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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 $$\lim_{x \to0^+} \frac{f(x)}{x}=0$$

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You might be able to use… Martin Sleziak's answer in particular. – David Mitra Jan 10 '12 at 0:05

1 Answer 1

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. $\diamondsuit$

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