# A problem of the limit of $\frac{\sup_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a}}$

Suppose that $$f$$ is a continuous function on $$[0,1]$$. For $$0, if $$\varlimsup_{\delta \rightarrow 0} \frac{\sup_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a}} = \infty,$$ then, given any $$\epsilon>0$$, is it true that $$\varliminf_{\delta \rightarrow 0} \frac{\sup_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a+\epsilon}} = \infty?$$

• What about $f(r)=r^{a-\epsilon}$ for some small $\epsilon>0$?
– Echo
Apr 8 at 7:56
• @Echo Its lower limit is infinite. Apr 8 at 8:00
• Ok, lets rename it, say $f(r)=r^{a-\delta}$ for some small $\delta>0$? As soon as $\epsilon>\delta$, you're in trouble. Finally, you let $\delta$ shrink with $r$.
– Echo
Apr 8 at 8:02
• @Echo Do you mean $f(r) = r^{a-r}$? Apr 8 at 8:07
• @Echo The dominator in the lower limit is $\delta^{a+\epsilon}$. Apr 8 at 8:09

This example is not continuous, but one can replace the jumps with linear pieces of fast growing slopes. Fix $$0<\epsilon. Define a sequence $$(r_j)_{j\in\mathbb N}$$ tending to zero inductively as follows: $$r_1=1$$ and $$0 so small that $$\frac{r_{j+1}^{a-\epsilon}}{r_j^{a+\epsilon}}\le 1.$$ Then set $$f(t)=\begin{cases}0&t=0,\\ r_{j+1}^{a-\epsilon}& r_{j+1}1. \end{cases}$$ We have $$\limsup\frac{f(t)}{t^a}=\lim_j\frac{f^+(r_{j+1})}{r_{j+1}^a}=\lim_j\frac{r_{j+1}^{a-\epsilon}}{r_{j+1}^a}=\infty$$ while on the other hand $$\liminf \frac{f(t)}{t^{a+\epsilon}}=\lim_j\frac{f(r_j)}{r_j^{a+\epsilon}}=\frac{r_{j+1}^{a-\epsilon}}{r_{j}^{a+\epsilon}}\le 1.$$
• Your sequence $r_j$ is kind of weird. Its ratio $r_{j+1}/r_j$ converges faster than $r_j$. Could you please example one sequence like this? Apr 8 at 12:11
• Say $r_{j+1}=r_j^{\frac{a-\epsilon}{a+\epsilon}}$.