Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Define the mixed Lebesgue space $l_{p,q}$ as the space of all doubly indexed sequences ${\bf a}= (a(i,j))_{i,j\in\mathbb{Z}}$ such that

\begin{equation} \|{\bf a}\|_{p,q} := \left( \sum_{i\in\mathbb{Z}} \left( \sum_{j\in\mathbb{Z}} |a(i,j)|^p \right)^{q/p} \right)^{1/q} <\infty. \end{equation}

Such spaces are sometimes called mixed Lebesgue spaces, Lebesgue-Bochner spaces or Strichartz spaces. I am interested in boundedness conditions for operators on $l_{p,q}$, namely I have the following concrete question:

Given a matrix operator ${\bf A}=\left(A(i,j;i'j')\right)_{i,j,i',j'\in\mathbb{Z}}$ acting on $l_{p,q}$ define

$$ C:=\max\left(\sup_{i,j}\sum_{i',j'}|A(i,j;i',j')|,\sup_{i',j'}\sum_{i,j}|A(i,j;i',j')| ,\sup_{i',j}\sum_{i,j'}|A(i,j;i',j')|,\sup_{i,j'}\sum_{i',j}|A(i,j;i',j')|\right). $$

Do we have

$$\|{\bf A}\|_{l_{p,q}\to l_{p,q}}\le C \mbox{ for }p,q \geq 1 ?$$

If $p=q$ this holds by Riesz-Thorin interpolation and considering the cases $p=q=1$ and $p=q=\infty$.

share|improve this question
    
Note that by interpolation, only the cases $p=1,q=\infty$ and $p=\infty,q=1$ need to be checked. And by duality, one is fact enough. –  Mikael de la Salle Jul 26 '11 at 9:52
1  
... and $A$ given by $A(i,j;i',j') = 1_{i=j'} 1_{j=i'}$ gives a counterexample. It indeed satisfies $C=1$, whereas it is not bounded on $\ell_{p,q}$ unless $p=q$. –  Mikael de la Salle Jul 26 '11 at 10:01
    
Thanks! I have modified the definition of C and the new question asks if the estimate holds with this definition. –  Philipp Jul 26 '11 at 10:21
    
Ok, my guess is that the answer is no. One has to replace the term $$ \sup_{i,j'}\sum_{i',j} \dots $$ with $$\sup_{i}\sum_{i'}\sup_{j'}\sum_{j}\dots,$$ and then it works. –  Philipp Jul 26 '11 at 11:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.