$\alpha \in L(L^p, L^{2p})\; \Rightarrow\; \alpha \in L(L^{2p}, L^{4p}), \, p \in [1,\infty)?$

Question

Let $p \in [1,\infty)$ and $\alpha \in \mathcal{L}(L^p\!, \,L^{2p}),$ meaning $\alpha\colon L^p \rightarrow L^{2p}$ is linear and bounded, where $L^p$ and $L^{2p}$ stand for the $L^p$- and $L^{2p}$-space.

Since $||\alpha||_{\mathcal{L}(L^p\!, \,L^{2p})} = \,\sup_{||x||_{L^p}\leq 1}||\alpha(x)||_{L^{2p}}$ and we know that $L^{2p} \subseteq L^p$ with $||\cdot||_{L^{2p}} \leq ||\cdot||_{L^p},$ the operator $\alpha$ is also an element of $\mathcal{L}(L^p\!, \,L^{p})$ and of $\mathcal{L}(L^{2p}\!, \,L^{2p}).$

Question1: $\;\;\alpha \in \mathcal{L}(L^p\!, \,L^{2p})\quad \stackrel{?}{\Longrightarrow}\quad\alpha \in \mathcal{L}(L^{2p}\!, \,L^{4p}).$

Question2: $\;\;$Do operators $\alpha \in \mathcal{L}(L^p\!, \,L^{2p})$ exist which have no representation as an (Hilbert-Schmidt) integral operator?

Question3: $\;\;$Do you have examples for operators $\alpha \in \mathcal{L}(L^p\!, \,L^{2p})?$ Even better: do you know classes of operators which belong to $\mathcal{L}(L^p\!, \,L^{2p})?$ Thank you in advance.

Answer 1

Pick a function in $g\in L^q$ where $q=p/(p-1)$, and a function $h\in L^{2p}$. Define $$ \alpha(f) = h\int fg $$ This is a rank-one operator from $L^p$ to $L^{2p}$. Its range consists of the constant multiples of the function $h$. In particular, it is not an operator from $L^{2p}$ to $L^{4p}$ unless $h$ happens to be in $L^{4p}$.

There are all kinds of operators between Lebesgue spaces. The problem with operators that don't have a nice integral form is that it's hard to describe them. For example, take a bounded sequence $b\in \ell^\infty$ and an unconditional basis $\{e_n\}$ in $L^{2p}$ such as the Haar basis. The map $\sum a_n e_n\mapsto \sum a_n b_n e_n$ is a bounded operator on $L^{2p}$ that is of a non-integral type. This can be composed with an integral operator from $L^p$ to $L^{2p}$.