Show that the following inequality, $\bigg\lVert\prod_{i=1}^{k}f_{i}\bigg\rVert_{L^{r}}\le\prod_{i=1}^{k}\lVert f_{i}\rVert_{L^{p_{i}}}$ holds

Question

Let $1\le p_{i},r\le\infty$ and $\sum_{i=1}^{k}\frac{1}{p_{i}}=\frac{1}{r}.$

Show that the following inequality holds: $$\bigg\lVert\prod_{i=1}^{k}f_{i}\bigg\rVert_{L^{r}}\le\prod_{i=1}^{k}\lVert f_{i}\rVert_{L^{p_{i}}}$$

Here is my working :

Firstly, we may assume each $\lVert f_{i}\rVert_{L^{p_{i}}}<\infty$, otherwise the inequality holds easily.

Now ,we restrict that $1< p_{i},r<\infty$

Note that,

$$\sum_{i=1}^{k}\frac{1}{p_{i}}=\frac{1}{r}\Longleftrightarrow\sum_{i=1}^{k}\frac{r}{p_{i}}=1\Longleftrightarrow\sum_{i=1}^{k}\frac{1}{\frac{p_{i}}{r}}=1$$Then we apply the Hölder's inequality as follows,

$$\bigg\lVert\prod_{i=1}^{k}f_{i}\bigg\rVert_{L^{r}}^{r}=\int\bigg|\prod_{i=1}^{k}f_{i}\bigg|^{r}dx=\int\prod_{i=1}^{k}|f_{i}|^{r}dx \le\prod_{i=1}^{k}\bigg(\int|f_{i}|^{p_{i}}dx\bigg)^{\frac{r}{p_{i}}}=\prod_{i=1}^{k}\lVert f_{i}\rVert_{L^{p_{i}}}^{r}$$,where the first inequality is holds by Hölder's inequality.

Next consider the case that $p_{j_{0}}=1$ and $p_{i}=\infty,i\ne j_{0}$ for some $j_{0}$ and hence in this case $r=1$

Then

$$\bigg\lVert\prod_{i=1}^{k}f_{i}\bigg\rVert_{L^{r}}=\int\bigg|\prod_{i=1}^{k}f_{i}\bigg|dx\le\prod_{i\ne j_{0}}\lVert f\rVert_{L^{p_{i}}}\int|f|dx=\prod_{i=1}^{k}\lVert f_{i}\rVert_{L^{p_{i}}} $$

Is there anyone checking my attempt for validity? Any suggestion or comment will be appreciated.Thanks for considering my request.

Answer 1

So you have missed the case that $r=\infty$ and all $p_{i}=\infty$. Of course this is easy because \begin{align*} |f_{1}(x)|&\leq\|f_{1}\|_{L^{\infty}},~~~~\text{a.e.}\\ |f_{2}(x)|&\leq\|f_{2}\|_{L^{\infty}},~~~~\text{a.e.}\\ \vdots\\ |f_{k}(x)|&\leq\|f_{k}\|_{L^{\infty}},~~~~\text{a.e.}, \end{align*} taking union of all the null sets, and for the points in the compliment of the union, we have $|f_{1}(x)\cdots f_{k}(x)|\leq\|f_{1}\|_{L^{\infty}}\cdots\|f_{k}\|_{L^{\infty}}$, and the union is still a null set, so \begin{align*} |f_{1}(x)\cdots f_{k}(x)|\leq\|f_{1}\|_{L^{\infty}}\cdots\|f_{k}\|_{L^{\infty}},~~~~\text{a.e.}, \end{align*} so \begin{align*} \|f_{1}\cdots f_{k}\|_{L^{\infty}}\leq\|f_{1}\|_{L^{\infty}}\cdots\|f_{k}\|_{L^{\infty}}. \end{align*}