rjawbreakerca

2022-07-03

Given $\underset{x\to 0}{lim}\frac{f\left(2x\right)-f\left(x\right)}{x}=0$ and $\underset{x\to 0}{lim}f\left(x\right)=0$ show that $\underset{x\to 0}{lim}\frac{f\left(x\right)}{x}=0$

Melina Richard

Expert

We could simplify all this. Let
$\rho \left(x\right)=\underset{|y|⩽|x|}{sup}|\frac{f\left(2y\right)-f\left(y\right)}{y}|$
We have for $n⩾0$
$\begin{array}{rcl}|f\left(x\right)-f\left(\frac{x}{{2}^{n+1}}\right)|& =& |\sum _{k=0}^{n}\left(f\left(\frac{x}{{2}^{k}}\right)-f\left(\frac{x}{{2}^{k+1}}\right)\right)|\\ & ⩽& \sum _{k=0}^{n}\rho \left(x\right)\frac{|x|}{{2}^{k+1}}\\ & ⩽& |x|\rho \left(x\right)\left(1-\frac{1}{{2}^{n+1}}\right)\end{array}$
By taking the limit of this inequality when $n\to \mathrm{\infty }$, it follows that$|\frac{f\left(x\right)}{x}|⩽\rho \left(x\right)\underset{x\to 0}{⟶}0$

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