An integral Minkowski inequality for the quasi-Banach case?

Question

The so called integral Minkowski inequality claims that for suitable positive functions $f$ defined on the product of two measure spaces $ (X\times Y, \mu \times \nu) $ and $p\geq 1$ we have that $$ \bigg( \int_Y \bigg( \int_{X}f(x,y) d\mu(x) \bigg)^p d\nu(y) \bigg)^{1/p} \leq \int_X \bigg( \int_Y f(x,y)^p d\nu(y) \bigg)^{1/p} d\mu(x). $$ The idea is that if $(X,\mu) $ is a finite space with the counting measure this is exactly the Minkowski (triangle) inequality for the Banach space $(Y,\nu)$.

When $0<p<1$ of course even the ordinary Minkowski inequality fails, but we have the property that $$ \bigg\Vert \sum_{n=0}^\infty f_n\bigg\Vert^p_{L^p(Y,\nu)} \leq \sum_{n=0}^\infty \Vert f_n \Vert_{L^p(Y,\nu)}^p. $$ Of course this can be written in integral form as;

$$ \int_Y \bigg( \int_{X}f(x,y) d\mu(x) \bigg)^p d\nu(y) \leq \int_X \int_Y f(x,y)^p d\nu(y) d\mu(x), $$ where $(X,\mu)$ countable measure space with the counting measure. Unfortunatelly there is no hope that this inequality holds for all measures $\mu$ because of the wrong homogeneity with respect to $\mu$. But is there a way, given a measure $\mu$ to define a measure $\mu_p$ such that this inequality holds ? For example if $\mu$ is a weighted sum of Dirac measures $ \mu = \sum{c_n \delta_{x_n}} $, then I guess the inequality holds for $\mu_p = \sum c_n^p \delta_{x_n}$, but I do not see how to carry out this proceedure for other measures $\mu$.

For starters the case where both measures live in the interval $(0,1)$ would be more than enough.

Answer 1

In my opinion the best formulation of the integral Minkowski is \begin{equation} \|u(x,y)\|_{L^{s}_{y}L^{r}_{x}}\le \|u(x,y)\|_{L^{r}_{x}L^{s}_{y}}. \end{equation} with obvious meaning of the notations (I hope), valid for all $0< r\le s\le \infty$ and arbitrary measures. Once you see it, the proof is an easy exercise: set $v=|u|^r$ and reduce to the usual case of exponents 1 and $s/r\ge1$.