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
$$ \varliminf\_{\delta \rightarrow 0} \frac{\sup\_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a+\epsilon}} = \infty? $$



A problem of the limit of $\frac{\sup_{0<\lvert y\rvert\leq \delta}\lvert f(x+y)-f(x)\rvert}{\delta^{a}}$
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
$$
\liminf \frac{f(t)}{t^{a+\epsilon}}=\lim\_j\frac{f(r\_j)}{r\_j^{a+\epsilon}}=\frac{r\_{j+1}^{a-\epsilon}}{r\_{j}^{a+\epsilon}}\le 1.
$$


