Mean value theorem for vector valued function (not integral form)

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 $$f(x+h) - f(x) = Ah.$$

• Where did you find the first two statements ? – Tony Piccolo Aug 13 '15 at 5:21
• @TonyPiccolo fundamental theorem of calculus? most book dealing with Lebesgue integration. mean value theorem for integral? a consequence of Hahn-banach separation theorem. you can find the proof even on math se. – user251257 Aug 13 '15 at 5:57
• I know the matter, but wanted to know if you had in mind some particular source. Usually this information helps in answering. – Tony Piccolo Aug 13 '15 at 6:18
• @TonyPiccolo I tried to adapt the proof for the MVT for integral, like in that answer. It should work. However, I would like an opinion by someone else. – user251257 Aug 13 '15 at 6:26
• The usual formula that I know only says $f(x+h)-f(x)=\int_0^1 f'(x+th)dt h$. I don't see where the $f'(x) h$ on the left hand side of your equation comes from. – PhoemueX Aug 13 '15 at 6:31

Under the given (finite dimensional) assumption, we can even obtain $$f(x+h) - f(x) \in \operatorname{conv}\{ f'(x+th)h \mid 0 < t < 1 \} = \operatorname{conv}\{ f'(x+th) \mid 0 < t < 1 \}h.$$
• The result is valid for general normed linear spaces. If the codomain of $f$ has finite dimension (and yours is the case), then one can avoid considering the closure (of the convex hull) as was shown by McLeod Mean value theorems for vector valued functions (1965). The fact depends on a property of the so called relative interior in the finite dimension case. – Tony Piccolo Aug 13 '15 at 18:38
• Why did you edit by "only obtain ... instead of ..." ? There is an equivalence: if you take the inner product side by side ...That is, if $C$ is convex, so is $C \cdot (y-x)$ with obvious meaning of the symbol. – Tony Piccolo Aug 13 '15 at 18:54
• @TonyPiccolo inner product with what? If the vector $Ah$ is a limit of convex combinations of the derivatives times $h$, why should $A$ be a limit of convex combination of the derivatives? If we can let go of the closure, then it is true. – user251257 Aug 13 '15 at 18:58
• I'm saying another thing. Don't think of your post now. If $C$ is a convex subset of the space, then $\{c \cdot h : c \in C \}$, where $h$ is an element of the space, is a convex set. – Tony Piccolo Aug 13 '15 at 19:05