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Let $\Bbb Q^+$ be the set of positive rational numbers. Find all solutions $f:\Bbb Q^+ \to \Bbb R$ of the functional equation $$ f(xy)=f(x)f(y), \quad x, y\in \Bbb Q. $$

Is $f(x)=x^a$ the only solution? If not, is it true if we assume that $f$ is continuous?

Chung. J
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  • Your question has an answer at http://math.stackexchange.com/questions/43964/if-fxy-fxfy-then-show-that-fx-xt-for-some-t. –  Mar 13 '14 at 01:59
  • The answer is for the equation on real numbers which is very well known, current question is for functions defined in rationals, and continuity assumed. Thank you. – Chung. J Mar 13 '14 at 04:32

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Fundamental theorem of Arithmetic can be interpreted as saying that the set of positive rational numbers under multiplication is a free abelian group with prime numbers as basis.

So one can define an endomorphism of the group by specifying it arbitrarily on primes. (An endomorphism is also a function from rationals to real numbers as required by you)

For example define $f(1)=1, f(p) = p$ for primes less than 100, and $f(p) = 1$ for primes $ p > 100$, and extend it multiplicatively. This is in particular not a 1-1 function and satisfies your functional equation and is not of the form $f(x) = x^a$.

P Vanchinathan
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