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Two real-valued functions on a real vector space

  1. Apr 14, 2013 #1
    I have a fixed function [itex]U[/itex] and a function [itex]f[/itex]
    that I want to know something about, both from
    [itex]\mathbb{R}^{n}\rightarrow\mathbb{R}[/itex], for which I know
    condition (G) that [itex]U(x)\geq{}U(y)[/itex] implies
    [itex]f(x)\geq{}f(y)[/itex]. [itex]f=U[/itex] and [itex]f=x_{0}[/itex]
    for any [itex]x_{0}\in{}\mathbb{R}[/itex] trivially fulfill (G). Which
    [itex]f[/itex] fulfills (G) non-trivially? Are there additional
    assumptions (continuity for example) which would allow me to claim
    that there is no [itex]f[/itex] that non-trivially fulfills (G)?

    This question will help me solve a set-theoretic puzzle based on
    Debreu's theorem -- that's why I've put the thread in this category.
    [itex]U[/itex] stands for the utility function, [itex]f[/itex] for a less discriminating
    utility function that is not trivially [itex]f=U[/itex] or constant.
  2. jcsd
  3. Apr 15, 2013 #2

    Stephen Tashi

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    Science Advisor

    I think we need a definition of "non-trivially". Is [itex] f(x) = 3 U(x) + 6 [/itex] a trivial or non-trivial fullfillment?
  4. Apr 15, 2013 #3
    Sorry, yes, I left out something important. I need [itex]f[/itex] to
    be LESS discriminating than [itex]U[/itex]. So, I want
    [itex]\exists{}x,y\in\mathbb{R}^{n}[/itex] with [itex]x\neq{}y[/itex]
    and [itex]f(x)=f(y)[/itex], whereas [itex]U(x)>U(y)[/itex].

    I think I've answered my own question. Yes, there is a continuous
    [itex]f[/itex] (given a continuous [itex]U[/itex]) which fulfills (G)
    is neither [itex]U[/itex] nor constant. The only constraint is that,
    if the functions are differentiable, [itex]f'(x)U'(x)[/itex] is
    non-negative. For my less-discriminating condition, the function must
    be zero somewhere where [itex]U[/itex] is monotonically increasing.

    Thanks, I think I can take it from here.
    Last edited: Apr 15, 2013
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