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.
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$\begingroup$ This question needs some work before it makes sense. Do you mean linear operators? You also presumably mean $DF(h)$, and in the linear case the derivative should be $F$ interpreted correctly, I believe. $\endgroup$– Otis ChodoshCommented Nov 15, 2011 at 5:40
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$\begingroup$ F is not linear and by D[F] i meant the differential operator $\endgroup$– NimaCommented Nov 15, 2011 at 6:43
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$\begingroup$ actually what i am interested in is a bit different. mean value theorem of this form might help me to prove that. I want to find a function $c(x,y)$ (or prove it exists) such that $[F(f)] (x) -[F(g)] (x) = \int c(x,y) (f(y) - g(y))$ for all x. $\endgroup$– NimaCommented Nov 15, 2011 at 6:51
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2 Answers
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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.)
answered Nov 15, 2011 at 8:10
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$\begingroup$ +1 for the remark about the inequality. $\endgroup$ Commented Nov 15, 2011 at 19:23
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$\begingroup$ One application for which you need the full MVT (and not just the inequality you mention) is in the proof of l'Hopital's rule (for which you actually need to strengthen the MVT a bit). However, I don't think there's any generalization of l'Hopital's rule to higher dimensions, so I guess that's nothing to be worried about. $\endgroup$– MarkCommented Nov 15, 2011 at 21:10
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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 $(1,0)$.
answered Nov 15, 2011 at 6:03
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$\begingroup$ this is not correct. for vector valued functions the integral form of mean value theorem does hold $\endgroup$– NimaCommented Nov 15, 2011 at 6:45
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5$\begingroup$ I think Robert Israel refers to the form of the MVT you quoted in the question, and that certainly does not hold for vector valued functions (e.g. $\exp(it)$ takes the same values at $0$ and $2\pi$) $\endgroup$ Commented Nov 15, 2011 at 7:44