Mathematical Reflections 2017, Issue 3 - Problem U414

Problem:
Let $p < q < 1$ be positive real numbers. Find all functions $f: \mathbb{R} \to \mathbb{R}$ which satisfy the
conditions:

(i) $f(px+f(x))=qf(x)$ for all real numbers $x$,
(ii) $\lim_{x \to 0} \dfrac{f(x)}{x}$ exists and its finite.

Proposed by Florin Stanescu, Gaesti, Romania

Solution:
Clearly, $f(x)=0$ for all $x \in \mathbb{R}$ is a solution to the problem. Let $f \neq 0$. Observe that by condition (ii) it must be $f(0)=0$. We have two cases.

(i) $\displaystyle\lim_{x \to 0} \dfrac{f(x)}{x}=\ell \in \mathbb{R}\setminus\{0\}$. Then, from condition (i), we get
$$\dfrac{f(px+f(x))}{px+f(x)}=\dfrac{qf(x)}{x(p+\frac{f(x)}{x})}$$ and if $x \to 0$ we get $$\ell=\dfrac{q \ell}{p+\ell} \iff \ell=q-p.$$ So, $f(x)=(q-p)x$ is a solution to the problem.

(ii) $\displaystyle\lim_{x \to 0} \dfrac{f(x)}{x}=0$. Then, $f(x) \sim Cx^{1+\delta}$ if $x \to 0$, where $C \neq 0$ and $\delta>0$.
Then, from condition (i), we get
$$\dfrac{f(px+f(x))}{(px+f(x))^{1+\delta}}=\dfrac{qf(x)}{x^{1+\delta}(p+\frac{f(x)}{x})^{1+\delta}}$$ and if $x \to 0$ we get
$$C=\dfrac{qC}{p^{1+\delta}} \iff p^{1+\delta}=q,$$ contradiction.

We conclude that the only solutions are $f(x)=0$ and $f(x)=(q-p)x$.