# What can I say about the form of an invariant function?

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?

$$g(b,c)$$ 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.
• 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. – Mr Puh Mar 6 at 20:22
• 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. – Andrei Mar 6 at 20:35
• 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: $$g(b,c)=\sum_{p\in \mathbb R}h_p\left(\frac bc\right)b^pc^{-1-p}$$ – Andrei Mar 6 at 20:45
• 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? – Mr Puh Mar 6 at 20:50