Reverse Loomis-Whitney Inequality for functions

Question

I was wondering if the reverse Loomis-Whitney inequality holds for general functions:

Let $n\geq 2$. Let $(X_i,\mu_i)$, $1\leq i\leq n$ be measure spaces. Write $x=(x_1,\dots,x_n)$ and for each $1\leq i\leq n$, write $$ \pi_i(x)=x_i'=(x_1,\dots,x_{i-1},x_{i+1},\dots,x_n)\in X'_i:=X_1\times \cdots X_{i-1}\times X_{i+1}\times\cdots X_n. $$ Let $f_i$ be nonnegative measurable functions defined on $X'_i$. Then do we have the following inequality: $$ \gamma_n\prod_{i=1}^n \left\| f_i\right\|_{L^{n-1}(X'_i)}\leq \int_{X_1}\cdots \int_{X_n}\prod_{i=1}^n f_i(x_i')dx. $$ where $0<\gamma_n<1$ is an absolute constant?

It is known that if $f_i=1_{\pi_i(K)}$ for some compact set $K$, then the reverse inequality holds. (See, for example, https://arxiv.org/pdf/1607.07891.pdf)

Answer 1

This conjecture is trivially true for $n=2$, but false for any $n\ge3$. Indeed, take any $n\ge3$ and for all $i$ let $X_i=\mathbb R$ and $\mu_i=\lambda$, the Lebesgue measure. Next, let $$f_i(x'_i):=\prod_{j:\ j\ne i}1_{i<x_j<i+1}. $$ Then $$\prod_i f_i(x'_i)=\prod_i\prod_{j:\ j\ne i}1_{i<x_j<i+1} =\prod_j\prod_{i:\ i\ne j}1_{i<x_j<i+1}=0 $$ for all $x$, but $$\|f_i\|_1=\int f_i=\prod_{j:\ j\ne i}\int_i^{i+1}dx_j=1 $$ for all $i$. So, the right side of your inequality is $0$, whereas its left side is $\gamma_n>0$.