Morphism that conserve sums Let $\{ a_i \}_{1,n} \in \mathbb{(R⁺)}^n$  verifying $\sum_i a_i = 1$ and $a_1 < a_2 < \dots < a_n$. 
Is there any function (or family of function) $f$ that verifies :


*

*$\sum_i f(a_i) = 1$

*$f(a_1) > f(a_2) > \dots > f(a_n$) 


?
If there is, can you give an exemple. If not, why ?
From 2., I know that f is decreasing over $\mathbb{R⁺}$ but I can't find any of those functions...
 A: We have such functions only for each $\ n>1\ $ separately but not any function which would work for any two different values of $n>1$ at the same time.
For each natural $\ n>1,\ $ such function
$$ f_n : \mathbb R_+\rightarrow \mathbb R\ $$
is unique; here $\ \mathbb R_+\ $ can be interpreted as $\ (0;\infty)\ $ or $\ [0;\infty),\ $ either way.



For $\ n>1,\ $ we let 
$$\ f_n(x) := \frac 1{n-1} - x $$



REMARK:  Such functions $\ f_n\ $ don't exist for $\ n>2\ \ $
if   we want
     $\ f_n:\mathbb R_+\rightarrow\mathbb R_+.$



The uniqueness of each function $\ f_n\ $
(for $\ n>1)\ $ follows from its affine property:
$$ f_n(a)+f_n(d)\ =\ f_n(b)+f_n(c) $$
whenever $\ a\ b\ c\ d\,\in\,\mathbb R_+\ $ and
$$ a+d\ =\ b+c\ \le\ 1 $$
This is easy to prove (a good exercise), and it shows
that there is a constant $\ C_n\ $ such that
$ f_n(x) = C_n-x.\ $ Obviously, it must be
$\ C_n = \frac 1{n-1}.$

A final remark.   The case $\ n=1\ $ is different, there is no uniqueness due to its triviality. The decreasing condition is meaningless for $\ n=1.\ $ Thus different arguments are not glued together, they are totally independent. Hence no uniqueness.

