For $1 \le p< \infty$, show that if the absolutely continuous function $F$ on $[a,b]$ is an indefinite integral of an $L^p[a,b]$ function, then there exists an $M>0$ such that for any partition $\{x_0,...,x_n\}$ of $[a,b]$, $$ \sum_{k=1}^{n}\dfrac{|F(x_k)-F(x_{k-1}|^p}{|x_k-x_{k-1}|^{p-1}}\le M$$
Im having trouble finding a $M$ that holds for any partition. Heres what I have done so far. Also, here is a corollary that I used:
If $E$ is measurable with finite measure and $1\le p_1 < p_2 < \infty $ then $L^{p_2}[a,b] \subseteq L^{p_1}[a,b]$ and $||f||_{p_1}\le m(E)^{\frac{p_2 - p_1}{p_1 \cdot p_2}}||f||_{p_2}$
First since $F$ is a indefinite integral of an $L^P[a,b]$ function then $F(x)-F(a)=\int_a^xf(s)ds$ for some $f \in L^P[a,b]$
So, $$ |F(x_k)-F(x_{k-1})| \le \int_{x_{k-1}}^{x_k}|f(s)| ds \le (x_k-x_{k-1})^{\frac{p-1}{p}}\big \vert \big \vert f_{[x_{k-1},x_k]}\big \vert \big \vert _p $$ $\Rightarrow$ $$ |F(x_k)-F(x_{k-1})|^p \le (x_k-x_{k-1})^{{p-1}}\big \vert \big \vert f_{[x_{k-1},x_k]}\big \vert \big \vert _p^p $$ $\Rightarrow$ $$ \dfrac{|F(x_k)-F(x_{k-1})|^p}{(x_k-x_{k-1})^{{p-1}}} \le \big \vert \big \vert f_{[x_{k-1},x_k]}\big \vert \big \vert _p^p $$ $\Rightarrow $ $$ \sum_{k=1}^n \dfrac{|F(x_k)-F(x_{k-1})|^p}{(x_k-x_{k-1})^{{p-1}}} \le \sum_{k=1}^n \big \vert \big \vert f_{[x_{k-1},x_k]}\big \vert \big \vert _p^p < \infty$$
But $\sum_{k=1}^n \big \vert \big \vert f_{[x_{k-1},x_k]}\big \vert \big \vert _p^p $ depends on the partition which is a problem. Any help is appreciated
I think your own notation is obscuring something simple:
You've shown
$$\sum_{k=1}^n \dfrac{|F(x_k)-F(x_{k-1})|^p}{(x_k-x_{k-1})^{{p-1}}} \le \sum_{k=1}^n \int_{x_{k-1}}^{x_k} |f|^p.$$
The sum on the right is just $\int_a^b |f|^p.$