Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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,

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.