# Sup of a Functional ?

1. Jul 6, 2012

### cris(c)

Hi guys,

I need some help please! Consider the following expression:

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

where $F:[0,1]\rightarrow [0,1]$ is a continuously differentiable function with $F'=f, x∈[0,1]$, and n>2. Suppose that $\rho$ 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 $\rho \in C$ such that (with $x<1$ fixed):

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

for all $\rho \in C$. 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. Jul 8, 2012

### haruspex

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
$\int_{x}^{1}F(\rho^*(\xi))f(\xi)d\xi \leq \int_{x}^{1}F(\rho(\xi))f(\xi)d\xi$
Anyway, I'm going to assume that.
Normally for calc of var one would consider $\rho(\xi) = \rho^*(\xi) + δh(\xi)$, small δ > 0, but I don't see how to incorporate the non-decreasing aspect.
Another approach I tried was integration by parts:
$\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi = [F(\rho(\xi))F(\xi)]_{x}^{1} - \int_{x}^{1}F(\xi)F'(\rho(\xi))d\rho(\xi)$
At least here we might be able to use $d\rho(\xi)$ ≥ 0, but I'm just as stuck.
Just posting this in case it gives you a useful idea.