1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mean value inequality?

  1. Jun 10, 2008 #1


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The statement of the mean value inequality (MVI) is as follows:

    "Let A be an open convex subset of R^n and let f:A-->R^m be continuously differentiable and such that ||Df(x)(y)||<=M||y|| for all x in A and y in R^n (i.e. the family
    [itex](Df(x))_{x \in A}[/itex] is uniformly lipschitz of constant M on R^n). Then for any x_1, x_2 in A, we have ||f(x_2)-f(x_1)||<=M||x_2-x_1||."

    If m=1, then this is just the mean value theorem (MVT) plus the triangle inequality. But otherwise, the MVT applied to each component of f separately only leads ||f(x_2)-f(x_1)||<=mM||x_2-x_1||. So the proof suggested by the book I'm reading is that we write f(x_2)-f(x_1) using the fondamental theorem of calculus (FTC) as


    and then use the triangle inequality for integrals to get the result.

    But notice that the integrand is an element of R^m. So by the above, they certainly mean


    which does not, to my knowledge, allows for a better conclusion than ||f(x_2)-f(x_1)||<=mM||x_2-x_1||.

    Am I mistaken?

  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: Mean value inequality?