The following is Exercise 22 from Royden $\S$7.2:
For $1 \leq p < \infty$, if the absolutely continuous function $F : [a, b] \to \mathbb{R}$ is the indefinite integral of an $L^p[a, b]$ function $f$, then there is a constant $M > 0$ such that for any partition $\{a = x_0 < x_1 < \cdots < x_n = b\}$ of $[a, b]$, $$ \sum_{k = 1}^n \frac{|F(x_k) - F(x_{k - 1})|^p}{|x_k - x_{k - 1}|^{p - 1}} \leq M $$
So far, what I have gotten is that for each $k$, $$ \frac{|F(x_k) - F(x_{k - 1})|^p}{|x_k - x_{k - 1}|^{p - 1}} = \frac{|\int_{x_{k - 1}}^{x_k} f|^p}{|x_k - x_{k - 1}|^{p - 1}} \leq \frac{\int_{x_{k - 1}}^{x_k} |f|^p}{|x_k - x_{k - 1}|^{p - 1}} $$ Now, I can certainly bound the numerator $\int_{x_{k - 1}}^{x_k} |f|^p$ with $\|f\|_p^p$ since $f$ is $L^p$ but I don't know what to do with the denominator. Is there a trick to apply Holder's Inequality here? Any hints are welcome.
Hint: Suppose $p >1$ By Holder's inequality $|\int_{x_{k-1}}^{x_k} f| \leq (\int |f|^{p})^{1/p} (\int _{x_{k-1}}^{x_k} 1^{q})^{1/q}$ where $\frac 1 p+\frac1 q=1$. Note that $q=\frac p {p-1}$. For $p=1$ the proof is much simpler.