Let f be bounded on $X\times Y$ measure space with $\mathbb{P}\times\mathbb{Q}$ probability measure, show that for $0<p\leq q$:
$\left \| \left \| f \right \|_{L^{p}(\mathbb{P})} \right \|_{L^{q}(\mathbb{Q})} \leq \left \| \left \| f \right \|_{L^{q}(\mathbb{Q})} \right \|_{L^{p}(\mathbb{P})} $.
For example, $\left \| \left \| f \right \|_{L^{p}(\mathbb{P})} \right \|_{L^{q}(\mathbb{Q})}=(\int (\int |f(x,y)|^{p}d\mathbb{P}(x))^{q/p}d\mathbb{Q}(y))^{1/q}$.
Only hints please.
Attempts
a)Because $\frac{q}{p}\geq 1$ by Jensen's:
$\left \| \left \| f \right \|_{L^{p}(\mathbb{P})} \right \|_{L^{q}(\mathbb{Q})}^{q} \leq (\int \int |f(x,y)|^{q}dQ dP)^{p/q} $. But because $p/q<1$, Jensen's does not work again.
b)For simple function $f=\sum a_{i}1_{A_{i}\times B_{i}}$ and $p=1$ and $q=2$ we get
$\left \| \left \| f \right \|_{L^{p}(\mathbb{P})} \right \|_{L^{q}(\mathbb{Q})} =[\sum a^{2}_{i}P(A_{i})^{2}Q(B_{i})+2\sum_{i<j}a_{i}a_{j}P(A_{i})Q(B_{i}\cap B_{j})P(A_{j})]^{1/2}$
and on the other hand
$\left \| \left \| f \right \|_{L^{q}(\mathbb{Q})} \right \|_{L^{p}(\mathbb{P})}=\int[\sum a^{2}_{i}1_{A_{i}}Q^{2}(B_{i})+2\sum_{i<j}a_{i}a_{j}1_{A_{i}\cap A_{j}}Q(B_{i}\cap B_{j})]^{1/2}dP.$
3)Say $f(x,y)=f(y,x)$ or more generally $\left \| f \right \|_{L^{p}(\mathbb{P})}=\left \| f \right \|_{L^{p}(\mathbb{Q})}$ and that $p>1$, then $\left \| f \right \|_{L^{p}(\mathbb{P})}\leq \left \| f \right \|_{L^{q}(\mathbb{Q})}$.
Hint: First prove Minkowski's Integral Inequality $$ || || f || _{L^1(P)} || _{L^r(Q)} \leq || || f ||_{L^r(Q)} || _{L^1(P)} $$ using Holder's Inequality(writing $F(y) = \int|f(x,y)| d \mathbb{P}(x)$ may make things more apparent for you). The result to your question follows by writing $q/p=r$ and using Minkowski's integral inequality.