I recently saw this post which discussed different understandings of the open set $X \subset \mathbb{R}^n$ that allows us to use the mean value theorem on several variables when $f: X \to \mathbb{R}$ is differentiable. I looked into the mean value theorem on several variables and found two different definitions. One from the Wikipedia page which uses the gradient and another from my lectures which uses the jacobian. It seemed like the second definition is a special case when $X$ is convex and the first definition is more general. Is this an accurate way to describe the difference between this two definitions
$\mathbf{Definition}-\mathbf{1}:$ Let $X \subset \mathbb{R}^n $ be an open set and let $f: X \to \mathbb{R}$ be differentiable on $X$. Then for all $a \in X$ and $h$ such that the segment joining $a$ to $a + h$ is contained in $X$. Then
there exists a point $c$ on that segment such that…
$ f(a+h) – f(a) = Df(c)\cdot h$
$\mathbf{Definition}-\mathbf{2}:$ If $b = a + h$ and $X$ is convex then
$f(b) – f(a) = \nabla f((1 – c)a + cb) \cdot (b – a)$ where $ \nabla$ is the gradient and is the dot product.
I also I have included the tag "paramaterization" because that process helps us find the mean value theorem for several variables and I suspect that if I am missing some key point it is related to that topic.
Best Answer
Hint: define $g:\mathbb [0,1]\to \mathbb R$ by $g(t)=f((1-t)a+t(a+h)),$ which makes sense in $(1)$, because the segment is contained in $X$ and in $(2)$, because $f$ is defined in a convex set. Note that $g$ is differentiable on $(0,1)$ and continuous on $[0,1]$ so by the mean value theorem, there is a $c\in (0,1)$ such that $g'(c)=g(1)-g(0)$. Can you finish by applying the multivariable chain rule, and using the definition of derivative of $f$, noting its relation to the gradient?