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I have a question concerning Littlewood-paley-theory. Suppose we have test functions $\psi_k$ supported in annuli $\{2^{k-1}\leq\vert\xi\vert\leq2^{k+1}\}$ such that $\sum_{k\in\mathbb{Z}}\psi_k(\xi)=1$ for $\xi\neq0$. Define the homogeneous Littlewood-Paley projections $P_k$ of a tempered distribution by $\widehat{P_ku}=\psi_k\hat u$. I'd like to show that the equality $$ u=\sum_{k\in\mathbb{Z}}P_ku\qquad \text{in} \qquad \mathcal{S}'$$ holds true modulo a polynomial. Unfortunately I'm not able to show this. That's why I'm hoping for your help.

Greets Lukas

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2 Answers 2

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You need to show that the series converges in the sense of distributions, i.e. for arbitrary Schwartz function $\phi\in \mathcal S$ you have (modulo polynomial distribution) $$\lim _{n\to+\infty}\sum_{k=-n}^n \langle P_k u,\phi\rangle=\langle u,\phi\rangle = \langle F[u],F^{-1}[\phi]\rangle.$$

Given that thet Fourier transform is an ismorphism of tempered distributions and of the space of Schwartz functions, we can write by definition $$\langle P_k u,\phi\rangle = \langle F[P_k u],F^{-1}[\phi]\rangle = \langle \psi_k F[u],F^{-1}[\phi]\rangle=\langle F[u],\psi_k F^{-1}[\phi]\rangle,$$ therefore, we need to consider the sum $$\sum_{k\in\Bbb Z}\psi_k F^{-1}[\phi].$$

By the hypothesis on $\psi_k$ this sum converges to $F^{-1}[\phi]$ modulo the value in $\xi=0$. Therefore, the resulting sum of Fourier transfroms of projections differs from $F[u]$ at most on the sinlgeton $\{\xi=0\}$ (in other words, by a finites sum of Dirac delta and its derivatives). By taking the inverse Fourier transfrom we obtain that the sum of projections differ from the initial distribution by the inverse Fourier transform of that finite sum, which yields you that polynomial.

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    $\begingroup$ If $u$ is a general tempered distribution (e.g. not a Schwartz function), I don't see why considering $\sum_{k\in\Bbb Z}\psi_k F^{-1}[\phi]$ and its weak convergence (to $F^{-1}[\phi]$ + some finite sum of Dirac) helps us when computing $\lim _{n\to+\infty}\langle F[u],\sum_{k=-n}^n \psi_k F^{-1}[\phi]\rangle$. $\endgroup$
    – Desura
    Commented May 11, 2021 at 20:41
  • $\begingroup$ @Desura it's been a while since I written this answer=) On the first sight, we need the convergence of $\sum_{k\in\Bbb Z}\psi_k F^{-1}[\phi]$ in $\mathcal S$ (up to something that happens in zero), not the weak convergence. $\endgroup$ Commented May 12, 2021 at 17:25
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This is not true in general.

Let $C$ be the annulus $\{\xi \in \mathbb R^d : 3/4 \leq |\xi| \leq 8/3\}$ and $B$ be the ball $B(0,4/3)$. Consider radial functions $\chi, \eta \in C^{\infty}(\mathbb R^d, [0,1])$ such that $\text{supp}(\chi) \subset B$, $\text{supp}(\phi) \subset C$,

$$1 = \chi(\xi) + \sum_{j \geq 0}\phi(2^{-j} \xi) \ \forall \xi \in \mathbb R^d$$

$$1 = \sum_{j \in \mathbb Z}\phi(2^{-j} \xi) \ \forall \xi \in \mathbb R^d \setminus \{0\}$$

and the sums are locally finite respectively on $\mathbb R^d$ and $\mathbb R^d \setminus \{0\}$. For a proof of existence of $\chi$ and $\phi$, look at H. Bahouri Fourier Analysis and Nonlinear PDE book.

In particular, for all $n \in \mathbb Z$ :

$$1 = \chi(2^n \xi) + \sum_{j \geq -n}\phi(2^{-j} \xi) \ \forall \xi \in \mathbb R^d$$

and since this sum is still locally finite :

$$u = \chi(2^n D)u + \sum_{j \geq -n}\phi(2^{-j} D)u$$

for any $n \in \mathbb Z$ and any tempered distribution $u \in S'(\mathbb R^d)$.

Thus, the homogeneous Littlewood-Paley decomposition $$\sum_{j \in \mathbb Z}\phi(2^{-j} D)u$$ converges weakly* to $u$ + some polynomial if and only if $$\chi(2^n D)u$$ converges weakly* to a polynomial. However, there is no guarantee at all that $\chi(2^n D)u$ will converge. Indeed, take :

$$d = 1, u \in S'(\mathbb R), \ \mathcal{F}[u] = \sum_{j\leq 0} j^{-2}\cdot \delta'_{2^j}\in S'(\mathbb R)$$ Then for $\psi \in S(\mathbb R)$ such that $\check{\psi} = \mathcal{F}^{-1}[\psi] \gt 0$ in $B(0,2)$ :

$$\begin{align*} \langle \chi(2^n D)u, \psi \rangle = \langle \mathcal{F}[u], \chi(2^n \xi)\check{\psi} \rangle &= \sum_{j\leq 0} j^{-2}\cdot 2^n \chi'(2^{n+j}) \check{\psi}(2^{j}) + \sum_{j\leq 0} j^{-2}\cdot \chi(2^{n+j}) \check{\psi}'(2^{j}) \\ &= (1) + (2) \end{align*}$$

The second sum is bounded. The first sum satisfies

$$\sum_{j\leq 0} 2^n j^{-2}\cdot \chi'(2^{n+j}) \check{\psi}(2^{j}) \geq \frac{2^n}{n^{2}} \chi'(1) \check{\psi}(2^{-n}) $$

which makes $\langle \chi(2^n D)u, \psi \rangle$ a non-convergent sequence.

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