Show that if $f$ is a nonzero function satisfying the equation below for all real numbers $a,b,c, f(0) = 0,$ and if $f(x)\neq x^\tau$ for any $\tau > 0$ (i.e. $f$ is not identically equal to $x^\tau$ as functions of $x$), then $f$'s graph is dense in $(0,\infty)^2$.
$(f(a)-f(b))(f(b)-f(c)) (f(c)-f(a)) = f(ab^2 + bc^2 + ca^2)-f(a^2 b + b^2 c + c^2 a)$.
One may use the following properties of $f$ without proof:
- $f$ is injective
- $f(ax)=f(a)f(x)$ for all real numbers a,x, and as a corollary $f(a)\ge 0$ for all $a\ge 0$
- $f$ is odd.
One may assume $f(1) = 1$ (by replacing $f$ with $-f$).
I need to show that for any $\epsilon > 0, (a,b) \in (0,\infty)^2$, there exists $x>0$ so that $\lVert (x,f(x)) - (a,b)\rVert_2 < \epsilon.$ Assume the graph is not dense. Then there exists $\epsilon > 0$ and $(a,b)\in (0,\infty)^2$ so that for all $x>0, \lVert (x,f(x)) - (a,b)\rVert_2 \ge \epsilon.$ We may obviously ignore all $x$ that differ from $a$ by at least $\epsilon$. Using property 2, we have $f(a^2) = f(a)^2$ and $f(ax)=f(a)f(x)$. If $f$ is bounded on the positive reals, with a supremum of $M > 0$, $M$ must be at most $1$ because otherwise we can choose $\epsilon$ sufficiently small so that $(M-\epsilon)^2 > M$ and $f(x^2) = f(x)^2 = (M-\epsilon)^2 > M.$ $M$ must actually equal $1$ because $f(1)=1$ by assumption. So $f$ must map the positive real numbers to the interval $(0,1]$. But one can prove that $f(x^{m/n}) = f(x)^{m/n}$ for any positive integers $m,n$, since $f(x^n) = f(x)^n$ for any positive integer n by multiplicativity (property 2). So if each value of $f$ is less than one, then for all positive x, $f(x^{1/n}) = \sqrt[n]{f(x)}$ tends to 1 as $x\to \infty.$ But since this applies for any real number $x$, this contradicts the injectivity of $f$. However, this proof is informal in some sense as the last statement requires more justification. Hence $f$ cannot be bounded.