Suppose $T>0$ . Does anyone know if there exists a sequence of partitions $(\pi_n)_{n\in\mathbb{N}}$ of the interval $[0,T]$ such that the mesh size goes to $0$, and such that it is of bounded $\alpha$-variation with $\alpha\in(0,1)$?
For those of you unfamiliar with $\alpha$-variation, a sequence of partitions $\pi_n=(0=t_0,t_1,\ldots,t_{N_{\pi_n}}=T),n\in\mathbb{N}$ is said to be of $\alpha$-variation iff
$\sup_{n\in\mathbb{N}}\sum_{j=0}^{N_{\pi_n}-1}(t_{j+1}-t_j)^\alpha<\infty$
A similar question has been asked here: $\alpha$ variation of identity map. But it does not solve the issue whether there exists such sequence of partitions or not. I have tried partitions such as $\pi_n:=(0,(\frac{1}{n})^p T,(\frac{2}{n})^p T,\ldots,T) $, with $p>\lceil\frac{1}{\alpha}\rceil$. The mesh goes to $0$ as $n$ goes to $\infty$, but to prove the $\alpha$-variation, I can only obtain estimate such as the following
$(k+1)^p-k^p=\sum_{j=0}^{p-1} C^p_j\,.\,k^j\leq 2^p k^{p-1}$
for every $k\in\mathbb{N}$ which implies
$T\sum_{k=0}^{n-1}((\frac{k+1}{n})^p-(\frac{k}{n})^p)^\alpha\leq T2^{p\alpha}\sum_{k=0}^{n-1}\frac{k^{(p-1)\alpha}}{n^{p\alpha}}\leq CT2^{p\alpha}\frac{n^{\lceil(p-1)\alpha\rceil+1}}{n^{p\alpha}}$
for some $C>0$, which is not useful.
Any help to improve my proof or to show existence/nonexistence of such sequence will be very appreciated. Thank you.
The following can be fleshed out, but I'll leave it in "hint" status, since I think the details are routine. I also use the interval $[0,1]$ rather than $[0,T]$, which is only a matter of scale.
For positive $x,y$ and positive exponent $\alpha<1$ the inequality $(x+y)^\alpha<x^\alpha+y^\alpha$ holds. This means that for any partition $\pi$, dropping a point in $\pi$ causes the sum of $\alpha$ powers of lengths of subintervals to decrease (or remain the same, if a division point occurs twice in $\pi$ and one removes a copy).
Now suppose what you say can be done. Then since one already knows the sum of $\alpha$ powers of differences must diverge to $+\infty$ as the mesh size goes to zero in the case of the "equipartitions" $0,1/n,...,(n-1)/n,1$, it follows that if we partition $[0,1]$ finely enough by a partition $P$, and then drop all the points excepting one each near the points $k/n$, we will only have decreased the $\alpha$ power difference sum of $P$, so that it must be that the sum for $P$ actually exceeds the sum for the equipartition into $n$ parts. Thus the power sum for such $P$ must also diverge.