English

When does the norm of a Fourier multiplier dominate its $L^\infty$ norm?

Classical Analysis and ODEs 2017-12-21 v1

Abstract

One can define Fourier multipliers on a Banach function space by using the direct and inverse Fourier transforms on L2(Rn)L^2(\mathbb{R}^n) or by using the direct Fourier transform on S(Rn)S(\mathbb{R}^n) and the inverse one on S(Rn)S'(\mathbb{R}^n). In the former case, one assumes that the Fourier multipliers belong to L(Rn)L^\infty(\mathbb{R}^n), while in the latter one this requirement may or may not be included in the definition. We provide sufficient conditions for those definitions to coincide as well as examples when they differ. In particular, we prove that if a Banach function space X(Rn)X(\mathbb{R}^n) satisfies a certain weak doubling property, then the space of all Fourier multipliers MX(Rn)\mathcal{M}_{X(\mathbb{R}^n)} is continuously embedded into L(Rn)L^\infty(\mathbb{R}^n) with the best possible embedding constant one. For weighted Lebesgue spaces Lp(Rn,w)L^p(\mathbb{R}^n,w), the weak doubling property is much weaker than the requirement that ww is a Muckenhoupt weight, and our result implies that aL(Rn)aMLp(Rn,w)\|a\|_{L^\infty(\mathbb{R}^n)}\le\|a\|_{\mathcal{M}_{L^p(\mathbb{R}^n,w)}} for such weights. This inequality extends the inequality for n=1n=1 from \cite[Theorem~2.3]{BG98}, where it is attributed to J.~Bourgain. We show that although the weak doubling property is not necessary, it is quite sharp. It allows the weight ww in Lp(Rn,w)L^p(\mathbb{R}^n,w) to grow at any subexponential rate. On the other hand, the space Lp(R,ex)L^p(\mathbb{R},e^x) has plenty of unbounded Fourier multipliers.

Keywords

Cite

@article{arxiv.1712.07609,
  title  = {When does the norm of a Fourier multiplier dominate its $L^\infty$ norm?},
  author = {Alexei Karlovich and Eugene Shargorodsky},
  journal= {arXiv preprint arXiv:1712.07609},
  year   = {2017}
}

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41 pages