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Sup of a Functional ?

  1. Jul 6, 2012 #1
    Hi guys,

    I need some help please! Consider the following expression:

    [itex]\left[1-\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi\right]^{n-1} [/itex]

    where [itex]F:[0,1]\rightarrow [0,1][/itex] is a continuously differentiable function with [itex]F'=f, x∈[0,1][/itex], and n>2. Suppose that [itex]\rho[/itex] belongs to the set of continuous and nondecreasing functions defined on [0,1]. Let C denote this set and endow it with the sup norm. I want to find a function [itex]\rho \in C[/itex] such that (with [itex]x<1[/itex] fixed):

    [itex] \left[1-\int_{x}^{1}F(\rho^*(\xi))f(\xi)d\xi\right]^{n-1}\geq \left[1-\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi\right]^{n-1}[/itex]

    for all [itex]\rho \in C[/itex]. Does this make any sense at all? if so, how can be sure I can find this function?

    Thank you so much for your help! I truly need it!
  2. jcsd
  3. Jul 8, 2012 #2


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    Looks like a calculus of variations problem to me, and I have only a passing acquaintance with that. But first, I don't understand the role of n here. I think the integral is never more than 1. If so, it simplifies to
    [itex] \int_{x}^{1}F(\rho^*(\xi))f(\xi)d\xi \leq \int_{x}^{1}F(\rho(\xi))f(\xi)d\xi[/itex]
    Anyway, I'm going to assume that.
    Normally for calc of var one would consider [itex]\rho(\xi) = \rho^*(\xi) + δh(\xi)[/itex], small δ > 0, but I don't see how to incorporate the non-decreasing aspect.
    Another approach I tried was integration by parts:
    [itex]\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi = [F(\rho(\xi))F(\xi)]_{x}^{1} - \int_{x}^{1}F(\xi)F'(\rho(\xi))d\rho(\xi)[/itex]
    At least here we might be able to use [itex]d\rho(\xi)[/itex] ≥ 0, but I'm just as stuck.
    Just posting this in case it gives you a useful idea.
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