# Mean Value Theorem for Several Variables

Please see the mean value theorem stated here: MVT.

Note that the MVT stated there requires that the segment connecting $x$ and $y$ lies in $G$. So clearly, MVT works for any pair of points if $G$ is made a convex open set.

Here are my questions:

1. What happens if $G$ is not convex, but connected? Looking at the proof of MVT, I think the MVT will still work as long as there is a path (a continuous map) from $x$ to $y$. And this is the case in a connected set. Am I right in thinking this?
2. Suppose the answer to (1) is yes. (Of course, if I am wrong in (1), please ignore this question.)Assume that $G$ is open and connected and let $x,y \in G$. Then there is some path (not necessarily a line segment) connecting $x$ and $y$ in $G$. MVT guarantees that there is a $z$ on this path such that $$f(x) - f(y) = \nabla f(z) \cdot (x - y).$$ Can we say something about the distance between $z$ and the points $x$,$y$? If the path is a line segment, then certainly, $$|x-z|\leq |x-y| \quad \text{and} \quad |y-z| \leq |x-y|.$$ If the path is not a line segment, what happens?
• If you look at the proof of MVT as sketched in Wikipedia (your source) it is important that the parametrization of the 'path' joining $x$ and $y$ is linear in the variable $t$. Otherwise (in wikipedia notation) $g'(c) \neq \nabla f((1-c)x + cy)$ there will be extraneous factors involved. If $G$ is not convex the best you can do is partition $G$ into finitely many convex sets $G_{i}$ such that $G_{i} \cap G_{i+1}$ is nonempty and $x \in G_{0}$ and $y \in G_{n}$. Then apply MVT to each partition separately. – DBS Jul 26 '13 at 5:25
• But a path $g$ joining $x$ and $y$ (even if it's not linear) still satisfies: $g(0) = x, g(1) = y,$ and of course, $g$ is continuous since it's a path. – Alex Strife Jul 26 '13 at 6:15
• Cosider a path $\gamma(t) = (1-t^{2})x + t^{2}.y, 0 \leq t \leq 1$ then $g^{'}(c^{2}) \neq \nabla f((1-c^{2})x + c^{2}y)$ it will be $\nabla f((1-c^{2})x + c^{2}y). < -2c, 2c>$. So it no longer has a nice form. – DBS Jul 26 '13 at 6:20
• I see. I get it now. Thanks :) – Alex Strife Jul 26 '13 at 7:30