2
$\begingroup$

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)

$\endgroup$
1
$\begingroup$

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 $\gamma_n>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.