A counterexample showing $BV_p \neq AC_p$

Question

I am trying to work through a supposedly simple counterexample given in papers by Love and Gehring regarding a $p$-power generalization of bounded variation and absolute continuity.

Let $p > 1$. For a function $f: [0,1] \to \mathbb{R}$, the $p$-variation is defined by $$ \text{Var}_p f = \sup \left(\sum\limits_{j=1}^n |f(u_j)-f(u_{j-1})|^p \right)^{\frac{1}{p}}, $$ where the supremum is taken over all partitions $0 = x_0 < x_1 < \dots < x_n = 1$ of $[0,1]$. Likewise, we can define the $p$-variation on a smaller subinterval $[a,b] \subseteq [0,1]$ which we denote $\underset{[a,b]}{\text{Var}_p} f$. The space of all functions with finite $p$-variation is denoted $BV_p([0,1])$ or just $BV_p$ since we are fixing the domain.

Likewise, we can define a generalized version of absolute continuity. We will say $f \in AC_p$ if for every $\varepsilon > 0$, there exists $\delta > 0$ such that $\left( \sum\limits_{j=1}^n |f(b_j) - f(a_j)|^p \right)^{\frac{1}{p}} < \varepsilon$ whenever $\{(a_j,b_j)\}$ is a collection of non-overlapping subintervals of $[0,1]$ such that $\left( \sum\limits_{j=1}^n (b_j - a_j)^p \right) < \delta$. Much like in the classic $p = 1$ case, one can show $AC_p \subseteq BV_p$. However, a continuous function of bounded $p$-variation need not be $p$-absolutely continuous, like the Cantor staircase function demonstrates in the $p = 1$ case.

Both of the papers I mentioned cite $$ u(x) = \sum\limits_{k = 0}^\infty \frac{1}{2^{k/p}} \cos(2^k\pi x) $$ as a counterexample which is continuous, has finite $p$-variation, and is not $p$-absolutely continuous (which is what I'm struggling to show). I have tried to work directly with the definition in addition to some alternative characterizations I found, but I haven't been successful so far. For example, Love's paper shows that if $f \in AC_p$, then $\underset{[x,x+h]}{\text{Var}_p} f = o(h^{\frac{1}{p}})$ for almost all $x$. Likewise, Gehring's paper mentions that if $f$ has finite $p$-variation and finite derivative except on possibly a countable set, then $f$ is $p$-absolutely continuous. It's worth pointing out that naively differentiating the terms defining $u$ yields a diverging series for all $x$.

I still think working directly with the negation of the $AC_p$ condition is the way to go. A key observation I made is that for $N$ large, the dyadic partition $x_j = j2^{-N}, j = 0, \dots, 2^N$ can satisfy the $\delta$ requirement (which obviously can't happen if $p = 1$). More specifically, for any $\delta > 0$, we require $N$ to be large enough such that $2^{-N/q} < \delta$, where $q$ is the conjugate of $p$ satsifying $\frac{1}{p} + \frac{1}{q} = 1$. I then would need to show $$ \left( \sum\limits_{j=1}^{2^N} |u(x_j) - u(x_{j-1})|^p \right)^{\frac{1}{p}} \geq 1 $$ or some other constant independent of $\delta$ (but not necessarily $p$). Using a dyadic partition allows for some nice reindexing of the series defining $u$, but I haven't figured out a good use yet. Part of the difficulty is that I need to directly work with cosines and bound the differences from below, rather than above.

Answer 1

So first of all, what is claimed in Love's paper is slightly different, It says that for a sufficient large choice of the parameter $c>0$ the function $$ g(x): = \sum_{n=0}^\infty c^{-n/p}\cos(c^n \pi x) $$ is of vounded $p$-variation but not $AC_p$. One way to see it is that to notice that the Weierstrass type function $$ W_{a,b}(x) = \sum_{n=1}^\infty b^{-na}\cos(b^n \pi x), $$ where $0<a<1, b \in \mathbb{N}$ is Holder continuous of order $\alpha: = -\log_b(a)$ but not Higher. I think the best source for this is Hardy's paper "Weierstrass's Non-Differentiable Function" Theorem 1.32. Although you can also read it off the Littlewood Paley characterization of Holder functions. Hence if $a=1/p, b=c$, this shows that for $c$ sufficiently large your function $g$ cannot satisfy $$ |g(x)-g(x+h)| \leq Var^p_{[x,x+h]} g = o(h^{1/p}) $$ suffices to take $ p\ln(p)<\ln(c)$.

Answer 2

For anyone who isn't aware, this question appeared as an exercise in Leoni's A First Course in Sobolev Spaces, Second Edition as Exercise 3.32. Xinrui Zhao and I wrote up a solution that goes into the explicit details of why $u$ isn't $p$-absolutely continuous, which can be found here on the supplemental resources page for the textbook.

Essentially it boils down to determining how to break up the series defining $|u(x+h) - u(x)|$ to get a lower bound via the reverse triangle inequality. The term at which the series is broken depends on $p$.

This turned out to be a more subtle problem than originally anticipated by the author!