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

Question

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

Answer 1

This example is not continuous, but one can replace the jumps with linear pieces of fast growing slopes. Fix $0<\epsilon<a$. Define a sequence $(r_j)_{j\in\mathbb N}$ tending to zero inductively as follows: $r_1=1$ and $0<r_{j+1}<r_j$ so small that $$ \frac{r_{j+1}^{a-\epsilon}}{r_j^{a+\epsilon}}\le 1. $$ Then set $$ f(t)=\begin{cases}0&t=0,\\ r_{j+1}^{a-\epsilon}& r_{j+1}<t\le r_j,\\ 1&t>1. \end{cases} $$ 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 $$ \liminf \frac{f(t)}{t^{a+\epsilon}}=\lim_j\frac{f(r_j)}{r_j^{a+\epsilon}}=\frac{r_{j+1}^{a-\epsilon}}{r_{j}^{a+\epsilon}}\le 1. $$