Link between Riesz p-variation and absolute continuity Fix $p\in(1,+\infty)$ and for  $f:[0,1]\to\mathbb{R}$ define
$$
V_p(f)=\sup \sum_{i=0}^{n-1}\frac{|f(t_{i+1})-f(t_i)|^p}{(t_{i+1}-t_i)^{p-1}} 
$$
where the sup is taken over all $n\in\mathbb{N}$ and among all possible partitions $0=t_0<t_1<\dots<t_n=1$.
I have to prove that $V_p(f)<+\infty$ if and only if $f$ is absolute continuous with weak derivative in $L^p([0,1])$.
If $f$ is absolutely continuous with weak derivative in $L^p([0,1])$, that it suffices to apply Holder inequality and easily it follows that
$$
\sum_{i=0}^{n-1}\frac{|f(t_{i+1})-f(t_i)|^p}{(t_{i+1}-t_i)^{p-1}}
\leq
\|f'\|_p^p.
$$
However I have thought a lot about the other implication but I really do not know how to do it. Does anyone know how to prove that?
 A: The variation defined in the OP is known as the Riesz $p$-variation, which is typically denoted as $V^R_p(f;[a,b])$, where $f:[a,b]\rightarrow\mathbb{C}$ is measurable.
Here I prove that if  $V^R_p(f;[a,b])<\infty$, then $f$ is absolutely continuous in $[a,b]$.
Let  $[\alpha_1,\beta_1],\ldots, [\alpha_n,\beta_n]$ be subintervals in $[a,b]$. Let $q$ be the conjugate exponent to $p$, that is $\frac1p+\frac1q=1$.  Then
$$\begin{align}
\sum^n_{j=1}|f(\beta_j)-f(\alpha_j)|&=\sum^n_{j=1}\frac{|f(\beta_j)-f(\alpha_j)|}{|\beta_j-\alpha_j|^{1/q}}|\beta_j-\alpha_j|^{1/q}\\
&\leq \Big(\sum^n_{j=1}\frac{|f(\beta_j)-f(\alpha_j)|^p}{|\beta_j-\alpha_j|^{p-1}}\Big)^{1/p}\Big(\sum^n_{k=1}|\beta_k-\alpha_k|\Big)^{1/q}\\
&\leq \big(V^R_p(f;[a,b])\big)^{1/p}\Big(\sum^n_{k=1}|\beta_k-\alpha_k|\Big)^{1/q}\end{align}$$
The absolute continuity of $f$ follows immediately. Then, by Lebesgue’s fundamental theorem of Calculus,  we have that $f'$ exists $\lambda$-almost surely ($\lambda$= Lebesgue measure), $f'\in L_1([a,b])$ and $f(x)=f(a)+\int^x_a f'\,d\lambda$ for all $a\leq x\leq b$.
The proof that $f'\in L_p[a,b]$ is a little more involved. One may check the original paper of F. Riesz, Untersuchungen über systems intergrierbarer funktionen, Mathematische annalen, 1910 vol 69, pp 449-497
The sketch of the proof is as follows:

*

*Consider the uniform partition $\{t_{n,k}=a+k(b-a)/n:k=0,\ldots,n\}$ of $[a,b]$ and consider the step functions
$$g_n(x)=\sum^{n-1}_{k=0}\frac{f(t_{n,k+1})-f(t_{n,k})}{t_{n,k+1}-t_{n,k}}\mathbb{1}_{[t_{n,k},t_{n,k+1})}(x)$$


*Show that $g_n\xrightarrow{n\rightarrow\infty}f'$ at any point $t\in[a,b]$ at which $f$ is differentiable and $t\notin\{t_{n,k}:n\in\mathbb{N}, 0\leq k\leq n\}$.


*An application of Fatou's lemma them implies that
$$\begin{align}
\int^b_a|f'|^p\,d\lambda&\leq\liminf_n\int^b_a|g_n|^p\,d\lambda\\
&=\liminf_n\sum^n_{k=1}\frac{|f(t_{n,k+1})-f(t_{n,k})|^p}{|t_{n,k+1}-t_{n,k}|^p}|t_{n,k+1}-t_{n,k}|\leq V^R_p(f;[a,b])\tag{1}\label{one}
\end{align}
$$

Observation: the inequalities obtained in the prove of necessity and sufficiency imply that
$$\|f'\|^p_{L_p[a,b]}=V^R_p(f;[a,b])$$
Indeed, if $V^R_p(f;[a,b])<\infty$, then $f$ is AC and so it is a.s. differentiable $f(\beta)-f(\alpha)=\int^\beta_\alpha f'(t)\,dt$ for all $a\leq \alpha\leq\beta\leq b$. Hence, for any partition $a=x_0<\ldots<x_n=b$, and application of Jensen's inequality yields
$$\begin{align}
\sum^n_{k=1}\frac{|f(x_k)-f(x_{k-1})|^p}{|x_k-x_{k-1}|^{p-1}}&=\sum^n_{k=1}\Big|\frac{1}{(x_k-x_{k-1})}\int^{x_k}_{x_{k-1}}f'\Big|^p|x_x-x_{k-1}|\\
&\leq\sum^n_{k=1}\int^{x_k}_{x_{k-1}}|f'|^p=\int^b_a|f'|^p
\end{align}$$
This shows that $V^R_p(f;[a,b])\leq\int^b_a|f'|$, which along with \eqref{one} proves the claim.
