If $f(x)$ and $\frac{f(2x)-f(x)}{x}$ have limit $0$ as $x\to 0$, then $\frac{f(x)}{x}\to 0$

Question

Prove that if $\lim_{x\to 0} f(x) = 0$ and $\lim\limits_{x\to 0} \frac{f(2x)-f(x)}{x}= 0$, then $\lim\limits_{x\to 0} \frac{f(x)}{x} = 0.$

I try to solve it in this way:

$f(x)$ is infinitesimal because $\lim\limits_{x\to 0} f(x) = 0,$

$\lim\limits_{x\to 0} \frac{f(2x)-f(x)}{x}= 0,\Rightarrow {f(2x)-f(x)}=o({x})\Rightarrow {f(2x)}=f(x)+o({x}).$

Well

$$\lim_{x\to 0} \frac{f(2x)-f(x)}{x}= \lim_{x\to 0} \frac{f(x)+o({x})}{x}= \lim_{x\to 0} \frac{f(x)}{x}+\lim_{x\to 0}\frac{o({x})}{x}=0$$

$\lim_{x\to 0}\frac{o({x})}{x}=0$, of course; then $$\lim_{x\to 0} \frac{f(x)}{x} = 0$$

Answer 1

As pointed out by Ilya the argument of the OP is incorrect. Here is my proposal:

There is a function $x\to g(x)$ with $g(x)\to 0$ $\ (x\to 0)$ such that

$$f(2x)-f(x)=x\ g(x)\ .$$

As $f(x)\to 0$ $\ (x\to 0)$ we can set up the following telescopic series:

$$f(x)\ =\ \sum_{k=1}^\infty \bigl(f(2\cdot 2^{-k}x)- f(2^{-k}x)\bigr) \ =\ x\ \sum_{k=1}^\infty 2^{-k} g(2^{-k}x)\ .$$

Since $\sum_{k\geq 1} 2^{-k}=1$ it follows that

$$\left|{f(x)\over x}\right|\ \leq\ \sup_{0<|t|<|x|}\ |g(t)|\to 0\qquad (x\to 0)\ .$$

Answer 2

This is Problem 3.2.7 from Problems in real analysis by Teodora-Liliana T. Rădulescu, Vincentiu D. Radulescu, Titu Andreescu; p.121-122. The only difference is that they use $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$, which is obviously equivalent to the condition from the question, and they work with $x>0$, which can be easily modified. I'll give a sketch of their solution here.

The basic idea is to notice $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$ implies that for given $\varepsilon>0$ there is $\delta>0$ such that $$|f(x)-f(x/2)|<\varepsilon |x|$$ whenever $x<\delta$. For any fixed $x<\delta$ we have $|f(x/2^n)-f(x/2^{n+1})|<\varepsilon |x|/2^n$ and by triangle inequality we get $$|f(x)-f(x/2^n)| \le 2\varepsilon |x|.$$ Now taking the limit $n\to\infty$ and using $\lim\limits_{x\to 0} f(x)=0$ gives $$|f(x)|\le 2\varepsilon |x|$$ $$\frac{|f(x)|}{|x|} \le 2\varepsilon$$ for any $x<\delta$.

Answer 3

Here is a slight variation. Define $\varphi$ on a punctured neighborhood of $x = 0$ by

$$ \varphi(x) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}, \qquad x \neq 0.$$

We wish to show that $\varphi(x)$ has a limit as $x \to 0$. To this end, we begin by rephrasing the limit condition in terms of $\varphi$:

$$ \lim_{x \to 0} \left[ 2\varphi(2x) - \varphi(x) \right] = 0, $$

or equivalently,

$$ \lim_{x\to0} \lambda(x) = 0 \qquad \text{where} \quad \lambda(x) = \varphi(x) - \frac{\varphi(x/2)}{2}. $$

Since we are interested in the limit of $\varphi$, we wish to write $\varphi(x)$ in terms of $\lambda(x)$. From this, we obtain

\begin{align*} \left| \varphi(x) \right| = \left| \lambda(x) + \frac{\varphi(x/2)}{2} \right| \leq \left| \lambda(x) \right| + \left| \frac{\varphi(x/2)}{2} \right|. \end{align*}

Taking $\limsup$ as $x \to 0$, this gives

$$ \limsup_{x \to 0} \left| \varphi(x) \right| \leq \frac{1}{2} \limsup_{x \to 0} \left| \varphi(x/2) \right| = \frac{1}{2} \limsup_{x \to 0} \left| \varphi(x) \right|, $$

from which we deduce

$$ \limsup_{x \to 0} \left| \varphi(x) \right| = 0. $$

This is equivalent to $\varphi(x) \to 0$ as $x \to 0$, hence we conclude the desired claim.