> [!definition] > > Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]] and $f \in L^1(\mu)$, then denote > $ > \mu \cdot f = \angles{\mu, f} = \int fd\mu > $ > If $p > 0$, define the operator > $ > P_k: \real^+ \to \real \quad x \mapsto x^k > $ > and its inverse > $ > P_{k}^{-1} = P_{1/k}: \real^+ \to \real \quad x\mapsto x^{1/k} > $ > Let > $ > \Pi_{j}: \real^n \to \real \quad x \mapsto \prod_{j = 1}^n x_j > $ > Write $\Pi_{j \ne k}$ to exclude the $k$-th index from the product. Note that this product commutes with the power operators. > [!theorem] [[Hölder's Inequality]] > > Let $(X, \cm, \mu)$ be a measure spaces. Let $\seqf{f_j} \in L^n \cap L^+$, and $r = n$, then > $ > (\mu \circ \Pi_j) \cdot f_j \le (P_{r}^{-1} \circ \Pi_j \circ \mu \circ P_r) \cdot f_j > $ > [!theorem] > > Let $\seqf{(X_j, \cm_j, \mu_j)}$ be a family of $\sigma$-finite measure spaces, and let $\nu_j = \prod_{k = 1}^j \nu_k$. Let $r = n - 1$, and $f, g \in (L^1 \cap L^+)(\nu_n)$ such that > $ > \abs{f}^n \le \prod_{j = 1}^n \angles{\mu_j, \abs{g}} = \braks{\Pi_j \circ \mu_j} \cdot \abs{g} > $ > [[Almost Everywhere|a.e.]], then $\norm{f}_{n/r} \le \norm{g}_1$. > > *Proof*. We claim that for each $1 \le k < n$, > $ > \angles{\nu_k, \abs{f}^{n/r}} \le (P_{r}^{-1} \circ \Pi_j \circ \eta_{k, j}) \cdot \abs{g} \quad > \eta_{k, j} = \begin{cases} > \nu_k & 1 \le j \le k \\ > \nu_k \times \mu_j & k < j \le n > \end{cases} > $ > If $k = 1$, then by assumption > $ > \begin{align*} > \abs{f}^n &\le \braks{\Pi_j \circ \mu_j} \cdot \abs{g}\\ > \angles{\nu_1, \abs{f}^{n/r}} = \angles{\mu_1, \abs{f}^{n/r}} &\le (\mu_1 \circ P_{r}^{-1} \circ \Pi_j \circ \mu_{j}) \cdot \abs{g} > \end{align*} > $ > Since $\eta_{1, 1} = \mu_1$, > $ > \angles{\mu_1, \abs{f}^{n/r}} \le (P_r^{-1} \circ \eta_{1, 1} \cdot \abs{g}) \cdot \braks{(\mu_1 \circ P_r^{-1} \circ \Pi_{j \ne 1} \circ \mu_{j}) \cdot \abs{g}} > $ > As $(n - 1)/r = 1$, applying [[Hölder's Inequality]] yields, > $ > \begin{align*} > (\mu_1 \circ \Pi_{j \ne 1}) \cdot (P_r^{-1} \circ \mu_{j} \cdot \abs{g}) &\le > (P_r^{-1} \circ \Pi_{j \ne 1} \circ \mu_1 \circ P_r) \cdot (P_r^{-1} \circ \mu_{j} \cdot \abs{g}) \\ > &= (P_r^{-1} \circ \Pi_{j \ne 1} \circ \mu_1 \circ \mu_{j}) \cdot \abs{g} \\ > &= (P_r^{-1} \circ \Pi_{j \ne 1} \circ \eta_{1, j}) \cdot \abs{g} > \end{align*} > $ > Combining with the prior equation gives > $ > \angles{\nu_1, \abs{f}^{n/r}} \le (P_r^{-1} \circ \Pi_j \circ \eta_{1, j}) \cdot \abs{g} > $ > Now suppose that the proposition holds for $k$. By [[Fubini-Tonelli Theorem|Tonelli's Theorem]], > $ > \angles{\nu_{k + 1}, \abs{f}^{n/r}} = \angles{\mu_{k + 1}, \angles{\nu_{k + 1}, \abs{f}^{n/r}}} \le \mu_{k + 1} \circ (P_{r}^{-1} \circ \Pi_j \circ \eta_{k, j}) \cdot \abs{g} > $ > Since $\eta_{k + 1, k + 1} = \eta_{k, k + 1}$, > $ > \angles{\nu_{k + 1}, \abs{f}^{n/r}} \le (P^{-1}_r \circ \eta_{k + 1, k + 1} \cdot \abs{g}) \cdot \braks{(\mu_{k + 1} \circ P_r^{-1} \circ \Pi_{j \ne k + 1} \circ \eta_{k, j}) \cdot \abs{g}} > $ > As $(n - 1)/r = 1$, applying Hölder's inequality yields > $ > \begin{align*} > &(\mu_{k + 1} \circ \Pi_{j \ne k + 1}) \cdot (P_r^{-1} \circ \mu_{j}) \cdot \abs{g} \\ > &\le (P_{r}^{-1} \circ \Pi_{j \ne k + 1} \circ \mu_{k + 1} \circ P_{r}) \circ (P_{r}^{-1} \circ \mu_j \cdot \abs{g}) \\ > &= (P_{r}^{-1} \circ \Pi_{j \ne k + 1} \circ \mu_{k + 1} \circ \eta_{k, j} \cdot \abs{g}) \\ > &= (P_{r}^{-1} \circ \Pi_{j \ne k + 1} \circ \eta_{k + 1, j} \cdot \abs{g}) > \end{align*} > $ > > Combining the equations gives > $ > \angles{\nu_{k + 1}, \abs{f}^{n/r}} \le (P_r^{-1} \circ \Pi_j \circ \eta_{k + 1, j}) \cdot \abs{g} > $ > By induction, the proposition holds for $k = n$. In which case, $\eta_{n, j} = \nu_n$ for all $n$, so > $ > \begin{align*} > \angles{\nu_n, \abs{f}^{n/r}} &\le (P_r^{-1} \circ \Pi_j \circ \nu_n) \cdot \abs{g} \\ > \norm{f}_{n/r}^{n/r}&= \norm{g}_1^{n/r} > \end{align*} > $