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Let $a\in\mathbb{C}^*$ with $|a|\not=1$. Let $m\in\mathbb{Z}$. Find all functions $g:\mathbb{C}^*\to\mathbb{C}^*$ and constants $c\in\mathbb{C}^*$ such that $g(x)=g(a^mx)c^m$.

I know one possibility is to let $n\in\mathbb{Z}$ then $c=a^n$ and $g(x)=x^{-n}$. My desired result is to have this be the only possible solution.

Mohsen Shahriari
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user44322
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  • As it turns out, it was also required for $g$ to be holomorphic. So, when we write $g$ as a Taylor series and compare coefficients, the result follows. – user44322 Nov 13 '13 at 19:13

1 Answers1

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Let $g(x) = kx^r$. Then $g(a^m x) c^m = k(a^m x)^r c^m$. From the functional equation,

$$ 1 = \frac{g(a^m x)}{g(x)} c^m,$$

$$ 1 = \frac{k(a^m x)^r}{kx^r} c^m,$$

$$ 1 = a^{mr} c^m, $$

$$ 0 = m(r\log a + \log c ),$$

so either $m=0$ (trivial) or $r= - \frac{\log c}{\log a}$. This gives us the slightly more general particular solution,

$$ g_0(x) = k x^{- \frac{\log c}{\log a}}, \space \text{where $k$ is an arbitrary constant.}$$

I doubt this is the most general solution, but I haven't thought of anything else yet.

David H
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