  differentiable on  U, and segment $[a,b]\subset U$.
Assume  that the set of derivatives $\{ f'(x)\in L(R^{^{n}},R^{^{m}}):x\in [a,b] \}$ is convex.
Prove that there exist a $\theta$ in $ [a,b]$ such that $f(b)-f(a)=f'(\theta )(b-a)$.
I have started working on "Real mathematical analysis" of Charles Chapman Pugh. I tried to solve this problem which is number 17 in chapter 5,but I was totally stucked even how to start.Any help is appreciated.

REPLY [2 votes]: There is an easy proof if you assume $f$ is of class $C^1$ and use a result from the same chapter in Pugh's book:
By the $C^1$ Mean Value Theorem we have
$$
f(q) - f(p) = T(q - p),
$$
where $T$ is the average of the derivative of $f$ on the segment $[p, q]$,
$$
T = \int_0^1 f'(p+t(q-p))dt.
$$
This integral is simply the limit of Riemann sums
$$
\sum_k f'(p+t_k(q-p)) \Delta t_k,
$$
which are clearly convex combinations (since $\sum_k \Delta t_k = 1$) of elements in the set of derivatives
$$
D = \{f'(x): x \in [p, q]\}.
$$
By convexity of $D$,these Riemann sums are also in $D$. Since $D$ is closed (this follows from continuity of $f'$ and compactness of $[p, q]$), we have $T \in D$ as desired.