The Carleson–Hunt Time–Frequency Theorem

A transformed grid has spatial intervals $y+rI$ and frequency intervals $\eta +r^{-1}\omega$ for r>0 and fixed y,eta. Use the same packet formula at their actual centers and lengths. Let $Uf(x)=r^{-1\mathrm{/}2}{e}^{2\pi i\eta x}f((x-y)\mathrm{/}r)$. Substitution in the packet formula shows that U sends each original packet to its transformed packet times a unimodular constant; that constant cancels in its coefficient times packet. Thus the transformed model is $U{C}_{S,N^{\mathrm{\prime }}}{U}^{-1}$, where $N^{\mathrm{\prime }}(z)=r(N(y+rz)-\eta )$. This selector is measurable. The $L^p$ norm of U is the scalar factor $r^{1\mathrm{/}p-1\mathrm{/}2}$, and that of its inverse is reciprocal, so the strong norm B is preserved. The distribution change of variables gives the same assertion for the weak norm. Therefore the hypothesis covers every transformed grid used below.

A weak bound $m\{\mathrm{\mid }v\mathrm{\mid }>t\}\le (K\mathrm{/}t{)}^p$ implies ${\int }_E\mathrm{\mid }v\mathrm{\mid }\le p^{\mathrm{\prime }}Km(E{)}^{1-1\mathrm{/}p}$ for each finite-measure E: by F10 with exponent one, its left side is at most ${\int }_0^{\mathrm{\infty }}\min (m(E),(K\mathrm{/}t{)}^p)dt$, which equals the stated expression after splitting at $Km(E{)}^{-1\mathrm{/}p}$. The zero cases give zero directly. Consequently the weak model hypothesis implies the testing bound $\mathrm{\mid }⟨Cf,g⟩\mathrm{\mid }\le p^{\mathrm{\prime }}B\mathrm{\parallel }f{\mathrm{\parallel }}_{p}m(E{)}^{1-1\mathrm{/}p}$ for $\mathrm{\mid }g\mathrm{\mid }\le {\mathbf{1}}_E$. The strong hypothesis implies $\mathrm{\mid }⟨Cf,g⟩\mathrm{\mid }\le B\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$ by F9. We will preserve these two testing inequalities through probability averages, first for bounded g supported in a bounded interval.

Fix an integer K>=0 and r in [1,2]. Retain only spatial scales $l=r{2}^k$, -K<=k<=K. Average transformed grids over $0\le y<Y=r{2}^K$ and $0\le \eta <W=2^K\mathrm{/}r$, with normalized measures dy/Y and deta/W. At each fixed scale retain initially only spatial indices i and frequency indices j with |i|,|j|<=M, so the operator is a finite model covered by steps 1.1 and 1.2. Let M increase. For each x and each scale, the upper-half selector condition activates at most one frequency index. Every coefficient has magnitude at most $\mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }\varphi {\mathrm{\parallel }}_2$ by F9, and the sum of the packet absolute values over its spatial lattice is at most $C{l}^{-1\mathrm{/}2}$ by Schwartz decay. Thus all truncated outputs and their pointwise limits are bounded by $C_K\mathrm{\parallel }f{\mathrm{\parallel }}_2$, uniformly in x,y,eta,r. The spatial series is absolutely convergent, and the unique active frequency index is eventually retained, so the limits exist for every x and parameter. Coefficients are continuous in the packet parameters, by domination on each compact parameter set; the selector indicators are measurable in x and the parameters. Hence all the functions being averaged are product-measurable. For bounded compactly supported g, F7 and this uniform bound pass both testing inequalities through M to the full-index, finite-scale average.

We compute its spatial average exactly. Fix l and a frequency interval with lower center nu. Write $p_{l,c,\nu }(x)=l^{-1\mathrm{/}2}{e}^{2\pi i\nu x}\varphi ((x-c)\mathrm{/}l)$. The spatial centers are $c=y+l(i+1\mathrm{/}2)$. Since Y/l is an integer, averaging their lattice over y gives density dc/l on the line. Therefore the averaged rank-one sum has kernel $\frac{1}{l}{e}^{2\pi i\nu (x-z)}R((x-z)\mathrm{/}l),R(v)=\int \varphi (v+u)\overline{\varphi (u)}du\mathrm{.}$ Indeed insert the coefficient integral and replace the averaged sum over c by l^(-1) times its integral; then substitute c=z-lu. These exchanges are absolute: at fixed x, the integral over z,c of the absolute integrand after averaging is at most $l^{-1}\mathrm{\parallel }f{\mathrm{\parallel }}_1\mathrm{\parallel }\varphi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\parallel }\varphi {\mathrm{\parallel }}_1$. F5 establishes absolute product integrability and F6 exchanges the integrals and the counting sum. Since psi is real and even, its inverse phi is real and even by the cosine integral, so R=phi*phi and F3 gives $\widehat{R}={\psi }^2$. F3 and F4 now identify the averaged output as $\int \widehat{f}(\xi )e^{2\pi ix\xi }\psi (l(\xi -\nu ){)}^{2}d\xi \mathrm{.}$ All these are Schwartz convolution identities at a fixed frequency.

Average next over eta. For a fixed x let a=N(x). At spatial length l, the frequency intervals have length1/l. The interval containing a is active exactly when the fractional position $v=l(a-\eta )-j$ lies in [1/2,1). Its lower center is $\nu =\eta +(j+1\mathrm{/}4)\mathrm{/}l$, so $l(\xi -\nu )=l(\xi -a)+v-1\mathrm{/}4$. Because W is an integer multiple of1/l, the eta average makes v uniform on [0,1]. It follows that the frequency-averaged multiplier at scale l is ${\int }_{1\mathrm{/}2}^1\psi (l(\xi -a)+v-1\mathrm{/}4{)}^{2}dv=H(l(a-\xi ))\mathrm{.}$ The half-open endpoints make a unique interval convention; their eta measure is zero and they do not alter the integral. Since $0\le \psi \le 1$ and $\widehat{f}\in L^1$, the Fourier integral and finite eta average interchange absolutely by F6. Crucially, this calculation holds at every fixed x for its actual value N(x); N is not frozen to a constant across the spatial variable.

Finally average r over [1,2] with probability measure $dr\mathrm{/}(r\log 2)$. The resulting finite-scale average $A_{K,N}f$ has the exact representation $A_{K,N}f(x)=\int \widehat{f}(\xi )e^{2\pi ix\xi }{m}_K(N(x)-\xi )d\xi ,m_K(d)=\frac{1}{\log 2}{\int }_{2^{-K}}^{2^{K+1}}H(td)\frac{dt}{t}\mathrm{.}$ This follows by changing variables t=r2^k in the finite scale sum: the intervals [2^k,2^(k+1)] partition the displayed integration range. The measure in r and the normalized measures in y,eta form a probability measure; they can equivalently be parameterized by y/Y,eta/W in [0,1]. Thus step 2.1 passes the testing bounds to $A_{K,N}f$. All integrations here are on compact parameter ranges with the stated uniform majorant.

The function H is nonnegative, bounded by1/2, and supported in [1/8,7/8]. Moreover $H(t)\ge 2\mathrm{/}9$ for $3\mathrm{/}8\le t\le 5\mathrm{/}8$: the interval $[t-1\mathrm{/}9,t+1\mathrm{/}9]$ lies inside [1/4,3/4] and its integrand is one. Consequently $0<\frac{2\log (5\mathrm{/}3)}{9\log 2}\le \kappa \le \frac{\log 7}{2\log 2}<\mathrm{\infty }\mathrm{.}$ For d<=0, m_K(d)=0. For d>0, the substitution u=td and expansion of the integration range give $m_K(d)\uparrow \kappa$. Thus $0\le m_K\le \kappa$ for every K. Since $\widehat{f}\in L^1$, F7 gives, pointwise for every finite N(x), $A_{K,N}f(x)⟶\kappa {\int }_{-\mathrm{\infty }}^{N(x)}\widehat{f}(\xi )e^{2\pi ix\xi }d\xi =\kappa T_{N(x)}f(x)\mathrm{.}$ The isolated frequency xi=N(x) has zero Lebesgue measure, so the endpoint does not change this identity. This proves both convergence and a strictly positive reconstruction constant without classifying translation-invariant operators or invoking distributions.

Since $\mathrm{\mid }{A}_{K,N}f\mathrm{\mid }\le \kappa \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1$, F7 passes its testing inequalities to the limit against any bounded compactly supported g. Under the strong hypothesis this yields $\mathrm{\mid }⟨T_{N}f,g⟩\mathrm{\mid }\le {\kappa }^{-1}B\mathrm{\parallel }f{\mathrm{\parallel }}_p\mathrm{\parallel }g{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$ for every such g. To recover the norm, use the following bounded truncations: put $g_{n,R}={\mathbf{1}}_{[-R,R]}{\mathbf{1}}_{\{\mathrm{\mid }{T}_{N}f\mathrm{\mid }>1\mathrm{/}n\}}\mathrm{\mid }{T}_{N}f{\mathrm{\mid }}^{p-2}{T}_{N}f$ for integers $n,R\ge 1$, defined as zero where $T_{N}f$ vanishes. Each $g_{n,R}$ is bounded and compactly supported, so the testing inequality applies to it. Indeed its magnitude is at most $\mathrm{\mid }{T}_{N}f{\mathrm{\mid }}^{p-1}\le \mathrm{\parallel }\widehat{f}{\mathrm{\parallel }}_1^{p-1}$, since $p>1$, with value zero at zeros of $T_{N}f$. The same bound also holds without the lower level cutoff; the cutoff is a permissible convenience, not a remedy for a singularity. Writing $J_{n,R}={\int }_{[-R,R]\cap \{\mathrm{\mid }{T}_{N}f\mathrm{\mid }>1\mathrm{/}n\}}\mathrm{\mid }{T}_{N}f{\mathrm{\mid }}^p$, the definition gives both $⟨T_{N}f,g_{n,R}⟩=J_{n,R}$ and $\mathrm{\parallel }{g}_{n,R}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}=J_{n,R}^{1\mathrm{/}{p}^{\mathrm{\prime }}}$, because $(p-1)p^{\mathrm{\prime }}=p$. Hence $J_{n,R}\le {\kappa }^{-1}B\mathrm{\parallel }f{\mathrm{\parallel }}_p{J}_{n,R}^{1\mathrm{/}{p}^{\mathrm{\prime }}}$ and $J_{n,R}^{1\mathrm{/}p}\le {\kappa }^{-1}B\mathrm{\parallel }f{\mathrm{\parallel }}_p$. If $J_{n,R}=0$ the second inequality holds trivially; otherwise divide. The level sets increase with n and R, so F8 gives $\mathrm{\parallel }{T}_{N}f{\mathrm{\parallel }}_p\le {\kappa }^{-1}B\mathrm{\parallel }f{\mathrm{\parallel }}_p$. Under the weak hypothesis, take g to be the phase of T_Nf on $E=\{\mathrm{\mid }{T}_{N}f\mathrm{\mid }>\lambda \}\cap [-R,R]$, a bounded compactly supported function. The limiting testing inequality gives $\lambda m(E)\le p^{\mathrm{\prime }}{\kappa }^{-1}B\mathrm{\parallel }f{\mathrm{\parallel }}_{p}m(E{)}^{1-1\mathrm{/}p}$ and hence the stated weak bound for T_Nf after increasing R, also when m(E)=0. These bounds are uniform in the arbitrary measurable selector.

Enumerate the rational cutoffs and apply F1's least-index selector to each finite maximum. Step 7.1 gives the same strong or weak bound for every finite maximum. These maxima increase to $C_{\mathbb{R}}f$ because the cutoff is continuous in its parameter. F8 gives the strong norm estimate; continuity from below of their level sets gives the weak estimate, equivalently another use of F8 on their indicators. This proves the Statement. The finite-to-infinite index limits, parameter averages and cutoff limits were justified separately with explicit majorants. AC is inherited through F11; there is no selector for an unattained infinite supremum and no new arbitrary-index choice.

Write $d(t)={\int }_{E\cap \{N\in {\omega }_t\}}{\chi }_{I_t}$. Consider all tiles t dominating some s in S and satisfying $d(t)>\delta \mathrm{/}2$. They form a finite nonempty set: $d(t)\le m(E)\mathrm{/}\mathrm{\mid }{I}_t\mathrm{\mid }$ implies $\mathrm{\mid }{I}_t\mathrm{\mid }<2m(E)\mathrm{/}\delta$, while domination implies $\mathrm{\mid }{I}_t\mathrm{\mid }\ge {\min }_{s\in S}\mathrm{\mid }{I}_s\mathrm{\mid }>0$. Only finitely many dyadic scales lie in this range; at each scale there is one spatial ancestor of each fixed s and finitely many frequency subintervals of its frequency interval. Nonemptiness follows from the definition of positive supremum delta. Let Tops be the maximal members of this finite set. Assign each s lying below a member of Tops to one such top using a fixed finite ordering. This gives disjoint trees. Every s left over has density at most $\delta \mathrm{/}2$: otherwise a witness t with $d(t)>\delta \mathrm{/}2$ would dominate s and extend to a maximal candidate.

For a top t let $I=I_t$, $\omega ={\omega }_t$, and let $2^{k}I$ be the interval with the same center and length $2^k\mathrm{\mid }I\mathrm{\mid }$. There is an integer $k\ge 0$ with $m(E\cap \{N\in \omega \}\cap 2^{k}I)\ge c\delta 2^{2k}\mathrm{\mid }I\mathrm{\mid }$, where c is an absolute positive constant fixed below. Indeed, partition the line into I and the shells $2^{k}I\setminus 2^{k-1}I$ for k>=1. On the kth shell ${\chi }_I\le A\cdot 2^{-20k}\mathrm{/}\mathrm{\mid }I\mathrm{\mid }$ for $A=2^{40}$, and the same bound with k=0 dominates its value on I. If every proposed mass bound failed, then $d(t)\le Ac\delta {\sum }_{k\ge 0}{2}^{-18k}$. Taking $c=(8A{)}^{-1}$ makes this less than $\delta \mathrm{/}2$, contradicting the choice of t. Assign the least successful k to t, and call its assigned family ${\mathcal{T}}_k$.

Fix k. Greedily choose from ${\mathcal{T}}_k$ a tile with greatest spatial length, retain it, and remove every remaining tile whose enlarged rectangle $2^k{I}_s\times {\omega }_s$ intersects its enlarged rectangle. Continue until none remain; this is a finite procedure. Retained enlarged rectangles are disjoint, so their sets $E\cap \{N\in {\omega }_t\}\cap 2^k{I}_t$ are disjoint. Step 2.1 gives ${\sum }_{t\text{ retained}}\mathrm{\mid }{I}_t\mathrm{\mid }\le (c\delta 2^{2k}{)}^{-1}m(E)$.

A tile s removed by a retained t has $\mathrm{\mid }{I}_s\mathrm{\mid }\le \mathrm{\mid }{I}_t\mathrm{\mid }$ and overlapping frequency intervals; dyadic nesting and reciprocal lengths imply ${\omega }_t\subseteq {\omega }_s$. All tiles removed by this same t therefore have mutually overlapping, hence nested, frequency intervals. Since Tops is an antichain, their original spatial intervals must be pairwise disjoint: otherwise dyadic nesting of both coordinates would make two tops comparable. Also overlap of the enlarged spatial intervals implies each original $I_s$ lies in the interval centered at $c(I_t)$ of radius $(3\mathrm{/}2)2^k\mathrm{\mid }{I}_t\mathrm{\mid }$. Thus the sum of their lengths is at most $3\cdot 2^k\mathrm{\mid }{I}_t\mathrm{\mid }$. Including t itself in its removed family, and using step 3.1, yields ${\sum }_{s\in {\mathcal{T}}_k}\mathrm{\mid }{I}_s\mathrm{\mid }\le (3\mathrm{/}c)2^{-k}{\delta }^{-1}m(E)$. Summing over k gives ${\sum }_{t\in \mathrm{T}\mathrm{o}\mathrm{p}\mathrm{s}}\mathrm{\mid }{I}_t\mathrm{\mid }\le (6\mathrm{/}c){\delta }^{-1}m(E)$. Together with step 1.1 this proves the partition and count assertion. Empty assigned trees may be deleted, only reducing the count. AC is inherited; the selections here are finite greedy choices and least integers.

Call a plus tree with top t strict when ${\omega }_t\subseteq {\omega }_{s,+}$ for every member s, so its top is not a member. Any plus tree can be made strict by replacing its top t with $\tilde{t}=I_t^{\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{n}\mathrm{t}}\times {\omega }_{t,+}$. This is a tile, dominates all its members, and satisfies the strict condition even for s=t. Its top length is twice the old length. Consequently, whenever a stock R has size greater than $\sigma \mathrm{/}2$, it has a strict plus subtree U with top t and ${\sum }_{s\in U}\mathrm{\mid }{a}_s{\mathrm{\mid }}^2>{\sigma }^2\mathrm{\mid }{I}_t\mathrm{\mid }\mathrm{/}8$. Indeed the supremum in F1 then has a witness with normalized sum greater than ${\sigma }^2\mathrm{/}4$, and the replacement divides that sum by two.

For $l=\mathrm{\mid }{I}_s\mathrm{\mid }\ge l^{\mathrm{\prime }}=\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }$, packet decay implies $\mathrm{\mid }⟨{\varphi }_s,{\varphi }_{s^{\mathrm{\prime }}}⟩\mathrm{\mid }\le C\sqrt{l^{\mathrm{\prime }}\mathrm{/}l}(1+\mathrm{\mid }c(I_s)-c(I_{s^{\mathrm{\prime }}})\mathrm{\mid }\mathrm{/}l{)}^{-20}\mathrm{.}$ To prove it, use exponent 40 in F2. For each x, the factor $1+\mathrm{\mid }c(I_s)-c(I_{s^{\mathrm{\prime }}})\mathrm{\mid }\mathrm{/}l$ is at most $(1+\mathrm{\mid }x-c(I_s)\mathrm{\mid }\mathrm{/}l)(1+\mathrm{\mid }x-c(I_{s^{\mathrm{\prime }}})\mathrm{\mid }\mathrm{/}{l}^{\mathrm{\prime }})$. Extract its inverse twentieth power from the product of the two decay factors. The remaining integral is at most $\int (1+\mathrm{\mid }x-c(I_{s^{\mathrm{\prime }}})\mathrm{\mid }\mathrm{/}{l}^{\mathrm{\prime }}{)}^{-20}dx=(2\mathrm{/}19)l^{\mathrm{\prime }}$. Multiplying by $(l{l}^{\mathrm{\prime }}{)}^{-1\mathrm{/}2}$ proves the bound. By F3, a Gram entry is zero when the two lower frequency halves are disjoint.

Starting with R=S, consider all strict witnesses satisfying the threshold inequality of step 1.1. Only finitely many possible tops occur: if $B={\sum }_{s\in S}\mathrm{\mid }{a}_s{\mathrm{\mid }}^2$ and ${\mathrm{\ell }}_0={\min }_{s\in S}\mathrm{\mid }{I}_s\mathrm{\mid }$, their lengths lie between ${\mathrm{\ell }}_0$ and $8B\mathrm{/}{\sigma }^2$. There are finitely many dyadic scales in this range. Each top dominates some s in S, so its spatial interval is the unique ancestor of $I_s$ at its scale, and its frequency interval is one of finitely many dyadic subintervals of ${\omega }_s$ at its prescribed length. Choose a possible top of minimum frequency center, breaking ties by any fixed ordering of this finite list, and choose a strict witness U for that top. Remove the full tree $V=\{s\in R:s\le t\}$, record U and V with that same top, and repeat. The process stops after at most |S| removals because every witness is nonempty. Possible witnesses only disappear as R shrinks, so the selected top-frequency centers are nondecreasing. At termination the residual size is at most $\sigma \mathrm{/}2$ by step 1.1. The V are disjoint tile collections, as are their subsets U.

The strict witnesses have the following separation property. If $s\in U$, $s^{\mathrm{\prime }}\in U^{\mathrm{\prime }}$ belong to different recorded witnesses and ${\omega }_{s,-}⊊{\omega }_{s^{\mathrm{\prime }},-}$, dyadic nesting implies ${\omega }_s\subseteq {\omega }_{s^{\mathrm{\prime }},-}$. Thus the top-frequency interval of U lies in ${\omega }_{s^{\mathrm{\prime }},-}$, while that of U' lies in ${\omega }_{s^{\mathrm{\prime }},+}$. The former center is smaller, so U was selected earlier. If $I_{s^{\mathrm{\prime }}}$ met $I_U$, the inequalities $\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }<\mathrm{\mid }{I}_s\mathrm{\mid }\le \mathrm{\mid }{I}_U\mathrm{\mid }$ and dyadic nesting would give $I_{s^{\mathrm{\prime }}}\subseteq I_U$. Together with ${\omega }_U\subseteq {\omega }_{s^{\mathrm{\prime }}}$ this would have removed s' with the earlier full tree. Hence $I_{s^{\mathrm{\prime }}}\cap I_U=\mathrm{\varnothing }$. Within one strict plus tree, distinct lower frequency halves are disjoint: if two unequal such halves were nested, the full smaller frequency interval would lie in the larger lower half, whereas the common top frequency must lie in its upper half. Equal lower halves mean equal full frequencies and equal spatial scales; distinct tiles then have disjoint spatial intervals.

Let $\mathcal{U}$ be the recorded strict witnesses, $L={\sum }_{U\in \mathcal{U}}\mathrm{\mid }{I}_U\mathrm{\mid }$, $Q={\sum }_U{\sum }_{s\in U}\mathrm{\mid }{a}_s{\mathrm{\mid }}^2$, and $H={\sum }_U{\sum }_{s\in U}{a}_s{\varphi }_s$. The witness threshold and original size give ${\sigma }^{2}L\mathrm{/}8<Q\le {\sigma }^{2}L$ when $L>0$. The part of the Gram expansion of $\mathrm{\parallel }H{\mathrm{\parallel }}_2^2$ with equal lower frequency halves has absolute value at most CQ. In fact, for each fixed frequency the spatial intervals form a subset of the lattice of intervals of its fixed length l. Step 1.2 bounds the row sums and column sums by $C{\sum }_{j\in \mathbb{Z}}(1+\mathrm{\mid }j\mathrm{\mid }{)}^{-20}<\mathrm{\infty }$. This series is finite by comparison with the integral of $x^{-20}$ on $[1,\mathrm{\infty })$. Apply $2\mathrm{\mid }{a}_s{a}_{s^{\mathrm{\prime }}}\mathrm{\mid }\le \mathrm{\mid }{a}_s{\mathrm{\mid }}^2+\mathrm{\mid }{a}_{s^{\mathrm{\prime }}}{\mathrm{\mid }}^2$ and sum the row and column bounds.

For fixed $s\in U$ consider all witness tiles s' whose lower frequency halves properly contain ${\omega }_{s,-}$. They belong to other witnesses by step 3.1, and their spatial intervals lie outside $I_U$. These intervals are pairwise disjoint. To check this, two corresponding lower frequency halves both contain ${\omega }_{s,-}$, so are nested. If unequal, their owning witnesses must differ by step 3.1; applying its separation property to that pair makes the smaller spatial interval disjoint from the other one's entire top interval, hence from the other spatial interval. If equal, they have the same scale and different spatial intervals. Put ${\chi }_I(x)=\mathrm{\mid }I{\mathrm{\mid }}^{-1}(1+\mathrm{\mid }x-c(I)\mathrm{\mid }\mathrm{/}\mathrm{\mid }I\mathrm{\mid }{)}^{-20}$. Since $\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }\le \mathrm{\mid }{I}_s\mathrm{\mid }$, the values ${\chi }_{I_s}(x)$ and ${\chi }_{I_s}(c(I_{s^{\mathrm{\prime }}}))$ on $I_{s^{\mathrm{\prime }}}$ differ by at most a fixed factor: their denominators before taking the twentieth power differ by at most $\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }\mathrm{/}(2\mathrm{\mid }{I}_s\mathrm{\mid })\le 1\mathrm{/}2$. Therefore ${\sum }_{s^{\mathrm{\prime }}}\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }{\chi }_{I_s}(c(I_{s^{\mathrm{\prime }}}))\le C{\int }_{\mathbb{R}\setminus I_U}{\chi }_{I_s}(x)dx\mathrm{.}$ All sums here are finite.

Singleton trees in F1 give $\mathrm{\mid }{a}_s\mathrm{\mid }\le \sigma \sqrt{\mathrm{\mid }{I}_s\mathrm{\mid }}$. Combining this with step 1.2 bounds an oriented unequal-frequency Gram term by $C{\sigma }^2\mathrm{\mid }{I}_s\mathrm{\mid }\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }{\chi }_{I_s}(c(I_{s^{\mathrm{\prime }}}))$. Thus step 4.1 bounds all such terms by $C{\sigma }^2{\sum }_U{\sum }_{s\in U}\mathrm{\mid }{I}_s\mathrm{\mid }{\int }_{\mathbb{R}\setminus I_U}{\chi }_{I_s}(x)dx\mathrm{.}$ For a fixed top interval J and a fixed scale l<=|J|, at most one frequency interval can occur above a given spatial interval of that scale: it must be the unique dyadic ancestor of the top frequency of length 1/l. Thus the spatial intervals at that scale form a subset of the dyadic subintervals of J. Write $m=\mathrm{\mid }J\mathrm{\mid }\mathrm{/}l$ and index the full list from k=0 to m-1. Direct integration on the two exterior half-lines bounds $l{\int }_{\mathbb{R}\setminus J}{\chi }_{I_k}(x)dx\le Cl((1+k{)}^{-19}+(1+m-1-k{)}^{-19})\mathrm{.}$ Indeed the exact two denominators are $1+k+1\mathrm{/}2$ and $1+m-k-1\mathrm{/}2$, with factor $l\mathrm{/}19$. The sum over k is at most Cl by integral comparison with $x^{-19}$. Summing over the dyadic scales $l=\mathrm{\mid }J\mathrm{\mid }{2}^{-j}$, j>=0, gives at most C|J| by the geometric sum. The oriented contribution is consequently at most $C{\sigma }^{2}L$; its conjugate orientation satisfies the same bound.

Steps 3.2 and 5.1, and the vanishing of every remaining Gram entry in step 1.2, give $\mathrm{\parallel }H{\mathrm{\parallel }}_2^2\le C{\sigma }^{2}L$. By the first-variable-linear convention, $⟨f,H⟩=Q$, so F4 gives $Q\le \mathrm{\parallel }f{\mathrm{\parallel }}_2\mathrm{\parallel }H{\mathrm{\parallel }}_2$. If L>0, combine this with $Q>{\sigma }^{2}L\mathrm{/}8$ and divide by $\sigma \sqrt{L}$ to obtain $L\le C{\sigma }^{-2}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. If no witness was selected, L=0 and the bound is immediate. The removed full trees V have these same tops, so this is their forest count, while step 2.1 gives the required residual size. No assumption that distinct top intervals are disjoint was made. All selections are finite; AC is inherited solely from the Fourier interfaces identified in F5.

The proof establishes the stronger sum of absolute tile contributions. Choose complex numbers ${\epsilon }_s$ of modulus one so that ${\epsilon }_s{a}_s⟨{\mathbf{1}}_{{\omega }_{s,+}}(N){\varphi }_s,g⟩$ is nonnegative real, taking ${\epsilon }_s=1$ for a zero product. Finite sesquilinearity bounds that sum by ${\int }_E\mathrm{\mid }{\sum }_s{\epsilon }_s{a}_s{\mathbf{1}}_{{\omega }_{s,+}}(N(x)){\varphi }_s(x)\mathrm{\mid }dx$. Write $b_s={\epsilon }_s{a}_s$. Empty T and sigma=0 give zero. If delta=0, exponent20 packet decay bounds the integral of $\mathrm{\mid }{\varphi }_s\mathrm{\mid }$ on $E\cap N^{-1}({\omega }_{s,+})$ by $C\sqrt{\mathrm{\mid }{I}_s\mathrm{\mid }}\delta =0$ for each s, so all contributions vanish. Hence assume delta,sigma>0. Singletons give $\mathrm{\mid }{b}_s\mathrm{\mid }\le \sigma \sqrt{\mathrm{\mid }{I}_s\mathrm{\mid }}$. A possible member s=t contributes at most $C\sigma L{\int }_{E\cap N^{-1}({\omega }_t)}{\chi }_{J_0}\le C\sigma \delta L$, by packet decay with exponent20. Remove this one member. Each remaining tile has its top frequency contained in exactly one half of its frequency interval. Split them into the strict plus and strict minus trees according as that half is right or left, retaining the same top t and the bounds delta,sigma.

We record a kernel bound. If $K_r$ satisfies $\mathrm{\mid }{K}_r(z)\mathrm{\mid }\le C{r}^{-1}(1+\mathrm{\mid }z\mathrm{\mid }\mathrm{/}r{)}^{-20}$ and h is Schwartz, then for every interval J of length j<=r and every x,y in J, $\mathrm{\mid }(K_r\ast h)(x)\mathrm{\mid }\le CMh(y)$. Indeed $\mathrm{\mid }x-y\mathrm{\mid }\le r$ makes the displayed weight at x comparable to that at y. On $\mathrm{\mid }z-y\mathrm{\mid }<r$ its integral against |h| is bounded by twice the centered average times a constant. On the annulus $2^{n}r\le \mathrm{\mid }z-y\mathrm{\mid }<2^{n+1}r$, it is at most $C{2}^{-19n}Mh(y)$, using the centered interval of radius $2^{n+1}r$. The geometric sum proves the bound uniformly in x,y,r.

For either nonempty strict subtree U, partition the line into maximal dyadic intervals J such that 3J contains no $I_s$, s in U; 3J denotes the concentric interval of triple length. Such intervals cover the line: all sufficiently small dyadic intervals have this property because U is finite with positive minimum spatial length, and sufficiently large ancestors of any fixed interval fail it because their triples eventually contain every fixed bounded interval. The property passes to dyadic subintervals since their triples lie in the larger triple; maximal good intervals are therefore disjoint and cover every point. They form a countable family $\mathcal{J}$. If j=|J| and $J^p$ is its dyadic parent, maximality supplies an $s^{\mathrm{\prime }}\in U$ with $I_{s^{\mathrm{\prime }}}\subseteq 3J^p$. Since this latter interval has length6j, dyadic lengths imply $\mathrm{\mid }{I}_{s^{\mathrm{\prime }}}\mathrm{\mid }\le 4j$. In particular $\mathrm{dist}(J,J_0)\le 6j$. If j>=L and J meets $J_0$, or if j>=L+dist(J,J_0) when they are disjoint, then $J_0\subseteq 3J$, contradicting goodness. Consequently intervals with dist(J,J_0)<L have j<2L and lie in a fixed concentric enlargement of $J_0$; their total length is at most CL. When d=dist(J,J_0)>=L, one has $d\mathrm{/}6\le j<d+L\le 2d$.

For the strict plus subtree put $H={\sum }_{s\in U}{b}_s{\varphi }_s$, a Schwartz function. Then $\mathrm{\parallel }H{\mathrm{\parallel }}_2\le C\sigma \sqrt{L}$. Here are the needed orthogonality details. Lower frequency halves at unequal scales are disjoint: if nested, the full smaller frequency interval lies in the larger lower half, whereas the common top lies in its upper half. Plancherel kills those Gram entries. At equal frequency and spatial scale l, packet decay gives $\mathrm{\mid }⟨{\varphi }_s,{\varphi }_{s^{\mathrm{\prime }}}⟩\mathrm{\mid }\le C(1+\mathrm{\mid }c(I_s)-c(I_{s^{\mathrm{\prime }}})\mathrm{\mid }\mathrm{/}l{)}^{-20}$: extract the center-distance factor from the product of exponent40 decay bounds by the triangle inequality, and integrate the remaining normalized exponent20 weight, whose integral is2/19. The row and column sums over the length-l spatial lattice are bounded by $C{\sum }_{n\in \mathbb{Z}}(1+\mathrm{\mid }n\mathrm{\mid }{)}^{-20}$. Using $2\mathrm{\mid }{b}_s{b}_{s^{\mathrm{\prime }}}\mathrm{\mid }\le \mathrm{\mid }{b}_s{\mathrm{\mid }}^2+\mathrm{\mid }{b}_{s^{\mathrm{\prime }}}{\mathrm{\mid }}^2$ proves $\mathrm{\parallel }H{\mathrm{\parallel }}_2^2\le C{\sum }_s\mathrm{\mid }{b}_s{\mathrm{\mid }}^2\le C{\sigma }^{2}L$, the last bound being the defining plus-tree size inequality.

On each J split the signed tile sum into small tiles $\mathrm{\mid }{I}_s\mathrm{\mid }\le 4j$ and large tiles $\mathrm{\mid }{I}_s\mathrm{\mid }\ge 8j$. These are all the possibilities because lengths are dyadic. The integral of the absolute small sum is at most $C\sigma \delta {\sum }_{s\in U:\mathrm{\mid }{I}_s\mathrm{\mid }\le 4j}\mathrm{\mid }{I}_s\mathrm{\mid }(1+\mathrm{dist}(I_s,J)\mathrm{/}\mathrm{\mid }{I}_s\mathrm{\mid }{)}^{-20}\mathrm{.}$ To see this for each tile of length l, exponent40 packet decay and the singleton bound give $\mathrm{\mid }{b}_s{\varphi }_s(x)\mathrm{\mid }\le C\sigma l{\chi }_{I_s}(x)(1+\mathrm{\mid }x-c(I_s)\mathrm{\mid }\mathrm{/}l{)}^{-20}$. Take the supremum of the last factor on J and integrate over $E\cap N^{-1}({\omega }_{s,+})$, a subset of the full-frequency set controlled by delta. At a fixed scale l, there is at most one possible frequency for each spatial interval, the unique ancestor of the top frequency of length1/l. The spatial intervals thus form a subset of the length-l dyadic lattice inside $J_0$. If l<=j, goodness and aligned dyadic endpoints put these intervals outside 3J, at distance at least j from J. Summing the displayed weights over that lattice gives at most $Cl{\sum }_{n\ge j\mathrm{/}l}(1+n{)}^{-20}\le Cj(l\mathrm{/}j{)}^{20}$ by integral comparison. For l=2j or4j, the full lattice sum is at most Cl<=Cj, with no separation needed. Summing dyadic l<=4j bounds the displayed spatial sum by Cj.

Let $G_J=E\cap J\cap {\bigcup }_{s\in U:\mathrm{\mid }{I}_s\mathrm{\mid }\ge 8j}{N}^{-1}({\omega }_{s,+})$, the support of the large sum on E. Then $m(G_J)\le C\delta j$. If there is no large tile the claim is immediate. Otherwise 8j<=L. For the witness s' in step 2.1, let v have spatial interval the dyadic ancestor of $I_{s^{\mathrm{\prime }}}$ of length4j and frequency the dyadic ancestor of the top frequency of length1/(4j). Then v is a tile above s', so its weighted density integral is at most delta. The center of its spatial interval is within Cj of J, because $I_{s^{\mathrm{\prime }}}\subseteq 3J^p$ and its ancestor has length4j. Hence ${\chi }_{I_v}\ge c\mathrm{/}j$ on J for an absolute c>0. Every large tile frequency is contained in ${\omega }_v$, since its length is at most1/(8j) and it also contains the top frequency. This gives $G_J\subseteq E\cap J\cap N^{-1}({\omega }_v)$ and proves the measure bound. Moreover 8j<=L and the witness geometry in step 2.1 place all such J in one fixed enlargement of $J_0$. Thus their total length is at most CL and ${\sum }_{J}m(G_J)\le C\delta L$. The use of 4j for the density witness is essential: the witness in 3 times the parent need not have length at most j.

We construct exact smooth scale projections for this plus subtree. For each dyadic r<=L, let ${\mathrm{\Omega }}_r=[{\alpha }_r,{\alpha }_r+1\mathrm{/}r)$ be the unique dyadic ancestor of the top frequency of length1/r. Set $\mathrm{\Theta }(v)=\psi ((v-1\mathrm{/}2)\mathrm{/}5)$. The exact plateau a=1/9 and support b=1/8 in F2 give $\mathrm{\Theta }=1$ on [0,1] and $\mathrm{\Theta }=0$ outside [-1/8,9/8]. Its inverse Fourier transform is $k(x)=5e^{\pi ix}\varphi (5x)$, as follows by substituting $v=5u+1\mathrm{/}2$ in the absolutely convergent inverse integral. Thus the inverse transform of $\mathrm{\Theta }(r(\xi -{\alpha }_r))$ is $K_r(x)=r^{-1}{e}^{2\pi i{\alpha }_{r}x}k(x\mathrm{/}r)$, bounded by $C{r}^{-1}(1+\mathrm{\mid }x\mathrm{\mid }\mathrm{/}r{)}^{-20}$. For any member with $\mathrm{\mid }{I}_s\mathrm{\mid }\ge r$, its Fourier support lies in ${\mathrm{\Omega }}_r$, so this multiplier is one on its support. If $\mathrm{\mid }{I}_s\mathrm{\mid }<r$, its full frequency is a proper ancestor of ${\mathrm{\Omega }}_r$ and its lower packet support is separated from its upper half by at least $1\mathrm{/}(8\mathrm{\mid }{I}_s\mathrm{\mid })\ge 1\mathrm{/}(4r)$. Since the top, hence ${\mathrm{\Omega }}_r$, lies in that upper half, the enlargement of ${\mathrm{\Omega }}_r$ by1/(8r) on each side misses that packet support. The multiplier is therefore zero there. By F6 and F7, $K_r\ast H={\sum }_{s\in U:\mathrm{\mid }{I}_s\mathrm{\mid }\ge r}{b}_s{\varphi }_s$ everywhere, not merely as an unspecified truncation estimate.

The small terms sum to at most $C\sigma \delta L$ over all J. For dist(J,J_0)<L this follows from the Cj bound and the total-length bound in step 2.1. For d=dist(J,J_0)>=L, each scale l<=L has total spatial length at most L, so its contribution to the spatial sum of step 3.1 is at most $L(1+d\mathrm{/}l{)}^{-20}$. Summing the dyadic scales gives $CL(L\mathrm{/}d{)}^{20}$. For $2^{k}L\le d<2^{k+1}L$, step 2.1 gives $j\ge 2^{k}L\mathrm{/}6$ and j<2^{k+2}L. The disjoint J in this shell lie in an interval of length $C{2}^{k}L$, so their number is bounded by an absolute constant. Summing $CL{2}^{-20k}$ for k>=0 completes the estimate. Countable sums of these nonnegative integrals are justified by F8 applied to finite unions of the disjoint J.

For each x, the upper frequency halves of a strict plus tree form a nested family, with all tiles at a fixed spatial scale having the same frequency. If N(x) belongs to an upper half, it belongs to all the larger upper halves in this family. Thus the activated spatial scales form an initial segment of the scales present in U. Let tau(x) be their largest spatial length, setting it to zero when none is active. Every strict member has length at most L/2, so 2tau(x)<=L whenever tau(x)>0. On J, if tau(x)>=8j, the large sum equals the scale block $8j\le \mathrm{\mid }{I}_s\mathrm{\mid }\le \tau (x)$, which by step 3.3 is $((K_{8j}-K_{2\tau (x)})\ast H)(x)$. If tau(x)<8j the large sum is zero. Both kernel scales are at least j. Step 1.2 therefore bounds the magnitude of the large sum, for x in E intersect J, by $C{\mathbf{1}}_{G_J}(x){\inf }_{y\in J}MH(y)$. Denote that infimum by q_J; it is an ordinary fixed nonnegative number, not a measurability assertion about uncountably selected points. By F4 and step 3.2, ${\sum }_{J}m(G_J)q_J^2\le C\delta {\sum }_{J}j{q}_J^2\le C\delta {\int }_{\mathbb{R}}(MH{)}^2\le C\delta \mathrm{\parallel }H{\mathrm{\parallel }}_2^2\mathrm{.}$ The middle inequality holds because $MH(y)\ge q_J$ on each disjoint J. Countable sums are again obtained from F8; alternatively the inequalities first hold for any finite subfamily.