Modulus of continuity and Hölder's inequality

Question

I am having trouble seeing how $\omega_p(\delta) \le \omega_q(\delta)$ follows "easily" from Hölder's inequality:

Is this a general result for any $p\le q$? I associate Hölder with $p$ and $q$ satisfying $p^{-1}+q^{-1}=1$. Thanks in advance!

Answer 1

For probability measure $\mu(X)=1$, one has \begin{align*} \left(\int_{X}|f|^{p}d\mu\right)^{1/p}&\leq\left(\left(\int_{X}(|f|^{p})^{q/p}d\mu\right)^{p/q}\right)^{1/p}\left(\left(\int_{X}1^{r}d\mu\right)^{1/r}\right)^{1/p}\\ &=\left(\int_{X}|f|^{q}d\mu\right)^{1/q}\mu(X)^{1/(pr)}\\ &=\left(\int_{X}|f|^{q}d\mu\right)^{1/q}, \end{align*} where $1/(q/p)+1/r=1$ and $q>p$.

Now let $\mu=dx/2\pi$ defined on the interval $[0,2\pi]$.