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We say a transform is linear if $cf(x)=f(cx)$ and $f(x+y)=f(x)+f(y)$. I wonder if there is another definition.

If it's relevant, I'm looking for sufficient but possibly not necessary conditions.

As motivation, there are various ways of evaluating income inequality. Say the vector $w_1,\dots,w_n$ is the income of persons $1,\dots,n$. We might have some $f(w)$ telling us how "good" the income distribution is. It's reasonable to claim that $cf(w)=f(cw)$ but it's not obvious that $f(x+y)=f(x)+f(y)$. Nonetheless, there are some interesting results if $f$ is linear. So I wonder if we could find an alternative motivation for wanting $f$ to be linear.

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May I ask what motivates this question? It's hard to imagine any criteria simpler and more intuitive than the given definition. –  EuYu Aug 23 '12 at 0:21
@EuYu: Added my motivation. It's somewhat backwards reasoning (it would be nice if this thing were linear, therefore let's prove it is) but I think it's interesting. –  Xodarap Aug 23 '12 at 0:45
It follows from linearity that $f((x+y)/2)=(f(x)+f(y))/2.$ Suppose there are two people, $A$ and $B$, and it is equally bad for person $A$ to have all the money as it is for person $B$ to have all the money. Then $f(1,0) = f(0,1) = f(\frac12,\frac12)$, so it is just as bad for both people to share the money equally. I would say this is a poor way to measure inequality. –  Rahul Aug 23 '12 at 1:08
@Rahul: Yes, if the social welfare function is linear, then inequality is meaningless. (Which is a strong result.) –  Xodarap Aug 23 '12 at 1:48

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Assume that we are working over the reals. Then the condition $f(x+y)=f(x)+f(y)$, together with continuity of $f$ (or even just measurability of $f$) is enough. This can be useful, since on occasion $f(x+y)=f(x)+f(y)$ is easy to verify, and $f(cx)=cf(x)$ is not.

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This is interesting. Do you know if the converse is true? (i.e. we only need to verify $f(cx)=cf(x)$) –  Xodarap Aug 23 '12 at 1:00
@Xodarap: For some information, please see the Wikipedia article on the Cauchy Functional Equation. If the function $f$ is continuous (but much less is needed) then it is linear iff $f(x+y)=f(x)+f(y)$ for all $x$, $y$. –  André Nicolas Aug 23 '12 at 1:15
This is very interesting. Do you know if there's an analogue for $R^n$? –  Xodarap Aug 24 '12 at 13:47
Yes, there are analogues. Don't have an immediate reference. –  André Nicolas Aug 24 '12 at 14:33

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