> [!theorem] > > If $\mu$ is finite and $0 < p < q \le \infty$, then $L^p \supset L^q$ and > $ > \norm{f}_p \le \norm{f}_q \cdot \mu(X)^{1/p- 1/q} > $ > > *Proof*. If $q = \infty$, then we get the result directly. Otherwise, by [[Hölder's Inequality]] with $q/p$ and $q/(q - p)$, > $ > \begin{align*} > \norm{f}_p^p = \int \abs{f}^p &= \int \abs{f}^p \cdot 1 \\ > &\le \norm{\abs{f}^p}_{q/p} \cdot \norm{1}_{q/(q - p)} \\ > &= \norm{f}_q^{p} \cdot \mu(X)^{(q - p)/q} > \end{align*} > $