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The Gagliardo Inequality generalizes Fubini's Theorem: let $f_j$ be $d-1$ non-negative measurable functions over ${\mathbb R}^{d-1}$. Let us form the function $$f(x)=\prod_{j=1}^df_j(\widehat{x_j}),$$ where as usual $\widehat{x_j}$ stands for the point $(x_1,\ldots,x_{j-1},x_{j+1},\ldots,x_d)$. Then the GI is $$\int_{{\mathbb R}^d}f(x)\,dx\le\prod_{j=1}^d\|f_j\|_{L^{d-1}({\mathbb R}^{d-1})}.$$ (Remark the nice constant $\bf1$).

I suspect that the following natural extension is true. Let $k$ be a integer, $1\le k\le d-1$. For every subset $I\subset[\![1,d]\!]$ of cardinal $k$, let $f_I$ be a non-negative measurable function over ${\mathbb R}^k$. Denote $x_I=(x_i)_{i\in I}$ and form the function $$f(x)=\prod_{|I|=k}f_I(x_I).$$

Is it true that $$\int_{{\mathbb R}^d}f(x)\,dx\le\prod_{|I|=k}\|f_I\|_{L^q({\mathbb R}^k)},$$ where $q=\binom{d-1}{k-1}$ ?

The exponent $q$ above is determined by scaling invariance. The case $k=d-1$ is GI. The case $k=1$ is just Fubini, and then the equality holds true. The case $k=d$ is a tautology.

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  • $\begingroup$ Just to be clear: are you worried about measure theoretic difficulties? If you assume all functions are, say, in Schwartz space, then what you want follows from Holder in each variable taking into account that the number of subsets of $\{1, \ldots, d\}$ of length $k$ that contains a given element $\{1\}$ is exactly $d-1$ choose $k-1$. Am I missing something? $\endgroup$ Commented Mar 1, 2017 at 18:26
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    $\begingroup$ Have you checked Brascamp-Lieb inequality? en.wikipedia.org/wiki/Brascamp%E2%80%93Lieb_inequality $\endgroup$ Commented Mar 1, 2017 at 20:24
  • $\begingroup$ Just to mention that I have know a proof of the general inequality. The proof follows about the same ideas as that of Gagliardo. It is just more cumbersome. $\endgroup$ Commented Jul 6, 2017 at 15:23

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