mean value theorem for operators - MathOverflow

mean value theorem for operators

2011-11-15T05:32:19Z

This might be a trivial question but I am not very familiar with the subject matter. I was wondering if some sort of mean value theorem works for operators on function spaces. Say $F: \mathcal{S_1} \to \mathcal{S_2}$ is an operator on the function spaces $\mathcal{S_{1,2}}$ then for every $f,g \in \mathcal{S_1}$ there exist $h$ such that \begin{align*} F(f) - F (g) = [DF(h)] (f - g), \end{align*} or something like that! What is a good reference to look at if I want to learn more.

Answer by Robert Israel for mean value theorem for operators

2011-11-15T06:03:30Z

Even for functions from ${\mathbb R}$ to ${\mathbb R}^2$ the Mean Value Theorem fails.
Thus it is possible to go from $(0,0)$ to $(1,0)$ in time 1 and have the velocity vector never equal to $(1,0)$.

Answer by Pietro Majer for mean value theorem for operators

2011-11-15T08:10:12Z

Here is a nice list (by John H. Mathews) of articles of various authors on the theme of extending the validity of the Mean Value Theorem to vector values function.

However, as Dieudonné remarks (Foundations of Modern Analysis) the main point of the classical MVT, even in the case of a one variable real valued function, is not the identity $$f(b)-f(a)=f'(\xi)(b-a)\ ,$$ also because we usally can say nothing about the point $\xi$, apart the fact that it is strictly betweeen $a$ and $b$. Rather, it is the inequality it implies: $$|f(b)-f(a)|\le \sup_{a < \xi < b} |f'(\xi)| |b-a| \ ,$$ and this is also the statement that generalizes naturally in the Banach setting, and has the most important consequences, as it is the key tool of most fundamental theorems of differential calculus (to quote some: the symmetry of higher order differentials, the Lagrange's remainder form in Taylor's formula, the theorem of the total differential, the theorem of limit under the sign of derivative, the inverse and the implicit function theorem,...&c.)