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I have a general scalar function which has the properties: \begin{align} f(s\,a,b,c)&=s\,f(a,b,c)\\ f(s\,a,s\,b,s\,c)&=f(a,b,c) \end{align}

where $s$ can be any real number, so the invariance is continous. How far can I restrict the function $f$ now? My approach is: $f$ is linear in $a$ and can be decomposed by a function $g$ that transforms as: \begin{align} f(a,b,c)&=a\cdot g(b,c)\\ g(s\,b,s \,c)&=\frac{1}{s} g(b,c) \end{align}

Can I further restrict $g$ from that tranformation law?

Especially can I assume any analytical form for $g$? If yes, how?

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$g(b,c)$ will be a constant times $b^pc^q$, where $p+q=-1$

Note as mentioned in the comment, $g(b,c)=h(\frac bc)b^pc^q$. The function $h$ depends only on the ratio of the $b$ and $c$ numbers.

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  • $\begingroup$ Are you sure that this is true? Take the function $g(b,c)=\frac{b^2}{(b^2+c^2)^{3/2}}$. This transforms to $1/s$ if p and q $\rightarrow$s*q. But I cant find a $p$ such that the equality holds for all b and c. $\endgroup$ – Mr Puh Mar 6 at 20:22
  • $\begingroup$ You are right. The "constant" that I've mentioned is a number that does not depend on the magnitude of $b$ or $c$ directly, but it's a function of the ratio of the two. $\endgroup$ – Andrei Mar 6 at 20:35
  • $\begingroup$ okay. I try to understand this thank you. Can you give me a hint what brings you to this conclusion? $\endgroup$ – Mr Puh Mar 6 at 20:37
  • $\begingroup$ Write $b=r\cos\theta$ and $c=r\sin\theta$. Since you multiply with $s$, if you expand $g$ in powers of $b$ and $c$, you have all terms with $p+q=-1$. So one more correction to make: $$g(b,c)=\sum_{p\in \mathbb R}h_p\left(\frac bc\right)b^pc^{-1-p}$$ $\endgroup$ – Andrei Mar 6 at 20:45
  • $\begingroup$ I do not understand this notation of the sum running over all p in Reals. What do you mean by that? also is $h_p$ a function or a constant now? $\endgroup$ – Mr Puh Mar 6 at 20:50

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