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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TopNewestLet \(x_1\), \(x_2\), \(\dots\), \(x_n\) 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. – Jon Haussmann · 3 days, 1 hour ago

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– Josh Banister · 3 days, 1 hour ago

The idea behind the second property is what happens if the third one was never there? (What happens now if we only assume the first 2 ideas. Hence the linear part of the title)Log in to reply

That's very interesting! I believe there is a geometric interpretation of the result, as the (normalized) distance of the vector

xprojected ontoa. – Calvin Lin Staff · 3 days, 2 hours agoLog in to reply