Two real-valued functions on a real vector space

[noowutah]

I have a fixed function $U$ and a function $f$
that I want to know something about, both from
$\mathbb{R}^{n}\rightarrow\mathbb{R}$, for which I know
condition (G) that $U(x)\geq{}U(y)$ implies
$f(x)\geq{}f(y)$. $f=U$ and $f=x_{0}$
for any $x_{0}\in{}\mathbb{R}$ trivially fulfill (G). Which
$f$ fulfills (G) non-trivially? Are there additional
assumptions (continuity for example) which would allow me to claim
that there is no $f$ 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.
$U$ stands for the utility function, $f$ for a less discriminating
utility function that is not trivially $f=U$ or constant.

 

[Stephen Tashi]

Which $f$ fulfills (G) non-trivially?

I think we need a definition of "non-trivially". Is $f(x) = 3 U(x) + 6$ a trivial or non-trivial fullfillment?

 

[noowutah]

Sorry, yes, I left out something important. I need $f$ to
be LESS discriminating than $U$. So, I want
$\exists{}x,y\in\mathbb{R}^{n}$ with $x\neq{}y$
and $f(x)=f(y)$, whereas $U(x)>U(y)$.

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

Thanks, I think I can take it from here.