Carleson tiles wave packets and tile order

Assume The Axiom of Choice for the countable-choice Fourier suppliers. A dyadic interval is $[j{2}^k,(j+1)2^k)$ with $j,k\in \mathbb{Z}$; its length is positive. A tile is $s=I_s\times {\omega }_s$ with both intervals dyadic and $\mathrm{\mid }{I}_s\mathrm{\mid }\mathrm{\mid }{\omega }_s\mathrm{\mid }=1$. Their centers are denoted c. Write ${\omega }_{s,-}$ and ${\omega }_{s,+}$ for the left and right half-open halves. The midpoint belongs only to the right half.

Here is an explicit nonzero packet convention. Define $\rho (t)=\exp (-1\mathrm{/}t)$ for $t>0$ and $\rho (t)=0$ for $t\le 0$. Here is the complete smoothness justification. Put $P_0(u)=1$ and recursively $P_{k+1}(u)=u^2(P_k(u)-P_k^{\mathrm{\prime }}(u))$. The exponential derivative, chain rule and product/quotient rules (The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$, Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$) show on $t>0$ that the kth derivative is $P_k(1\mathrm{/}t)\exp (-1\mathrm{/}t)$. For each polynomial P and each integer r>=0, $t^{-r}P(1\mathrm{/}t)\exp (-1\mathrm{/}t)\to 0$ as $t\downarrow 0$: expand P into finitely many monomials, put u=1/t, and apply The exponential dominates every fixed nonnegative integer power at $+\mathrm{\infty }$ with parameter one and The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$. Extend each displayed kth-derivative formula by zero on $t\le 0$. Each extension is continuous at zero (r=0); its difference quotient at zero tends to zero (r=1), so its derivative there equals the next extension's value. On the two open half-lines differentiation already gives the next extension. Induction therefore proves $\rho \in C^{\mathrm{\infty }}(\mathbb{R})$ and ${\rho }^{(k)}(0)=0$ for every k. Positivity of the exponential makes $\rho (t)>0$ exactly when $t>0$. Set $a=1\mathrm{/}9$, $b=1\mathrm{/}8$ and $\psi (\xi )=\frac{\rho (b^2-{\xi }^2)}{\rho (b^2-{\xi }^2)+\rho ({\xi }^2-a^2)}\mathrm{.}$ The denominator is positive: for $\mathrm{\mid }\xi \mathrm{\mid }\le a$ its first term is positive, for $\mathrm{\mid }\xi \mathrm{\mid }\ge b$ its second term is positive, and between a and b both are positive. Thus $\psi$ is smooth, $0\le \psi \le 1$, equals one on $[-a,a]$ and vanishes outside $[-b,b]$. Every derivative is bounded on its compact support, so it is Schwartz under Schwartz space and its seminorms. Define $\varphi (x)=\widehat{\psi }(-x)$. The transform theorem Fourier transform acts continuously on Schwartz space makes this Schwartz, and Fourier inversion on Schwartz space gives $\widehat{\varphi }=\psi$. In detail inversion applied to $\psi$ says $\int \widehat{\psi }(y)e^{2\pi iy\xi }dy=\psi (\xi )$; changing $y=-x$ gives the claimed transform. All integrals are absolute by Schwartz derivatives are integrable. In particular $\varphi$ is not zero. No unspecified zero packet is allowed.

Use ${\mathrm{Tr}}_{y}h(x)=h(x-y)$, ${\mathrm{Mod}}_{\xi }h(x)=e^{2\pi i\xi x}h(x)$ and ${\mathrm{Dil}}_{\mathrm{\ell }}^{2}h(x)={\mathrm{\ell }}^{-1\mathrm{/}2}h(x\mathrm{/}\mathrm{\ell })$ for $\mathrm{\ell }>0$. Define ${\varphi }_s(x)=\mathrm{\mid }{I}_s{\mathrm{\mid }}^{-1\mathrm{/}2}{e}^{2\pi ic({\omega }_{s,-})x}\varphi ((x-c(I_s))\mathrm{/}\mathrm{\mid }{I}_s\mathrm{\mid })\mathrm{.}$ The translation, modulation and dilation laws Translation, modulation, linear dilation and reflection laws give $\widehat{{\varphi }_s}(\xi )=\mathrm{\mid }{I}_s{\mathrm{\mid }}^{1\mathrm{/}2}{e}^{-2\pi ic(I_s)(\xi -c({\omega }_{s,-}))}\psi (\mathrm{\mid }{I}_s\mathrm{\mid }(\xi -c({\omega }_{s,-})))\mathrm{.}$ Its support lies strictly inside ${\omega }_{s,-}$, since its half-width is $1\mathrm{/}(8\mathrm{\mid }{I}_s\mathrm{\mid })$ whereas that half-interval has half-width $1\mathrm{/}(4\mathrm{\mid }{I}_s\mathrm{\mid })$. For every integer M there is a finite constant $C_M$ with $\mathrm{\mid }{\varphi }_s(x)\mathrm{\mid }\le C_M\mathrm{\mid }{I}_s{\mathrm{\mid }}^{-1\mathrm{/}2}(1+\mathrm{\mid }x-c(I_s)\mathrm{\mid }\mathrm{/}\mathrm{\mid }{I}_s\mathrm{\mid }{)}^{-M}$, directly from the Schwartz seminorms. The pairing is $⟨f,h⟩=\int f\bar{h}$, linear in f. The fixed packet is not assumed to have norm one; its fixed norm and seminorms enter constants.

Set $s\le t$ when $I_s\subseteq I_t$ and ${\omega }_t\subseteq {\omega }_s$. This is a partial order: reflexivity and transitivity follow from inclusions, and mutual comparability gives equality of both intervals. A finite tree is a set T with a designated top tile t such that $s\le t$ for every $s\in T$; the top need not belong to T and is part of the data. A plus tree has, in addition, ${\omega }_t\subseteq {\omega }_{s,+}$ for every $s\in T\setminus \{t\}$. The top alone is allowed. Empty trees have no contribution; any assigned top is retained only when a forest count is explicitly specified. For a finite tile set S, a measurable selector $N:\mathbb{R}\to \mathbb{R}$ and $f\in L^2$, define the finite model $C_{S,N}f(x)={\sum }_{s\in S}⟨f,{\varphi }_s⟩{\mathbf{1}}_{{\omega }_{s,+}}(N(x)){\varphi }_s(x)\mathrm{.}$ The coefficients exist by Cauchy–Schwarz The complex $L^2$ pairing is well-defined and satisfies Cauchy–Schwarz and Schwartz integrability. Each summand is measurable, the sum is finite and for fixed S,N it is linear in f. Its testing form for $\mathrm{\mid }g\mathrm{\mid }\le {\mathbf{1}}_E$, $m(E)<\mathrm{\infty }$, is the sum of $⟨f,{\varphi }_s⟩⟨{\mathbf{1}}_{{\omega }_{s,+}}(N){\varphi }_s,g⟩$; these integrals exist since the packets are integrable and g is bounded. These definitions do not assert orthogonality of overlapping packets.