Solutions to q(x y) = q(x) q(y)

This is a problem from Loomis and Sternberg's Advanced Calculus

Suppose the function $q: \R \to \R$ satisfies $q(x y) = q(x) q(y)$ for
all $x, y \in \R$. Note that $q(x) = x^n$ ($n$ a positive integer) and
$q(x) = |x|^r$ ($r$ any real number) satisfy this ``functional
equation''. So does $q(x) = 0$ ($r = - \infty$?). Show that if $q$
satisfies the functional equation and $q(x) > 1$ for $x > 1$, then
there is a real number $r > 1$ such that $q(x) = x^r$ for all positive
$x$.

I figured out so far that q(x^{m/n}) = q(x)^{m/n}, and that it suffices
to prove that q(x)^r = q(x^r), but now I'm stuck

Any help would be greatly appreciated

--Boaz

.

Relevant Pages

  • Re: Solutions to q(x y) = q(x) q(y)
    ... > This is a problem from Loomis and Sternberg's Advanced Calculus ... Prev by Date: ...
    (sci.math)
  • Re: Solutions to q(x y) = q(x) q(y)
    ... Timothy Little wrote: ... >> This is a problem from Loomis and Sternberg's Advanced Calculus ... by working backwards you can also show it is sufficient when given the ...
    (sci.math)