5

I saw this problem on my problem book:

IF $\lim\limits_{x\to0 }f(x)=0$ and $\lim\limits_{x \to 0 }\frac{f(2x)- f(x)}{x} =0$ prove that $\lim_{x \to 0} \frac{f(x)}{x}=0 $.

I tried to solve it but failed the only thing that I was able to prove its that if $\lim_{x \to 0} \frac{f(x)}{x}$ exist in $\overline {\mathbb{R}}$it it either $0$ or $\pm \infty$ and that it east to show $$\lim_{x \to 0} \frac{f(x)}{x}= \lim_{x \to 0} \frac{f(x)-f(2x)}{x} + 2\lim_{x \to 0} \frac{f(2x)}{2x}$$

i.e $L = 2L$

But I was not able to prove that the limit exist .

pie
  • 4,192
  • If $\lim_{x\to a} h_1(x)$ and $\lim_{x\to a} h_2(x)$ exists then $\lim_{x\to a} (h_1(x)\pm h_2(x))$ exists and it is equal to $\lim_{x\to a} h_1(x)\pm\lim_{x\to a} h_2(x)$ by properties of limits. Taking $h_1 = (f(x)-f(2x))/x$, $h_2 = f(2x)/2x$ and $a=0$, you should be done. – Sam Feb 03 '24 at 05:23
  • @Sam Why does $\lim h_2(x)$ exist? – geetha290krm Feb 03 '24 at 05:25
  • @Sam We don't know the limit of $h_2(x)$ exists here... it's equivalent to the question being asked. – Zarrax Feb 03 '24 at 05:25
  • Aah I see. Thank you. Is there a function f such that $\lim_{x \to 0} f(2x)/2x = 0$, but $\lim_{x \to 0} f(x)/x \ne 0$. Maybe it has something to do with the other condition. – Sam Feb 03 '24 at 05:29

1 Answers1

8

Since $\lim_{x \rightarrow 0} f(x) = 0$ you can write $f(x)$ as a telescoping sum $$f(x) = \sum_{n=0}^{\infty} (f(2^{-n} x) - f(2^{-n-1}x))$$ $$ = \sum_{n=0}^{\infty} (2^{-n - 1}x) {f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}$$ Thus one has $$|f(x)| \leq \sum_{n=0}^{\infty} 2^{-n - 1}|x| \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ $$\leq \bigg(\sum_{n=0}^{\infty} 2^{-n - 1}|x|\bigg)\sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ $$= |x| \sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ So we have $$\bigg|{f(x) \over x}\bigg| \leq \sup_{n \geq 0} \bigg|{f(2^{-n} x) - f(2^{-n-1}x) \over 2^{-n - 1} x}\bigg|$$ Now the fact that ${\displaystyle \lim\limits_{x \to 0 }\frac{f(2x)- f(x)}{x} =0}$ shows that the supremum here goes to zero as $x \to 0$ and we are done.

Zarrax
  • 44,950