Replication invariance of weighted sums Consider the following function:

*

*$\phi(\langle x_1,x_2,...,x_n \rangle) = \alpha_1 x_1 + \alpha_2 x_2
   + ... + \alpha_n x_n$,

where $x_1, x_2, ...$ are all real numbers such that $x_1≤x_2≤...$ and we also have $\alpha_1 ≥ \alpha_2 ≥ ... > 0$.
Now consider the following property:

*

*$\phi(\langle x_1,...,x_n \rangle) = \phi(\langle y_1,...,y_n
   \rangle)$
if and only if, for all $m≥1$,
$\phi(\langle \underbrace{x_1,\ldots,x_1}_{m\text{
   times}},...,\underbrace{x_n,\ldots,x_n}_{m\text{
   times}}\rangle)=\phi(\langle \underbrace{y_1,\ldots,y_1}_{m\text{
   times}},...,\underbrace{y_n,\ldots,y_n}_{m\text{ times}}\rangle)$.
My question is: Is it true that if this property holds, then $\alpha_1 = \alpha_2 =\dots$?
Or, to put it differently, if a weighted sum with positive coefficients $\alpha_1, \alpha_2, \dots$ is invariant to replication, as it were, does it follow that the coefficients are in fact equal?
I am inclined to think Yes. This seems to be a corollary of Proposition 7 in Ebert (1988), but I am sure a proof simpler than his is possible. Thanks for any pointers!
 A: The answer is affirmative. We do induction on $N$ to prove $\alpha_1 = \alpha_2 = \dots = \alpha_N$ as follows.
First of all, the base case $N = 2$ is proved in the bottom(the proof of this case actually automatically implies the case for arbitrary $N$). We next finish the process by showing that case $N = k$ implies case $N = k+1$(for $k \geq 2$):
When $k$ is odd, substitute $n = 2$ and $m = \frac{k+1}{2}$ in the property yields $$ \frac{k+1}{2} \alpha_1 = \frac{k-1}{2} \alpha_1 + \alpha_{k+1} .$$
Hence $\alpha_{1} = \alpha_{k+1}$.
When k is even,  substitute $n = 2$ and $m = \frac{k}{2} + 1$ in the property yields $$ ( \frac{k}{2} + 1 ) \alpha_1 = ( \frac{k}{2} - 1 )\alpha_1 + \alpha_{k+1} + \alpha_{k+2}.$$
Hence $2 \alpha_1 = \alpha_{k+1} + \alpha_{k+2}$. Note that $\alpha_1 \geq \alpha_2 \geq ... > 0$. We have $ \alpha_1 = \alpha_{k+1} = \alpha_{k+2}$.
The process of induction is therefore now finished. In conclusion, we have $\alpha_1 = \alpha_2 =\dots.$

Proof of the case $N = 2$:
Let $p = \frac{\alpha_1 + \alpha_2}{\alpha_1}$ and $q = \frac{\alpha_1 + \alpha_2 + \alpha_3}{\alpha_1}$. We prove $p = 2$ and $q = 3$ at one strike. By repeatedly applying the property, it's easy to show the following equalities:$$\alpha_1 + \dots + \alpha_{2^{n_1}} = p^{n_1} \alpha_1,$$and $$\alpha_1 + \dots + \alpha_{3^{n_2}} = q^{n_2}\alpha_1$$for arbitrary positive integers $n_1$ and $n_2$. Taking the quotient of these equalities yields$$ \frac{\alpha_1 + \dots + \alpha_{3^{n_2}}}{\alpha_1 + \dots + \alpha_{2^{n_1}}} = \frac{q^{n_2}}{p^{n_1}}.$$
Now taking sufficiently large $n_1$ and $n_2$ that satisfy $2^{n_1} < 3^{n_2} < 2^{n_1+1}$ yields $p = 2$ and $q = 3$ since we have estimation(recall $\alpha_1 \geq \alpha_2 \geq ... > 0$)$$1 < \frac{\alpha_1 + \dots + \alpha_{3^{n_2}}}{\alpha_1 + \dots + \alpha_{2^{n_1}}} < 2$$for arbitrary $n_1$ and $n_2$ that satisfy $2^{n_1} < 3^{n_2} < 2^{n_1+1}$. We're through.
(Note that this actually implies $\alpha_1 = \alpha_2 =\dots $ directly, since if $ \frac{\alpha_1 + \dots + \alpha_{3^{n_2}}}{\alpha_1 + \dots + \alpha_{2^{n_1}}} = \frac{3^{n_2}}{2^{n_1}}$, then $ \alpha_1 = \dots = \alpha_{3^{n_2}}. $)
