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Let $1<p,q<\infty$ and $\Omega$ some $\sigma$-finite measure space. Let $T$ denote a bounded convolution operator on $L^p(\mathbb{R}^d)$ with scalar valued kernel $K$ which is locally integrable on $\mathbb{R}^d\setminus\{0\}$. Define the associated $L^q(\Omega)$-valued extension $\vec{T}$ via $$(\vec{T}u)(x,\cdot)=\int_{\mathbb{R}^d}K(x-y)u(y,\cdot)dy$$ for sufficiently well-behaved $u\in L^p(\mathbb{R}^d;L^q(\Omega))$. Analogously, one can define a Banach space valued extension of $T$ for any Banach space $X$ via $$(\vec{T}u)(x)=\int_{\mathbb{R}^d}K(x-y)u(y)dy$$ for well-behaved $u\in L^p(\mathbb{R}^d;X)$. I'm looking for sufficient conditions for the $L^q(\Omega)$-valued extension $\vec{T}$ to extend to a continuous operator on $L^p(\mathbb{R}^d;L^q(\Omega))$.

The cases $q=p$ and $q=2$ work without any assumptions on $T$ besides continuity on $L^p(\mathbb{R}^d)$ by Fubini and Marcinkiewicz-Zygmund, respectively. The case $q$ between $2$ and $p$ then follows by interpolation. Not taking the convolution structure into account, that's the best one can get in general.

If $K$ is in fact in $L^1(\mathbb{R}^d)$ then $\vec{T}$ is continuous for any $p,q\in[1,\infty]$ by Young's inequality. This actually works for any Banach space valued extension of $T$, but $K\in L^1(\mathbb{R}^d)$ is quite restrictive.

If $K$ satisfies a Hörmander condition, then $\vec{T}$ is continuous for any $p,q\in (1,\infty)$ and again this actually works for any Banach space valued extension of $T$, provided $\vec{T}$ is continuous on $L^p(\mathbb{R}^d;X)$ for some $p\in (1,\infty)$.

Now the above conditions either exploit the structure of $L^q(\Omega)$ without exploiting the structure of $T$, i.e. that $T$ is a bounded convolution operator or they rely mainly on the structure of $T$ without taking the structure of $L^q(\Omega)$ into account. This leaves me with the vague feeling that one might be able to do better, i.e. that there are conditions on the convolution kernel $K$ which are less restrictive than the Hörmander condition or at least different but which still ensure that one can extend the bounded operator $T$ on $L^p(\mathbb{R}^d)$ to a bounded operator $\vec{T}$ on $L^p(\mathbb{R}^d;L^q(\Omega))$ for any $q\in (1,\infty)$. Any ideas or references would be highly appreciated. Of course the same holds true for counterexamples illustrating that one cannot do (significantly) better, if such counterexamples exist. In fact, I even wonder whether there is an example of a convolution operator which is bounded on $L^p(\mathbb{R}^d)$ for $1<p<\infty$ but does not admit a bounded extension to $L^p(\mathbb{R}^d;L^q(\Omega))$ for every $q\in(1,\infty)$.

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Also, I would like to know whether this question is considered appropriate for mathoverflow or not. –  lvb May 8 '11 at 1:40
    
Yes, I think it is. I know a bit of harmonic analysis and I can't answer it (yet). I would try it on mathoverflow. Please post the link here if you do. –  Jonas Teuwen May 8 '11 at 10:17
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