Let $f:U\to\mathbb R^m$ be differentiable with $U\subseteq \mathbb R^n$ being open and convex.
If $f$ is absolutely continuous, then by fundamental theorem of calculus we have following version of mean value theorem: For any $x,x+h\in U$ it follows
$$ f(x+h) – f(x) = \left( \int_0^1 f'(x+t h) \; d t \right) h. $$
By mean value theorem for integral, we obtain
$$ A = \int_0^1 f'(x+t h) \; d t \in \overline{\operatorname{conv}} \{ f'(x+th) \mid 0< t < 1 \}. $$
Is it possible to obtain that result without $f$ being absolutely continuous? That is, assume $f$ is only differentiable. Is the following statement correct?
For every $x,x+h\in U$ there exists some $A$ in the closure of the convex hull of $\{ f'(x+th) \mid 0< t < 1 \}$ with
$$ f(x+h) – f(x) = Ah. $$
Best Answer
Under the given (finite dimensional) assumption, we can even obtain $$ f(x+h) - f(x) \in \operatorname{conv}\{ f'(x+th)h \mid 0 < t < 1 \} = \operatorname{conv}\{ f'(x+th) \mid 0 < t < 1 \}h. $$
Thank @TonyPiccolo for the reference:
McLeod Mean value theorems for vector valued functions (1965).