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.


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 $L^{2p}$.

  • $\begingroup$ Interesting examples, thank you! $\endgroup$ – Obriareos Jan 15 '17 at 10:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.