Modulus of continuity and Hölder's inequality 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!

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 $[0,2\pi]$.
