When is $\Vert \Vert f \Vert_p \Vert_q \leq \Vert\Vert f\Vert_q\Vert_p$?

Question

Under which condition on two positive real numbers $p,q$ (not necessarily Hölder-conjugated) holds the following inequality for functions where the integral exists? $$\left(\int_{\Omega_1} \left( \int_{\Omega_2} \vert f(s,t)\vert^p d\mu(s)\right)^{\frac{q}{p}} d\nu(t)\right)^{\frac{1}{q}} \leq \left(\int_{\Omega_2} \left( \int_{\Omega_1} \vert f(s,t)\vert^q d\nu(t)\right)^{\frac{p}{q}} d\mu(s)\right)^{\frac{1}{p}}$$ For $p=q$ this is Fubini. Furthermore, if $f(s,t)=g(s)h(t)$ we also have equality. My conjecture is that this holds for $p\leq q$ but I have not found a way to prove it yet. I have in particular tried the case $p=1$ and then applying Hölder's inequality but I did not reach any conclusion. Any hint on how to get started is greatly appreciated.

Answer 1

Write $F(s, t) = |f(s, t)|^p$ and $\alpha=q/p \geq 1$. Then it boils down to showing

$$ \left( \int_{\Omega_1} \left( \int_{\Omega_2} F(s, t) \, \mu(ds) \right)^{\alpha} \, \nu(dt) \right)^{\frac{1}{\alpha}} \leq \int_{\Omega_2} \left( \int_{\Omega_1} F(s, t)^{\alpha} \, \nu(dt) \right)^{\frac{1}{\alpha}} \, \mu(ds). \tag{*}$$

Note that this is the content of Minkowski's integral inequality. One proof strategy is to utilize the duality between $L^p$-spaces. Indeed, let $\beta$ be such that $\frac{1}{\alpha}+\frac{1}{\beta}=1$ and consider the operator $T$ on $L^{\beta}(\nu)$ defined by

$$ T\varphi = \int_{\Omega_1} G(t)\varphi(t) \, \nu(dt), \qquad G(\cdot) = \int_{\Omega_2} F(s, \cdot) \, \mu(ds). $$

By the duality, the operator norm of $\|T\|$ is exactly

$$ \|T\| = \left\| G \right\|_{L^{\alpha}(\nu)} = \text{[LHS of (*)]}. $$

On the other hand, by Fubini's theorem followed by Hölder's inequality,

\begin{align*} |T\varphi| &=\left| \int_{\Omega_2} \int_{\Omega_1} F(s, t)\varphi(t) \, \nu(dt)\mu(ds)\right| \\ &\leq \int_{\Omega_2} \| F(s , \cdot)\|_{L^{\alpha}(\nu)} \|\varphi\|_{L^{\beta}(\nu)} \, \mu(ds) \\ &= \left( \int_{\Omega_2} \| F(s , \cdot)\|_{L^{\alpha}(\nu)} \, \mu(ds) \right)\|\varphi\|_{L^{\beta}(\nu)} \end{align*}

So it follows that

$$ \|T\| \leq \int_{\Omega_2} \| F(s , \cdot)\|_{L^{\alpha}(\nu)} \, \mu(ds) = \text{[RHS of (*)]} $$

and the inequality $\text{(*)}$ holds.

Answer 2

Assuming $p\le q$ and letting $g(s,t)=|f(s,t)|^p$ and $\frac1r+\frac pq=1$, this is equivalent to $$ \begin{align} \left(\int_{\Omega_1}\left(\int_{\Omega_2}g(s,t)\,\mathrm{d}\mu(s)\right)^{q/p}\,\mathrm{d}\nu(t)\right)^{p/q} &=\sup_{\|h\|_r=1}\int_{\Omega_1}\int_{\Omega_2}g(s,t)\,\mathrm{d}\mu(s)\,h(t)\,\mathrm{d}\nu(t)\tag1\\ &=\sup_{\|h\|_r=1}\int_{\Omega_2}\int_{\Omega_1}g(s,t)\,h(t)\,\mathrm{d}\nu(t)\,\mathrm{d}\mu(s)\tag2\\ &\le\int_{\Omega_2}\sup_{\|h\|_r=1}\int_{\Omega_1}g(s,t)\,h(t)\,\mathrm{d}\nu(t)\,\mathrm{d}\mu(s)\tag3\\ &=\int_{\Omega_2}\left(\int_{\Omega_1}g(s,t)^{q/p}\,\mathrm{d}\nu(t)\right)^{p/q}\,\mathrm{d}\mu(s)\tag4 \end{align} $$ Explanation:
$(1)$: Converse Hölder Inequality
$(2)$: Fubini's Theorem
$(3)$: an $h$ for each $s$ is at least as big as one $h$ for all $s$
$(4)$: Hölder's Inequality