# Mean value in higher dimensions

I need to prove this result:

Suppose that the set: $\{(Df)_x\in\mathbb{L}(\mathbb{R}^n,\mathbb{R}^m):x\in[p,q]\}$ is convex. Show that there exists an $\theta\in [p,q]$ such that $f(q)-f(p)=(Df)_{\theta}(q-p)$.

What I have done (and is probably incorrect) is writing:

$f(q)-f(p)=\int_p^q (Df)_x dx = \int_0^1 ((Df)_q t+(Df)_p (1-t))(q-p)dt$ The previous is $\dfrac{1}{2}((Df)_q+(Df)_p)(q-p)$, where there must exist an $\theta$ such that $\dfrac{1}{2}((Df)_q+(Df)_p)=(Df)_{\theta}$ since $\dfrac{1}{2}((Df)_q+(Df)_p)$ is the medium point of $[(Df)_p,(Df)_q]$ which is convex. This concludes the exercise, but I'm pretty sure that the integral part is wrong somewhere. Can someone help me?

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It looks right. Another question: Higher dimensions? – martini Mar 27 '12 at 18:33
Oops, there was a typo in the text. I have corrected it. – Gustavo Marra Mar 27 '12 at 19:09