Littlewood-Paley decomposition

Solution 1:

Not only is it possible to build Littlewood-Paley theory on a Torus, but this is even often used.

The basic idea is that you can build the Fourier transform on a Torus in the following way: Let $\mathcal{S}$ be the space of Schwartz functions, namely (periodic) $C^{\infty}$ functions on $\mathbb{T}^d = (\mathbb{R} / \mathbb{Z})^d.$

It's topological dual $\mathcal{D}$ is the space of "tempered" distributions. In this case tempered means that they are periodic whereas on the real line it meant that they could grow at most polynomially at infinity.

Then for $\phi \in \mathcal{S}$ you can define: $$ \mathcal{F} \phi (k) = \frac{1}{(2 \pi)^{d}}\langle \phi, e^{-ikx} \rangle. $$ where the inner product is the $L^2$ inner product.

Not surprisingly all the known results still hold, e.g. Plancherel forumla, inversion or even the fact that Schwartz functions are mapped bijectively to functions with rapid decay.

Then you can define Littlewood Paley block as before. Note that this time you may think that you do not need a smooth partition of the unity (the Fourier space is discrete). But as it turns out the smoothness of the partition of the unity is still important in order to have: $$ \| \Delta_j f\|_{\infty} \le C \|f\|_{\infty}. $$

As a last remark, you might wonder what the connection between the spaces on the Torus and the ones on $\mathbb{R}^d$ is. Often the link is given by the so-called Poisson formula.

For a detailed reference I can only give the "EBP lecture notes on singular stochastic PDEs" by Gubinelli and Perkowski, pages 37 and onwards (and some parts of the introduction).

ADDENDUM:

1. The crucial thing in the estimate I provided is that the constant $C$ is independent of $j$.

2. In the periodic setting you can define for example: $$ \widetilde{\Delta}_j f : = \mathcal{F}^{-1} (1_{[2^j, 2^{j+1})} \hat{f}) $$ This object, which could be considered a Littlewood-Paley block, is a smooth function, unlike in the case on the real line. Indeed in the real line Schwartz functions are mapped into Schwartz functions, and the indicator function is definitely not such a function. On a torus, though, the Fourier transform is a function on $\mathbb{Z}^d.$ So the indicator function is perfectly fine (think of interpolating it on the integer points with some smooth function).

3. For the Fourier transform on $\mathbb{R}^d$ and a given Schwartz function $\rho$ we know that for $\rho_j (\cdot) = \rho (2^{-j} \cdot)$ the following holds. $$ \mathcal{F}_{\mathbb{R}^d}^{-1} \rho_j (\cdot) = 2^{jd} \mathcal{F}_{\mathbb{R}^d}^{-1} \rho (2^{j} \cdot) $$ which implies that $\mathcal{F}^{-1}_{\mathbb{R}^d}\rho_j$ has $L_1$ norm uniformly bounded in $j$. Through the Poisson summation formula the same holds for the Fourier transform on the torus. But you need that the function $\rho$ is smooth. For this reason the construction at the point 2) is not enough to give us the required bound. For more details consult the reference I gave. This problem is explicitly addressed.

Solution 2:

Why not? Just look at a certain range of fourier coefficients - say $2^k \le |n| < 2^{k+1}$ for $n \in \mathbb{Z}$. And do an analogous thing in higher dimensions. Look up Haar functions (one reference is Muscalu Schlag volume 1 chapter 8).

Solution 3:

Yes sure of course, there is analogy and here is a link to Yves Meyer's beautiful book https://books.google.tn/books?id=y5L5HVlh3ngC&pg=PA1&hl=fr&source=gbs_toc_r&cad=3#v=onepage&q&f=false see page 6 and beyond.