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When does the norm of a Fourier multiplier dominate its $L^\infty$ norm? (1712.07609v1)

Published 20 Dec 2017 in math.CA

Abstract: One can define Fourier multipliers on a Banach function space by using the direct and inverse Fourier transforms on $L2(\mathbb{R}n)$ or by using the direct Fourier transform on $S(\mathbb{R}n)$ and the inverse one on $S'(\mathbb{R}n)$. In the former case, one assumes that the Fourier multipliers belong to $L\infty(\mathbb{R}n)$, 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 $X(\mathbb{R}n)$ satisfies a certain weak doubling property, then the space of all Fourier multipliers $\mathcal{M}{X(\mathbb{R}n)}$ is continuously embedded into $L\infty(\mathbb{R}n)$ with the best possible embedding constant one. For weighted Lebesgue spaces $Lp(\mathbb{R}n,w)$, the weak doubling property is much weaker than the requirement that $w$ is a Muckenhoupt weight, and our result implies that $|a|{L\infty(\mathbb{R}n)}\le|a|{\mathcal{M}{Lp(\mathbb{R}n,w)}}$ for such weights. This inequality extends the inequality for $n=1$ 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 $w$ in $Lp(\mathbb{R}n,w)$ to grow at any subexponential rate. On the other hand, the space $Lp(\mathbb{R},ex)$ has plenty of unbounded Fourier multipliers.

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