Find all the functions $f:\left[0,1\right]\rightarrow\mathbb{R}$ so that ${x}'f\left(x\right)\geq 2xf\left(x^{2}\right)$

Question

Find all the functions $f:\left [ 0, 1 \right ]\rightarrow\mathbb{R}$ on the interval $\left [ 0, 1 \right ]$ so that $${x}'f\left ( x \right )\geq {\left ( x^{2} \right )}'f\left ( x^{2} \right ),$$ in other words, that $f\left ( x \right )\geq 2xf\left ( x^{2} \right ).$

I simplify the original problem, I think we need to find the increasing function $f\left ( x \right )= {g}'\left ( x \right )$ since $x\geq x^{2}$ but are these all ? And how to find at least one function $g\left ( x \right )$ like that.. is it necessary to add the condition that the function $f$ is continuous ? Thanks for helping.

Answer 1

A discontinuous solution is $f:[0,1]\to[0,\infty)$ given by $$f(x) =\begin{cases} \frac{-1}{\log x} & x\in(0,1), \\0 &x=0\text{ or } 1.\end{cases}$$Then there is equality at $x=0$ and $x=1$, and for $x\in(0,1)$, $f(x^2) = \frac{-1}{2\log x} = \frac12 f(x)$, so it is an exact solution to $$ f(x) = 2f(x^2)$$ since for $x\in (0,1)$, $2\ge 2x$ and $f\ge 0$, it solves the inequality $$ f(x) \ge 2xf(x^2).$$ The same proof also shows that for any $a,b\ge 0$, the following function is also a solution: $$f(x) =\begin{cases} \frac{-a}{\log x} & x\in(0,1), \\b &x=0, \\ 0 &x=1.\end{cases}$$

In fact, we can create many more solutions by using the extra space in the inequality $2x\le2$ better. Given a positive solution to $f(x)\ge 2f(x^2)$ if $h$ is a positive solution to $h(x)\ge x h(x^2)$ then $\tilde f:= hf$ is another solution to the inequality in the question. It is easy to show that we can use $h(x)=x^{r}$ for $r\ge -1$ and $h(x)=(1-x)^s$ for $s\le0$.

Furthermore, it is also similarly easy to show that $1/\log(1-x)$ is a negative solution to $2f(x^2)\le f(x)$. Due to the sign change, the relations for $r$ and $s$ in determining if $x^r$ and $(1-x)^s$ can be used are flipped.

Finally the inequality is additive in the function $f$. Thus, if $f_1$ and $f_2$ are solutions then so is $f_1+f_2$.

So for instance, another solution is $$\frac{x^{-3}(1-x)^{100}}{\log(1-x)}-\frac{(1-x)^{-2}}{\log x}$$ This one blows up as you approach both endpoints (after suitable redefinition at the endpoints so that they are indeed functions on $[0,1]$).

Answer 2

The only continuous function on $[0,1]$ such that $f(x)\ge 2xf(x^2)$ is identically zero.

Proof.

We can immediately note that $f(0)\ge 0$ and $f(1)\le 0$.

Suppose $f(x)<0$ for some $x\in [0,1]$. By continuity we can choose $x\in (0,1)$. Then $f(x^2)\le \frac{f(x)}{2x}$ and so $f$ is negative on $x,x^2,x^4,...$ where the arguments are tending to $0$. Furthermore, $f$ is strictly monotonic decreasing on this sequence as soon as the argument becomes less than a half. Therefore, by continuity $f(0)<0$, a contradiction.

Now suppose $f(x)>0$ for some $x\in [0,1]$. By continuity we can choose $x\in (0,1)$. Then $f(\sqrt x)\ge 2x{f(x)}$ and so $f$ is positive on $x,x^{1/2},x^{1/4},...$ where the arguments are tending to $1$. Furthermore, $f$ is strictly monotonic increasing on this sequence as soon as the argument becomes more than a half. Therefore, by continuity $f(1)>0$, a contradiction.