Suppose that \[ 1\leq p_j \leq \infty, && \sum_{j=1}^n {1\over p_j} = {1\over r} \leq 1 .\]
Show that if \(f_j \in L^{p_j}\) for each \(1\leq j \leq n\), then \[ \prod f_j \in L^r, && \norm{ \prod f_j }_r \leq \prod \norm{f_j}_{p_j} .\]
Solution
1. Reduce to the case where all \(p_j < \infty\).
if some \(p_j = \infty\), then \(|f_j| \le \norm{f_j}_\infty\) a.e., so \(\norm{\prod f_j}_r \le \norm{f_j}_\infty \norm{\prod_{k\ne j} f_k}_r\), reducing to the \((n-1)\)-term case. Iterating, assume \(1 \le p_j < \infty\).
2. Set \(q_j = p_j/r\); then \(q_j \ge 1\) and \(\sum_j 1/q_j = 1\).
\(\sum_j 1/p_j = 1/r \le 1\) gives \(r \le p_j\) for each \(j\) (as \(1/p_j \le 1/r\)), so \(q_j = p_j/r \ge 1\); and \(\sum_j \frac{1}{q_j} = \sum_j \frac{r}{p_j} = r \cdot \frac1r = 1\).
3. Apply Hölder’s inequality to the functions \(|f_j|^r\) with exponents \(q_j\).
by the generalized (multiplicative) Hölder inequality — iterated from the two-factor case — 2 gives \[ \int \prod_j |f_j|^r \le \prod_j \Big(\int |f_j|^{r q_j}\Big)^{1/q_j} = \prod_j \Big(\int |f_j|^{p_j}\Big)^{r/p_j} = \prod_j \norm{f_j}_{p_j}^r . \]
4. Conclude.
taking the \(r\)-th root in 3: \[ \norm{\prod_j f_j}_r = \Big(\int \prod_j |f_j|^r\Big)^{1/r} \le \prod_j \norm{f_j}_{p_j} , \] so \(\prod f_j \in L^r\) with the claimed norm bound. (When \(r = \infty\), i.e. \(\sum 1/p_j = 0\), all \(p_j = \infty\) and the product is bounded by \(\prod \norm{f_j}_\infty\).)
5. Q.E.D.