Prove $\exists \alpha \in ]0,1]$ such as : $|F(x)-F(y)| \leq ||f||_p |x-y|^{\alpha}$ (Holder)

Question

Let $1 < p \leq +\infty$ and $f \in L^p ([0,1])$, thus $f \in L^p([0,1])$. Then, $\forall x \in [0,1]$, the function $f 1_{[0,x]}$ is integrable. Its integral is $F(x) = \int_0^x f(t) dt$.

How to show that $\exists \alpha \in ]0,1]$ such as : $|F(x)-F(y)| \leq ||f||_p |x-y|^{\alpha}$, $\forall (x,y) \in [0,1]$.

Could someone help me ?

Answer 1

$$|F(x)-F(y)|=| \int_0^x f(t) dt -\int_0^yf(t) dt|= |\int_0^x f(t) dt+\int_y^0 f(t) dt|=|\int_y^x f(t) dt| \leq |x-y|||f(t)||_\infty \leq |x-y|||f(t)||_p$$

the last inequality from here https://math.stackexchange.com/a/2683042/311112