Finding a dimension-free bound for a certain multiplier on Euclidean space

The following question is indirectly motivated by strong type maximal function estimates. Let $f\in L_{p}(\mathbb{R}^{n})$. For $\xi=(\xi_{1},\ldots,\xi_{n})\in\mathbb{R}^{n}$ define $m(\xi)$ so that

$$m(\xi)=(\xi_{1}^{2}+\cdots+\xi_{n}^{2})^{1/2}\prod_{i=1}^{n}\frac{\sin\pi\xi_{i}}{\pi\xi_{i}}$$

Define $\widehat{f}(\xi)=\int_{\mathbb{R}^{n}}e^{-2\pi ix\cdot\xi}f(x)dx$, and define the multiplier $T$ by $$\widehat{Tf}(\xi)=m(\xi)\widehat{f}(\xi)$$

It is known that $T$ is bounded as a map from $L_{p}(\mathbb{R}^{n})$ to $L_{p}(\mathbb{R}^{n})$ for $1<p<\infty$. That is, $||T||_{L_{p}(\mathbb{R}^{n})\to L_{p}(\mathbb{R}^{n})}=c(p,n)<\infty$. However, for $p$ fixed, it may be the case that $c(p,n)\to\infty$ as $n\to\infty$. Some evidence suggests that $c(p,n)\to\infty$ as $n\to\infty$, but I am not certain. My question is therefore:

For $1<p<\infty$, is $||T||_{L_{p}(\mathbb{R}^{n})\to L_{p}(\mathbb{R}^{n})}$ bounded, independently of $n$?

If the answer is yes, one would get dimension-free $L_{p}$ bounds for the maximal operator associated to the cube. See page 306 of MÃ¼ller, "A geometric bound for maximal functions associated to convex bodies."

If this problem is too hard, perhaps one can ask: are there any dimension-free approaches to proving boundedness of multipliers that could possibly work here? Can one find any $L_{p}$ estimate for these multipliers at all? Concerning the first question, I am not aware of any multiplier theorem that gives $L_{p}$ bounds in a dimension-free manner. Even if the conditions are rather strong, such a theorem would be nice to know. One can get dimension-free Littlewood-Paley estimates, but as far as I know this can only be done when the Littlewood-Paley projections are associated to a semigroup. Concerning the second question, see the aforementioned reference.

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It seems this is true for $p=2$ by a simple computation. But I am somewhat sleepy so I might have made mistakes ... (but it should be simple to check it). – Helge Jul 20 '11 at 5:32
Yes, the $p=2$ case is true, as far as I can tell. And Müller interpolates the $p=2$ estimate with a $p=\infty$ estimate. However, his $p=\infty$ estimate grows with dimension, so all of the interpolated estimates (for $2<p<\infty$) also grow with dimension. – Steven Heilman Jul 21 '11 at 1:19
It seems that this problem is essentially solved here: arxiv.org/pdf/1212.2661v1.pdf – Steven Heilman Dec 13 '12 at 17:38