Showing $f(x)/x \to 0$ when $\lvert f(x) - f(\lambda x)\rvert/x \to 0$ I would like to solve this problem, but I do not know how ...
Let  $f:(0;1) \rightarrow \mathbb{R}$ be a function such that:
$$\lim_{x \to0^+}f(x)=0$$
and such that there exists  $0<\lambda<1$ such that:
$$\lim_{x \to0^+} \frac{ \left [ f(x)-f(\lambda x) \right ]}{x}=0$$
prove that
$$\lim_{x \to0^+} \frac{f(x)}{x}=0$$
 A: Since
$$ \frac{f(x) - f(\lambda x)}{x} \to 0,$$
for any $\epsilon > 0$, we can restrict $x$ near enough $0$ so that we have $\lvert f(x) - f(\lambda x)\rvert \leq \epsilon \lvert x \rvert$. Since $0 < \lambda < 1$, this means that we also have $\lvert f(\lambda^n x) - f(\lambda^{n+1} x) \rvert \leq \epsilon \lvert x \rvert\lambda^n$ for each $n \geq 0$. By using the triangle inequality, we get that
$$ \begin{align}
\lvert f(x) - f(\lambda^n x) \rvert &= \lvert f(x) - f(\lambda x) + f(\lambda x) + \cdots - f(\lambda^n x)\vert \\
&\leq \epsilon \lvert x \rvert ( 1 + \lambda + \lambda^2 + \cdots + \lambda^{n-1}) \\
&\leq \epsilon \lvert x \rvert \frac{1 - \lambda^n}{1 - \lambda} \\
&\leq \epsilon \lvert x \rvert \frac{1}{1 - \lambda}.
\end{align}$$
Notice the final expression on the right is independent of $n$. By letting $n \to \infty$, the right hand side does not change, while the term $f(\lambda^n x) \to 0$ on the left hand side. This leads to an expression of the form
$$\lvert f(x) \rvert \leq \epsilon \lvert x \rvert \frac{1}{1 - \lambda},$$
or equivalently
$$ \frac{\lvert f(x) \rvert}{\lvert x \rvert} \leq \epsilon \frac{1}{1 - \lambda}$$
for all $\epsilon > 0$. Choosing $\epsilon \to 0$ completes the proof. $\diamondsuit$
