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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?

Martin Sleziak
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  • I tried solving it with a Taylor series about $0$. I got a reccurance relation for the constants, where $c_{k+2}$ is a function of $c_k$, all the odd coefficients are zero, and $c_0$ is arbitrary. Unfortunately the series doesn't seem to converge so I won't bother reproducing that here.

    You could try something along the same lines but more sophisticated. Perhaps a series about $a$ or $1-a$, or a Laurent series.

     – Paul Castle Jun 04 '15 at 03:38
  • Should $f$ be continous or not? What's its domain? Codomain? please complete the question. Thanks. – Mohsen Shahriari Jun 11 '15 at 07:20

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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.

Ivan Neretin
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