## Boundedness criterion for operators on mixed Lebesgue spaces

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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 $p=q=\infty$.

-
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 2011 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 2011 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 2011 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 2011 at 11:49
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