$\lim_{h \to 0} \frac{F(x+h)-F(x)}{h^{1/q}} = 0$ for $F(x)=\int^{x}_{0}f(t)dt$ for $f \in L^{p}[0,1]$

Question

Let $F(x) = \int^{x}_{0} f(t)dt$ for $f \in L^{p}([0,1])$ for $1 <p< \infty$, then $$\lim_{h \to 0} \frac{F(x+h)-F(x)}{h^{1-1/p}} = 0.$$

This looks somewhat similar to the Lebesgue differentiation theorem, but the exponent in the denominator is throwing me off a bit.

Answer 1

Hint: Apply Holder's inequality to $$F(x+h)-F(x)=\int_x^{x+h}f(t)\,dt\quad(h>0).$$

Hmm...

Maybe I should note that there's more than one way to apply Holder here. The most naive application gives $$|F(x+h)-F(x)|\le||f'||_ph^q,$$which is not quite good enough. But instead write $\int_x^{x+h}f'=\int(f'\chi_{[x,x+h]})(\chi_{[x,x+h]})$ and apply Holder to that product...