0
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ What about $f(r)=r^{a-\epsilon}$ for some small $\epsilon>0$? $\endgroup$
    – user473423
    Apr 8, 2022 at 7:56
  • $\begingroup$ @Echo Its lower limit is infinite. $\endgroup$
    – Watheophy
    Apr 8, 2022 at 8:00
  • $\begingroup$ 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 $r$. $\endgroup$
    – user473423
    Apr 8, 2022 at 8:02
  • $\begingroup$ @Echo Do you mean $f(r) = r^{a-r}$? $\endgroup$
    – Watheophy
    Apr 8, 2022 at 8:07
  • $\begingroup$ @Echo The dominator in the lower limit is $\delta^{a+\epsilon}$. $\endgroup$
    – Watheophy
    Apr 8, 2022 at 8:09

1 Answer 1

1
$\begingroup$

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 $$ \liminf \frac{f(t)}{t^{a+\epsilon}}=\lim_j\frac{f(r_j)}{r_j^{a+\epsilon}}=\frac{r_{j+1}^{a-\epsilon}}{r_{j}^{a+\epsilon}}\le 1. $$

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Watheophy
    Apr 8, 2022 at 12:11
  • $\begingroup$ Say $r_{j+1}=r_j^{\frac{a-\epsilon}{a+\epsilon}}$. $\endgroup$
    – user473423
    Apr 8, 2022 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.