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Let $\rho \in S(\mathbb R^{n})= \text{Schwartz space}, \ \rho:\mathbb R^{n}\to [0,1]$ be smooth radial function verifying $\rho(\xi)=1$ for $|\xi|_{\infty}\leq \frac{1}{2}$ and $\rho(\xi)=0$ for $|\xi|_{\infty}\geq 1$ [For $\xi=(\xi_{1},...,\xi_{n}), |\xi|_{\infty}:= \max_{i=1,2,...,n}|\xi_{i}|$ ]. Let $\rho_{k}$ be a translation of $\rho,$ $\rho_{k}(\xi):=\rho(\xi -k), k\in \mathbb Z^{n}.$ Let $Q_{k}$ be be the unit cube with center at $k,$ $\{Q_{k}\}_{k\in\mathbb Z^{n}}$ constitutes a decomposition of $\mathbb R^{n}.$ We notice, $\rho_{k}(\xi)=1$ in $Q_{k}$, and so $\sum_{k\in \mathbb Z^{n}}\rho_{k}(\xi)\geq 1$ for all $\xi \in \mathbb R^{n}.$ Denote, $\sigma_{k}(\xi):= \rho_{k}(\xi)\left(\sum_{k\in\mathbb Z^{n}}\rho_{k}(\xi)\right)^{-1}, \ k\in \mathbb Z^{n}.$ Then we have,

(1) $|\sigma_{k}(\xi)|\geq c, \forall z \in Q_{k}$,

(2) $\text{supp} \ \sigma_{k} \subset \{\xi: |\xi-k|_{\infty}\leq 1 \},$

(3) $\sum_{k\in \mathbb Z^{n}} \sigma_{k}(\xi)\equiv 1, \forall \xi \in \mathbb R^{n},$

(4) $|D^{\alpha}\sigma_{k}(\xi)|\leq C_{|\alpha|}, \forall \xi \in \mathbb R^{n}, \alpha \in (\mathbb N \cup \{0\})^{n}.$

Hence, the set $A_{n}:=\{ \{\sigma_{k}\}_{k\in \mathbb Z^{n}}:\{\sigma_{k}\}_{k\in \mathbb Z^{n}} \ \text{satisfies} (1) to (4) \}$ is non empty.

Let $\{\sigma_{k}\}\in A_{n}$ and define frequency- uniform decomposition operator $$\square_{k}^{\sigma}:= \mathcal{F}^{-1}\sigma_{k}\mathcal{F}, k\in \mathbb Z^{n};$$ where $\mathcal{F}-$ denotes Fourier transform and $\mathcal{F}^{-1}-$ inverse Fourier transform.

My Questions: (I)Why this operator is known as frequency-uniform decomposition operator ? (II) Can we expect, $$\square_{k}^{\sigma}=\sum_{|\ell|_{\infty}\leq 1}\square_{k}^{\sigma}\square_{k+\ell}^{\phi};$$ for $\{\sigma_{k}\}_{k\in \mathbb Z^{n}}, \{\phi_{k}\}_{k\in \mathbb Z^{n}} \in A_{n}$; if yes, How ?

My Motivation:(Importance of frequency-uniform decomposition operator); It is well-known that $S(t)=e^{it\triangle}: L^{p}\to L^{p}$ if and only if $p=2.$ But the frequency-uniform decomposition has at least two advantages for the Shr\"odinger semi-group: (a) $\square_{k}e^{it\triangle}:L^{p'}\to L^{p}$ satisfies a uniform truncated decay, (b) $\square_{k}e^{it\triangle}$ is uniformly bounded on $L^{p}.$

Thanks,

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You should compare this construction to the dyadic (or Littlewood-Paley) decomposition.

The answer to (i) is simply that the volume of the cubes $Q_k$ stays $1$ while, e.g. the volume of the annuli in the Littlewood-Paley decomposition grows like $O(2^{nk})$.

The answer to (ii) is also very easy and follows from the property $\sum_{k\in \mathbb Z^{n}} \sigma_{k}(\xi)\equiv 1$. Just take this also for $\{\rho_k\}_{k \in \mathbb Z}$ and multiply $\square_k^\sigma$ by the sum $\sum_{k\in \mathbb Z^{n}} \rho_{k}(\xi)\equiv 1$ and see what remains.

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