i have a doubt about the following problem:
Lets define a multivariable function $u(x):\mathbb{R}^{n} \rightarrow \mathbb{R}$, which has the "additive separability" property, meaning we can writte $\displaystyle u(x) = \sum_{l=1}^{n}u_{l}(x_{l})$ We need to show that a monotonic transformation preserves this property ONLY if this transformation is affine.
The first step is to show that a affine function preserves this so called "additive separability" property:
Lets create the function $\hat{u}(x) = au(x) + b$, where $a>0$ and $b \in \mathbb{R}$. Then:
\begin{align} \hat{u}(x) = a \sum_{l=1}^{n}u_{l}(x_{l}) + b \\ \hat{u}(x) = \sum_{l=1}^{n} a u_{l}(x_{l}) + b \\ \hat{u}(x) = \sum_{l=1}^{n}( a u_{l}(x_{l}) + \frac{b}{n}) \\ \hat{u}(x) = \sum_{l=1}^{n} \tilde{u}_{l}(x_{l}) \end{align}
Now, my problem is to understand the second part of the proof:
We need to show that if some monotonic transformation preserves the property, it is necesarily a affine transformation: The proof that i found states: Lets create a monotonic transformation $u^{*}(x)=f(u(x))$ We want to show that if $u^{*}(x)=f(u(x)) = f(u(x)) = f(\sum_{l=1}^{n}u_{l}(x_{l})) = \sum_{l=1}^{n}f_{l}(x_{l})$ then $f()$ is a affine function.
First:
$\dfrac{\partial u^{*}(x)}{\partial x_{j}} = \dfrac{\partial f(\sum_{l=1}^{n}u_{l}(x_{l}))}{\partial x_{j}} = f'(\sum_{l=1}^{n}u_{l}(x_{l}))u_{j}'(x_{j}) \hspace{2cm} (1)$
Also, since we know that $u^{*}(x)=\sum_{l=1}^{n}f_{l}(x_{l})$, then:
$\dfrac{\partial u^{*}(x)}{\partial x_{j}} = \dfrac{\partial \sum_{l=1}^{n}f_{l}(x_{l}) }{\partial x_{j}} = f_{j}'(x_{j}) \hspace{2cm} (2)$
From (1) and (2):
$f'(\sum_{l=1}^{n}u_{l}(x_{l})) = \dfrac{f_{j}'(x_{j})}{u_{j}'(x_{j})} \hspace{2cm} (3)$
So, to show that $f()$ is affine is enough to show that $f'()$ is constant.
Lets use the following to vectors: $x^{1}$ and $x^{2}$, both in $\mathbb{R}^{n}$, such that:
$\sum_{l=1}^{n}u_{l}(x^{1}_{l}) \neq \sum_{l=1}^{n}u_{l}(x^{2}_{l})$ and $x^{1}_{j} = x^{2}_{j} = x_{j}$
So, because of (3):
\begin{align} f'(\sum_{l=1}^{n}u_{l}(x^{1}_{l})) = \dfrac{f_{j}'(x^{1}_{j})}{u_{j}'(x^{1}_{j})} = \dfrac{f_{j}'(x_{j})}{u_{j}'(x_{j})} \\ f'(\sum_{l=1}^{n}u_{l}(x^{2}_{l})) = \dfrac{f_{j}'(x^{2}_{j})}{u_{j}'(x^{2}_{j})} = \dfrac{f_{j}'(x_{j})}{u_{j}'(x_{j})} \end{align}
So, finally:
$f'(\sum_{l=1}^{n}u_{l}(x^{1}_{l})) =f'(\sum_{l=1}^{n}u_{l}(x^{2}_{l})) $
$\square$
I understand everything except why we can state without lost of generality that the j-dimension of $x^{1}$ and $x^{2}$ are equal? Is there any other way to prove this without this assumption? It seems a strong assumption.
Thanks.