Suppose $h$ is a nondecreasing function on real numbers with real values and satisfies $h(0) = 0, h(1) = 1, h(xy) = h(x)h(y)$ for all real numbers $x,y > 0$. Prove or disprove: , then $h(x^q) = h(x)^q$ for all $x > 0$ and rational $q$ and $h(x) = x^k$ for all $x > 0,$ where $k\ge 0.$
Edit: $h(x) > 0$ for all $x>0$.
It suffices to show the result for all positive rationals. One can show that $h(x^q) = h(x)^q$ for all positive rational numbers $q$ as follows: first note it suffices to show the claim for integers $a$ (i.e. $h(x^a) = h(x)^a$ for all integers $a$). It suffices to show the claim for positive integers since $1 = h(1) = h(1/x) h(x)\Rightarrow h(1/x) = 1/h(x)$ (so $h(x^{-a}) = 1/h(x)^a$ assuming the claim holds for positive integers $a$) and $h(x^0) = h(1) = x^0$. $h(x^1) = h(x)^1$. Suppose $h(x)^n = h(x^n)$ for some $n\ge 1$. Then $h(x)^{n+1} = h(x)^n h(x) = h(x^n) h(x) = h(x^{n+1}).$ Suppose $q > 0$ is rational and write $q = a/b, b > 0, a \in\mathbb{Z}$. Then $h(x)^a = h(x^a) = h(x^q)^b\Rightarrow h(x^q) = h(x)^{a/b}.$
Finally, let $r\in \mathbb{R}, r > 0$. Let $(q_n), (p_n)$ be decreasing and increasing sequences of positive rationals converging to $r$. Then since $h$ is nondecreasing, we have $h(x)^{p_n} \leq h(x^r) \leq h(x)^{q_n}$ for all $n$ and taking limits gives $h(x^r) = h(x)^r.$ So if $h(x^q) = x^q$ for all positive rationals q and $x>0,$ then $h(x^r) = x^r$ for all real numbers $x > 0, r > 0,$ and in particular $h(x) = x$. So how can I show the required result for this problem using only the given assumptions?