Find a continuous function such that$f:\mathbb{R}\to \mathbb{R}$, $f(0)=0$ and $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.

Question

Could you give me a hint for this problem?

I have to find all continuous functions $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and $f(2x)\geq x+f(x)$ and $f(3x)\leq 2x+f(x)$.

I think this might be all the $f(x)=|x|$, but can it be any other functions?

Answer 1

$f(2x)\ge x+f(x)$

rewrite the second as $f(x)+2x\ge f(3x)$

Take $$f(x) =x$$

we have $$2 x\ge x+x;\;x+2x\ge3x$$ verified for any $x\in\mathbb{R}$

Answer 2

From $f(2x)\ge x+f(x)$ we obtain, repeatedly replacing $x$ by $x/2$, $$f(x)\ge x/2+f(x/2)\ge x/2+x/4+f(x/4)\ge\ldots\ge\sum^n_{k=1}x/2^k+f(x/2^n).$$ By continuity and $f(0)=0$, we have $\lim_{n\to\infty}f(x/2^n)=0$, and thus, $$f(x)\ge\sum^\infty_{k=1}x/2^k=x.$$ From the second inequality, we get in the same way $$f(x)\le2x/3+f(x/3)\le2x/3+2x/9+f(x/9)\le\ldots\le\sum^\infty_{k=1}2x/3^k=x.$$ So $x\ge f(x)\ge x$, i.e. $f(x)=x$.

Answer 3

$f(x) = x$ is a valid solution.

To see this is the only one, define $g(x) = f(x) - x$. It suffices to show that $g$ is constant. Then $f(2x) \geq x + f(x) \Leftrightarrow g(2x) \geq g(x)$, so $g(x) \geq g(\frac{x}{2^n})$ by induction. Similarly, and $f(3x) \leq 2x + f(x) \Leftrightarrow g(3x) \leq g(x)$ so $g(x) \leq g(\frac{x}{3^n})$ by induction. For any $a$ we have $$g(0) = \lim_{n \rightarrow \infty} g(\frac{a}{2^n}) \leq g(a) \leq \lim_{n \rightarrow \infty} g(\frac{a}{3^n}) = g(0)$$ so $g(a) = g(0)$ and thus $g$ is constant. Plug $f(x) = x + g(0)$ back in to conclude $g(0) = 0$.