Find a strictly decreasing $f:(0,1) \to [0, \infty]$ such that $f(0^{+})= \infty$, $f(1)=0$ and $f(u^2)=2 f(u)$.

Question

Assume that $f$ is a function from $(0,1]$ to $[0, \infty]$ that is strictly decreasing and satisfies $f(1)=0$, $f(0^{+})= \infty$ and $f(u^{2})=2 f(u)$. I do not know if it helps, but let us assume that $f$ is convex. There exists one family of solutions given by the set $\lbrace x \to -k \ln x, k>0 \rbrace$. Is there another family of solutions that does not lie in the above set?

Answer 1

Let $g(x):=f\big(\exp(-x)\big)$ for $x\in\mathbb{R}_{>0}$. First, I shall prove that every solution $f:(0,1)\to\mathbb{R}$ to the functional equation $f(u^2)=2\,f(u)$ for all $u\in(0,1)$ takes the form $$f(u)=-\ln(u)\,h\left(\frac{\ln\big(-\ln(u)\big)}{\ln(2)}\right)\tag{*}$$ for every $u\in(0,1)$, where $h:\mathbb{R}\to\mathbb{R}$ is a periodic function of period $1$.

We have $$g(2x)=f\big(\exp(-2x)\big)=f\Big(\big(\exp(-x)\big)^2\Big)=2\,f\big(\exp(-x)\big)=2\,g(x)$$ for all $x>0$.

Let $h(t):=\dfrac{g\big(2^t\big)}{2^t}$ for all $t\in\mathbb{R}$. Then, $$h(t+1)=\dfrac{g\big(2^{t+1}\big)}{2^{t+1}}=\frac{g\big(2\cdot 2^t\big)}{2\cdot 2^t}=\frac{2\cdot g(2^t)}{2\cdot 2^t}=\frac{g(2^t)}{2^t}=h(t)\,.$$ Therefore, $h$ is a periodic function with period $1$. Ergo, $$f(u)=g\big(-\ln(u)\big)=-\ln(u)\,h\left(\frac{\ln\big(-\ln(u)\big)}{\ln(2)}\right)$$ for every $u\in(0,1)$.

With the additional assumption that $f$ is continuous and nonnegative, $h$ is also continuous and nonnegative. From the limit conditions on $f$, we have $$\lim\limits_{x\to 0^+}\,g(x)=0\text{ and }\lim\limits_{x\to\infty}\,g(x)=\infty\,.$$
This shows that $h$ is bounded below above $0$, namely, $h(t)\geq l$ for every $t\in\mathbb{R}$, where $l>0$ is fixed. These are all required properties of $h$. That is, all solutions $f$ with the continuity, nonegativity, and limit conditions must be of the form (*), where $h:\mathbb{R}\to\mathbb{R}_{>0}$ is a continuous periodic function with period $1$.

In particular, if $h\equiv k$ for some constant $k>0$, then we get the family of solutions that the OP proposed. While it has not been taken into account that $f$ is strictly decreasing, it is an easy exercise to show that there are nonconstant periodic functions $h$ that makes $g$ a strictly increasing function, which in turns makes $f$ a strictly decreasing function. I have not thought about the convexity condition, but it is highly probable that only convex solutions take the form proposed by the OP.

If you only need $f$ to be strictly decreasing, then $h(t):=1+\epsilon\,\sin(2\pi t)$ for all $t\in\mathbb{R}$ works for any real number $\epsilon$ such that $|\epsilon|\leq \dfrac{\ln(2)}{\sqrt{4\pi^2+\big(\ln(2)\big)^2}}\approx 0.10965$. This gives another family with parameter $\epsilon$.