> [!theorem] > > Let $A$ be a set and $0 < p < q \le \infty$, then $l^p \subset l^q$ with $\norm{f}_q \le \norm{f}_p$. > > *Proof*. If $q = \infty$, then > $ > \begin{align*} > \sup_{a \in A}\abs{f(a)}^p &\le \sum_{a \in A}\abs{f(a)}^p \\ > \norm{f}_\infty^p &\le \norm{f}_p^p \\ > \norm{f}_\infty &\le \norm{f}_p > \end{align*} > $ > If $q < \infty$, then using the previous theorem with $p, q, \infty$ yields > $ > \norm{f}_q \le \norm{f}_p^{p/q} \cdot \norm{f}_\infty^{1 - p/q} \le \norm{f}_p > $