Let $f:(-a,a) \rightarrow \mathbb{R}$, with $a>0$. Assume $f(x)$ is continuous at 0 and such that the limit $$ \lim_{x \rightarrow 0} \frac{f(x)-f(kx)}{x} = l $$ exists, where $0<k<1$. Show that $f'(0)$ exists. What happens to this conclusion when $k>1$?
Someone can explain me how to prove the last one?
$$\color{Red}{Assumption\, |k-1|=|1-k|\ge 1}$$ I start from any given $\varepsilon^*=\varepsilon|1-k|$ we have $$ \left | \frac{f(x)-f(kx)}{x}-l\right|\lt \varepsilon^* \,,\text{if }|x|\lt\delta $$After simplification we have : $$ \left | \frac{f(x)-f(kx)}{x-kx}-\frac{l}{1-k}\right|\lt \varepsilon \,,\text{if }|x|\lt\delta$$ The above inequality holds, whenever x is in the $\delta$ neighborhood of $\color{green}{0}$ i.e : $$|x|\lt\delta$$ And hence for all x in smaller $\frac{\delta}{|1-k|}$ neighborhood of $\color{green}{0}$ i.e : $$|x|\lt\frac{\delta}{|1-k|}\iff |x-kx|\lt\delta$$ Now our favourite substitution put $\color{brown}{x-kx=-h}$ for all x in$\frac{\delta}{|1-k|}$ neighborhood of $\color {Green} 0$, which further simplifies our inequalities to : $$\left|\frac{f(x+h)-f(x)}{h}-\frac{l}{1-k}\right|\lt \varepsilon \,,\text{if }|h|\lt\delta$$ From where, we can say function $f$ is derivable in$\frac{\delta}{|1-k|}$ neighborhood of $\color{green}{0}$ $\color{purple}{\text{if}}$ $f$ is continuous in $\frac{\delta}{|1-k|}$ neighborhood and hence at $\color{green}{0}$, provided $\color{red}{|k-1|\ge1}$.
$\color{purple}{It}$ could also be easily shown the results holds true for all $1<k<2$ using the same technique where we work in $\frac{\delta}{|k|}$ neighborhood.
$l - \varepsilon^{} < \frac{f(x)-f(kx)}{x} < l + \varepsilon^{}$
– Mar 16 '15 at 16:47