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I've been looking at functions $f:\mathbb{R}^n \to \mathbb{R}$ which necessarily satisfy the following 3 properties. Given $ a_1, a_2, \dots a_n \in \mathbb{R}^+ $

$\begin{array} { l l } 1. & f(x_1 + c, x_2 + c, \dots , x_n + c) = f(x_1, x_2, \dots , x_n) + c \\ 2. & f(cx_1, cx_2, \dots , cx_n) = cf(x_1, x_2, \dots, x_n) \\ 3. & \sum_{i=1}^n a_i x_i = 0 \Leftrightarrow f(x_1, x_2, \dots, x_3) = 0 \end{array}$

I believe that the the only function that satisfies this is $f(x_1,x_2, \dots, x_n) = \frac{\sum_{i=1}^n a_i x_i}{\sum_{i=1}^n a_i}$

I have the proof too while I'll write up later. I'm interested in all of your interpretations of this.

What happens if we only have 2 out of these 3 conditions?

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  • $\begingroup$ If $n>1$ let $f(x_1,,,,x_n)=(\sum_{j=2}^na_jx_j)/(\sum_{j=2}^n).$ Then $f$ satisfies (1,) and (2.) but not (3.). $\endgroup$ – DanielWainfleet Mar 3 '18 at 10:16
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Let $x_1$, $x_2$, $\dots$, $x_n$ be given real numbers. We seek a value of $c$ so that $a_1 (x_1 + c) + a_2 (x_2 + c) + \dots + a_n (x_n + c) = 0.$ Solving for $c$, we get $c = -\frac{a_1 x_1 + a_2 x_2 + \dots + x_n x_n}{a_1 + a_2 + \dots + a_n}.$

From the third property, $f(x_1 + c, x_2 + c, \dots, x_n + c) = 0$, so by the first property, $f(x_1, x_2, \dots, x_n) = -c = \frac{a_1 x_1 + a_2 x_2 + \dots + x_n x_n}{a_1 + a_2 + \dots + a_n}.$ Note that the second property is never used.

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