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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


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 Müller, "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: – Steven Heilman Dec 13 '12 at 17:38

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