4
$\begingroup$

Let me first recall a particular case of the classical Hörmander-Mikhlin multiplier theorem: Let $m$ be a bounded function on $\mathbb {R} ^{n}$ which is smooth except possibly at the origin, and such that the function $$\text{ $|\xi|^{k}|\nabla ^{k}m(\xi)|$ is bounded for all integers $0\leq k\leq \frac n2+1$}, \tag{$\ast$}$$ then the Fourier multiplier $m(D)$ is a bounded endomorphism of $L^p(\mathbb R^n)$ for all $1 < p < ∞$ and is bounded from $L^1(\mathbb R^n)$ into $L^{1,\infty}(\mathbb R^n)$.

There is no doubt in my mind that an analogous statement holds true on the torus, but precisely, what means analogous? A Fourier multiplier on the torus $\mathbb T^n$ is simply a function $m$ defined on $\mathbb Z^n$ and we have $$ (m(D) u)(x)=\sum_{k\in \mathbb Z^n} m(k) \hat u(k)e^{2π i x\cdot k}. $$ Assuming $m$ bounded is of course enough for $L^2(\mathbb T^n)$ boundedness (and also necessary), but is certainly not sufficient for $L^p(\mathbb T^n)$ boundedness or Marcinkiewicz' $L^1\rightarrow L^{1,\infty}$ continuity.

Question: What is the discrete relevant assumption on $m$ which could replace on the lattice $\mathbb Z^n$ the kind of homogeneity assumption given by $(\ast)$?

$\endgroup$

2 Answers 2

2
$\begingroup$

There are two ways of proceeding. You can impose $m$ to be the restriction to $\mathbb{Z}^n \subset \mathbb{R}^n$ of a function satisfying the condition $(\ast)$ or you can formulate an analogue of condition $(\ast)$ with discrete derivatives.

I can not give a reference from the top of my head for the second approach. But I remember that Ruzhansky and Turunen, "Pseudo-Differential Operators and Symmetries" contains a more general formulation of the Hörmander condition for pseudodifferential operators.

$\endgroup$
2
$\begingroup$

The authors cited by Adrián González Pérez published a paper on the subject. A precise example is also given in this more recent article. It seems simpler to rely on Leeuw's theorem (transference), though.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .