$\int_0^xf(t)dt$ for $f \in L^3$ is $\frac23$- Holder Continuous

Question

An exam question in Measure Theory:

[State and prove Holder's inequality... Prove $L_q(\mu) \subset L_p(\mu)$ for $1 \leq p \leq q < \infty$]

Which was fine, and then:

Let $f:\Bbb R \to \Bbb R$ be a Lebesgue measurable function in $L_3(\lambda)$. Show that the function

$F(x) := \int_0^xf(t)dt$

satisfies

$|F(x)-F(y)| \leq C|x-y|^{2/3}$

I can't think of how to proceed:

$|\int_y^xf(t)dt| \leq \int_y^x|f(t)|dt$

Answer 1

You can continue by using Holder's inequality: $$\int_x^y |f(t) dt| \leq \left( \int_x^y |f(t)|^3 dt \right)^\frac{1}{3} \left( \int_x^y 1^\frac{3}{2} \right)^\frac{2}{3}$$ as $3$ and $3/2$ are conjugate exponents.

With $C=\Vert f \Vert_{L^3}$ it give you the result.