Convex set of derivatives implies mean value theorem 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.
 A: (1) Recall MVT for vector valued function : If $f: S\subset {\bf
R}^n\rightarrow {\bf R}^m$ is differentiable and if
$\overline{xy}\subset S$, then for any $a\in {\bf R}^m$, $$ a\cdot
(f(y)-f(x))= a\cdot \{ f'(z)(y-x )\}$$ for some $z\in \overline{xy}$
(2) Define $$ F(z)=f(z)-Az,\ A:=f'(x)$$ so that $dF (x)=0 $ By
convexity, $$ dF (z)=c_z B,\ B:= (f'(y)-A) $$ for some $c_z$ where
$z\in \overline{xy} $ So we have that for $v$, $$ v\cdot
(F(y)-F(x))= v\cdot \{ F'(z)(y-x)\} =v\cdot k(v) B(y-x) \ \ast $$
for some $z$ and $k : {\bf R}^m\rightarrow {\bf R} $
Since $\{ F(y)-F(x) - k(v) B(y-x) |v\in {\bf R}^m \}$ is in a line
so there exists $v_0$ s.t. $v_0$ is orthogonal to the line So for
$\ast$ we have $$F(y)-F(x) - k(v_0) B(y-x) =0 $$
(3) [Add] For $z\in \overline{xy}$, by convexity, $$ f'(z)=\alpha f'(x)
+(1-\alpha ) f'(y) $$
By (1) $$ a\cdot \{ f(y)-f(x) - f'(z) (y-x) \} =0
$$
$$ a\cdot \{ f(y)-f(x) - \alpha f'(x) (y-x)
-(1-\alpha ) f'(y) (y-x) \} =0  $$
Note that $$f(y)-f(x) - \alpha f'(x) (y-x)
-(1-\alpha ) f'(y) (y-x)$$ is in some line $l$. When $a$ is
perpendicular to the line $l$, then $$f(y)-f(x) - \alpha f'(x) (y-x)
-(1-\alpha ) f'(y) (y-x) =0$$ i.e. $$f(y)-f(x) - f'(z)  (y-x) =0$$
A: 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.
