Solving an equation for a function: $1-\frac{f(x)}{f(ax)} = (1-a)^2x^2$ I am trying to do some proof and in connection with that this question arose:
Can you find a decreasing function so that
$$
1-\frac{f(x)}{f(ax)} = (1-a)^2x^2
$$
where $0\leq a \leq 1$ and $x$ is positive? I have tried to plug in various guesses for $f(x)$ like $\frac{1}{1+ax}$ but with no luck.
Any suggestions? 
 A: Let's see whether the function can really be decreasing.
From $\frac{f(x)}{f(ax)} = 1-(1-a)^2x^2$ we immediately see that for large enough $x$ there would be a change of sign of $f$ between $x$ and $ax$, and also between $ax$ and $a^2x$, which is incompatible with monotonic decrease (or increase, for that matter). Thus $f$ can't be monotonic on $\mathbb R_+$. Let's try and make it monotonic at least in some neighborhood of 0, maybe that would suffice for your purposes, whatever those might be.
Now, functional equations linking $f(x)$ to $f(ax)$ for some constant $a$ are "bad", in that they leave too much freedom. See, we may define $f(x)$ to be a totally arbitrary function (even nowhere continuous, if we'd like) for $x$ between $a$ and $1$, and then continue both left and right, repeatedly using the expression for $f(x)$ via $f(ax)$ or vice versa. That would be a solution defined on entire $\mathbb R_+$ and consistent with the equation.
If we want the function to be reasonably well-behaved (say, continuous at 0), that's another story. In that case we may rely on certain limit:
$$f(x) = \Big(1-(1-a)^2x^2\Big)f(ax) = \Big(1-(1-a)^2x^2\Big)\Big(1-(1-a)^2x^2a^2\Big)f(a^2x)=\Big(1-(1-a)^2x^2\Big)\Big(1-(1-a)^2x^2a^2\Big)\Big(1-(1-a)^2x^2a^4\Big)\cdot\ \dots\ \cdot f(0)$$
Wait, isn't this the q-Pochhammer symbol? Although not an elementary function, it is pretty well-known and indeed decreasing in the neighborhood of 0. So this is the thing you were after.
