Finding a Sup of a Functional?

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In summary, the conversation discusses a mathematical expression involving a continuously differentiable function, a set of continuous and nondecreasing functions, and a fixed value. The speaker is seeking to find a specific function within the set that satisfies a certain inequality. The conversation also touches on the potential use of calculus of variations to solve the problem.
  • #1
cris(c)
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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!
 
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  • #2
cris(c) said:
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].
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.
 

1. What is a Sup of a Functional?

A Sup (also known as a supremum) of a functional is the smallest upper bound of the set of values that the functional can take on. It is the highest value that the functional can reach.

2. How is the Sup of a Functional calculated?

The Sup of a Functional can be calculated by finding the maximum value of the functional over its entire domain. This can be done through various mathematical methods, such as taking derivatives and setting them equal to zero.

3. Why is the Sup of a Functional important?

The Sup of a Functional is important because it helps to determine the behavior and limits of the functional. It can also be used to prove the existence of certain values and to make predictions about the functional's behavior.

4. Can the Sup of a Functional be infinite?

Yes, the Sup of a Functional can be infinite if the functional has no upper bound. This can occur in certain situations, such as when dealing with unbounded functions or when the functional has a vertical asymptote.

5. How is the Sup of a Functional used in real-world applications?

The Sup of a Functional has various uses in real-world applications, such as in optimization problems, economics, and physics. It can be used to determine optimal solutions, predict future trends, and analyze the behavior of systems.

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