Comparing a short depth block with the same finite-depth constant

This chapter gives a close reading of Lemma 7.1 and Lemma 7.2 on pp. 52–57 of H, using the October 5, 2026 version.

The preceding chapter split the transfer errors for the two complementary arguments into far and near bands. Far bands yield a locally small error. Near bands still have to be estimated within a short depth block, where the inputs usually have only normalized square-integral bounds on terminal leaves. Such leafwise bounds do not automatically give the pointwise bounds we need, or even global $L^q$ bounds.

Two sampling steps reduce the quadrilinear sum over a short block to the original finite-depth constant $C_N(q)$. Between them, a shared quadratic randomization improves integrability in the fast variable, and only the single potentially heavy slot is clipped. We retain the explicit cost

Since $q>2$, this is no larger than the form allowed in the original paper, $C_q(1+C_N(q))(1+\log(K+2))^C(K+2)^{C(q-2)}$. The absolute power $11$ can therefore be fixed before choosing $q$. We have not replaced $C_N(q)$ by a constant independent of $N$. The later absorption argument still requires proof.

Why three direct approaches fall short

The following reconstruction explains the logic of the argument. It is not an account of how the authors actually discovered it.

A first natural approach is to upgrade each leafwise $L^2$ bound directly to an $L^q$ bound and then use $C_N(q)$. The simplest spike already rules this out: on a leaf of length one, take $z=\varepsilon^{-1/2}1_{[0,\varepsilon]}$. Then $\|z\|_2=1$, but $\|z\|_q=\varepsilon^{1/q-1/2}\to\infty$. Extra integrability must be produced, rather than inferred from the leaf average.

A second approach is to use an absolute estimate at each depth. This gives $CK|R|$, but its exponent does not become small as $q\downarrow2$. Even if all four slots are clipped first, crude interpolation from $L^2$ to $L^{p_j}$ leaves a product of clipping factors equal to $A^{\sum_j(1-2/p_j)}=A^2$, because $\sum_j1/p_j=1$. We therefore need to confine the potentially heavy input to at most one slot and obtain higher integrability for the other slots by a different argument.

A third approach is to randomize the frequencies independently in the four slots. This may improve each individual moment, but it generally destroys the original four-factor integral. The two linear constraints in (29) suggest looking for a shared phase whose total phase vanishes on their nullspace, while its single-slot fourth-moment expansion eliminates unpaired resonances. The quadratic phase in (28) does both.

The leaf-depth rectangles ensure that there is at most one heavy slot. Quadratic randomization supplies fourth-order integrability in the ordinary slots, leaving only the exception to be clipped. The two sampling steps place these operations on a fast torus and then return to the original kernel and the same $C_N(q)$. We now work through this argument.

1. The local form and the bounds not yet available

H, p. 4 · Coupled near-band errors in the preceding chapter

The slots are always indexed by $j=0,1,2,3$, with arithmetic-progression coordinates

We use $v$ for the real common difference, and after sampling use $u,v_1$ for the fast coordinates, to distinguish them from depth indices. For a dyadic cell $I$ of length $r_I$, set

The kernel is supported where all four $y_j$ lie in $I$ at distance at least $c_0r_I$ from its boundary. Its first derivatives are $O(r_I^{-1})$, and its higher derivatives satisfy the fixed bounds of the original kernel class. Its marginal integral in each slot direction vanishes. All constants may depend on this fixed kernel class.

An interval subscript denotes a normalized norm. For example,

Real-variable norms without an interval subscript use ordinary Lebesgue measure.

Let $M$ be the uncentered Hardy–Littlewood maximal operator on the real line:

Here $J$ ranges over bounded intervals. Fix $2<q<3$. On p. 4, H defines $C_N(q)$ to be the smallest constant for which

holds. Here $\mathcal I$ is any finite family of cells in a dyadic grid lying at at most $N$ consecutive depths, the kernels belong to the original class, and the inputs are compactly supported $L^q$ functions. The right-hand side may be infinite, although it will be finite in the application below.

At this stage we may use only $C_N(q)\lesssim N<\infty$. Indeed, the local absolute estimate gives

Integrate over the disjoint cells at each depth, then sum over at most $N$ depths. This does not give a depth-independent bound.

We also need the full range of the local absolute estimate: if $p_j\ge1$ and $\sum_j1/p_j\le2$, then

For any two distinct slots, the map $(x,v)\mapsto(y_i,y_j)$ has determinant $j-i\ne0$. This immediately gives the endpoint with two $L^1$ norms and two $L^\infty$ norms. Hölder's inequality for the positive integral supplies the remaining exponents. Section 7 gives the discrete-times-torus version used below. It does not depend on cancellation of the kernel.

2. Why the maximal estimate must preserve the norm order

H, p. 52 · The extrapolation theorem actually used

The norm that will arise is

The random variable $\xi$ occupies the middle level, and the fast variable $w$ the innermost level. There is generally no reason to interchange these three integrations, nor is there a common ordering of their exponents by size. We need an estimate in precisely this given order.

Let $(X,\mu)$ and $(W,\nu)$ be finite measure spaces. For $a>1$, a scalar weight $\omega\in A_a(\mathbb R)$ is defined by

where $\omega>0$ almost everywhere. The weighted norm replaces only the outermost $\mathrm dy$ by $\omega(y)\mathrm dy$. The weight is independent of $\xi,w$.

We use two classical inputs. For the first, we use the primary-source statement in Cruz-Uribe, Extrapolation and Factorization, Theorem 2.3, pp. 6–7. The scalar weighted maximal theorem states that for every $a>1$ and $\omega\in A_a$,

The second input is Cruz-Uribe–Martell–Pérez, Extensions of Rubio de Francia's extrapolation theorem, Theorem 3.1(a), published p. 206. We use its open-exponent range. In its formulation for nonnegative pairs, if one family $\mathcal F$ of nonnegative function pairs satisfies, for some $a_0>1$,

with a finite nondecreasing function controlling the dependence on the weight characteristic and a constant independent of the pair, then the corresponding estimate holds for every $a>1$ and $\omega\in A_a$. It does not require $U$ to be the image of $V$ under an operator. The conventions preceding this 2006 version allow the hypothesis and conclusion to be restricted to pairs with finite left-hand side. In the next section we first ensure this finiteness, then pass to a monotone limit. H cites Theorem 3.9 of the same authors' 2011 monograph. Its original book pages have not been directly verified here. The source appendix explains the correspondence and the quantitative dependence on the weight characteristic. These two results are external inputs, not theorems reproved in this chapter.

These inputs imply that for any $1<r,s,t<\infty$,

Here $M_y$ acts only in $y$, and the constant is independent of the numbers of atoms in the two measure spaces. We give both extrapolation steps next.

3. Two extrapolations without interchanging mixed norms

H, p. 52 · H, p. 53 · Explicit weight-characteristic dependence · Why interchanging norms can fail

First suppose that $g\ge0$ is bounded and supported, in $y$, on a bounded interval $J_0$. All the required left-hand sides are then finite: every inner norm is bounded by a fixed multiple of $M1_{J_0}$, which belongs to the appropriate weighted $L^a$ by (7). The total masses of the finite measure spaces are used only to establish finiteness. They do not enter the final inequality constant.

Fix $\xi$ and define

Apply (7) for each fixed $w$, then use Tonelli's theorem to obtain

As the admissible $g$ and $\xi$ vary, put all resulting $(U_\xi,V_\xi)$ into one family of pairs. The constant in (10) is independent of the pair, of $\xi$, and of the number of atoms. The first extrapolation therefore gives the estimate for every $a>1$ and every $A_a$ weight.

Take $a=s$ and integrate in $\xi$. With

we obtain, for every $\omega\in A_s$,

Extrapolate the new family of pairs $(U,V)$ a second time, now to $r$. This gives (9). The order $L_y^rL_\xi^sL_w^t$ has been preserved throughout. Tonelli is used only after matching powers have been expanded into nonnegative integrals.

For general $g\ge0$, approximate by $g_n=\min(g,n)1_{[-n,n]}(y)$. Then $g_n\uparrow g$. Monotone convergence on each interval, followed by the supremum over intervals, gives $M_yg_n\uparrow M_yg$. Apply monotone convergence at each level of integration as well. For complex-valued functions, first take absolute values.

Finally apply (9) to $|f|^q$ with

Whenever $p,b,h>q\ge1$, all three exponents are strictly greater than one. Direct calculation at each level gives

Hence

The constant may grow as an exponent approaches an endpoint. We need only that, once the exponents are fixed, it be independent of the tree depth, $K,N$, and the number of signs.

4. The precise short-block hypotheses

H, p. 53 · Why the eighth moment suffices

Let $\mathcal T$ be a finite dyadic tree fragment rooted at $R$. Complete it by adding all terminal leaves, so that the leaves partition $R$. Only internal cells enter the sum of forms. The internal cells lie at at most $K$ consecutive depths and remain within the original range defining $C_N(q)$, which has at most $N$ depths. Leaves lie within the block or one depth below its end. If there are no internal cells, the result is zero. Otherwise assume $K\ge1$.

In each slot there is a deterministic Hilbert-space-valued function $z_j$, supported on $R$, satisfying

The random scalar functions $z_j^\sigma$ are defined on one probability space and satisfy the pointwise moment bounds

The slots may be correlated. The constants are independent of the tree, $K,N$, and the vector dimension. This proof needs moments only up to order eight, all of which are covered by the original hypotheses.

For arbitrary measurable coefficients $|\epsilon_I(\sigma)|\le1$ and any subfamily of internal cells,

This is a version of H's equation (16) with an explicit power.

The short-block constant may also depend on the uniformly fixed eighth-moment bound in (14) and on the fixed kernel class. It is independent of the tree, the block length, the original depth range, the Hilbert-space dimension, and the number of random signs. The moment bound is fixed uniformly before choosing any tree, stack, or random scalarization. Lower moments are controlled by monotonicity of norms on a probability space.

If some $s_j=0$, (13) makes the corresponding deterministic input zero almost everywhere, and (14) does the same for its random scalarization, so the conclusion holds. Otherwise divide by each $s_j$ and henceforth assume $s_j=1$ for all slots. Every internal cell is a union of complete leaves, so we also have

The expected absolute value of the finite cell sum can be expressed as its expectation after multiplication by one common measurable complex number of modulus one. Absorb this number into every $\epsilon_I$. It then suffices to bound the absolute value of the expected signed sum. This common factor preserves the kernel class and does not require the $\epsilon_I$ to be independent.

5. Why leaf-depth quadruples are partitioned into rectangles

H, p. 54 · A concrete leaf-depth grouping

Let $k_j(y)$ be the depth of the leaf containing $y$. There are at most $K+1$ possible leaf depths. Place this integer interval inside a dyadic integer interval of length at most $4(K+1)$ and take its standard dyadic subdivision.

First suppose that at least two slots attain the minimum depth. Fix the minimum $t$ and the set $S$ of slots attaining it. These slots select leaves at depth exactly $t$, while the other slots select leaves deeper than $t$. By a fixed rule, choose from $S$ two slots $a,b$. This collection of product masks is called a rectangle.

If only slot $a$ attains the minimum, take the smallest dyadic integer interval containing both the minimum and the second-smallest depth. Denote its lower and upper halves by $L,U$. The minimum lies in $L$ and the second-smallest depth in $U$. Fix a nonempty set $S\subseteq\{0,1,2,3\}\setminus\{a\}$: its slots have depths in $U$, and all remaining slots have depths greater than $\max U$. Slot $a$ has depth in $L$. From $S$ choose $b$ by a fixed rule, and set $t=\min U$. This also defines a rectangle.

Each leaf-depth quadruple belongs to exactly one of these cases and positions, giving a disjoint partition. Group the rectangles by the length of their dyadic integer interval. There are at most $C(1+\log(K+2))$ such groups and a fixed number of slot assignments in each. For a fixed length and assignment, index the positions by $\iota=1,\ldots,J$, where $J\le C(K+1)$. Let $E_{j,\iota}$ be the union of all complete leaves selected in slot $j$ by that rectangle. Then

Indeed, at a fixed length the $L$ are disjoint, as are the $U$. The case of a nonunique minimum forms a separate group, with position indexed by $t$. The masks in the other two slots need not be disjoint.

After absorbing the common unit complex factor into the kernels, expand each progression integral using this finite disjoint partition, and then use $|\sum_{\text{rectangles}}\mathbb E S_{\text{rectangles}}|\le\sum_{\text{rectangles}}|\mathbb E S_{\text{rectangles}}|$. The partition by leaf depth does not depend on the random signs. Fix one rectangle from now on and suppress $\iota$. Except possibly in slot $a$, all selected leaves have depth at least $t$. These are the ordinary slots. In the unique-minimum case, slot $a$ is called the heavy slot.

This rectangle can contribute only to internal cells at depths $s<t$. When $s\ge t$, the nonunique-minimum case has a selected leaf at depth $t\le s$. In the unique-minimum case, slot $a$ has depth smaller still than $t\le s$. Dyadic nesting prevents such a terminal leaf from meeting the interior of an internal cell at depth $s$. The interior support margin of the kernel ensures that the small sampling shifts below do not reintroduce these contributions.

6. The respective roles of the slow lattice and the fast torus

H, p. 54

At depth $s$, denote the interval length by $r_s$. Choose a sufficiently small fixed dyadic number $\beta>0$, depending only on the kernel's support margin, and set

This scale is fixed throughout the current rectangle. It cannot change with the summation depth $s$. For the first two progression coordinates, write

Then

Here $n_j$ is the integer generated from the first two integer coordinates. It need not be the floor of $y_j/\delta$. Retaining the possibly negative values of $w_j$ is precisely what preserves the exact progression relation.

Use the fixed torus $\mathbb T_0=\mathbb R/(16\mathbb Z)$ with normalized Haar measure. Initially, the fast variables in slots $0,1$ are restricted to $[0,1)$. For slots $2,3$, fixed lift intervals containing $[-1,2]$ and $[-2,3]$, respectively, suffice. Both can be chosen as $[-4,4]$. On the corresponding intervals define

and set the functions to zero elsewhere on the torus. The lift intervals do not overlap themselves modulo the period. The first two factors restrict the original $u,v_1$ to $[0,1)$, so there is no wraparound ambiguity for the last two factors.

Replace $w_I(x,v)$ by its value at the slow coordinates

The first-derivative bound makes the kernel difference $O(\delta/r_s)$. The change of variables has Jacobian $\delta^2$, so each normalized form has discrete normalization factor $(\delta/r_s)^2$. Using normalized Haar measure introduces only the fixed factor $16^2$.

The sampled fast integral is

Its kernel coefficient depends only on the slow lattice point. The original marginal cancellations need not survive this replacement, so we cannot yet apply $C_N(q)$ to the sampled kernel.

7. How two-coordinate absolute estimates control both sampling errors

H, p. 54 · H, p. 55 · Deriving the full exponent range · The same mass calculation for both sampling steps

Fix a depth $s<t$ and an internal cell $I$. Every slow index that can enter the original kernel, the sampled kernel, or the later narrow-pulse comparison lies in an integer interval of length $O(r_s/\delta)$. In each slot use the local counting measure $(\delta/r_s)\sum_n$ and Haar measure in the fast variable. Their total masses are uniformly bounded.

For distinct slots $i,j$, the slow-index map

is injective on the integer lattice, though not necessarily surjective. On $\mathbb T_0^2$, the same integer matrix defines a surjective group homomorphism and therefore pushes normalized Haar measure forward to normalized Haar measure. The estimate with two $L^1$ norms and two $L^\infty$ norms consequently remains valid on the product of the local slow counting space and the fast torus. Bounded local mass and Hölder give the full exponent range of (4). In particular, we may use both the four-$L^2$ estimate and the $(1,3,3,3)$ estimate.

For all fast sampling windows associated with $I$, denote their deterministic square mass in slot $j$ by $M_{j,I}$, counting their bounded overlap. If $\beta$ is sufficiently small, the support margin places all these complete windows inside $I$. Thus (16) and disjointness of cells at each depth give

The second estimate also uses the fact that $E_j$ is a union of complete leaves, so its deterministic square mass is at most $|E_j|$.

The absolute $L^2$ estimate in all four slots, followed by four-factor Hölder in the shared variable $\sigma$, bounds one sampling error by

The fourth moments of the random local $L^2$ norms needed here follow from (14) and Minkowski. Independence between slots is not required.

For the two slots other than $a,b$, first use $M_{j,I}/|I|\le C$, then apply Cauchy–Schwarz over the cells at one depth. The depthwise sum of (24) is at most

Since $\delta/r_s=\beta2^{s-t}$,

the total sampling error for one rectangle is $C|E_a|^{1/2}|E_b|^{1/2}$. At a fixed interval length, (17) further gives

This is why each of the two sampling steps incurs only the logarithmic number of length groups, with no extra factor of $K$.

8. Why shared quadratic randomization preserves the fast progression integral

H, p. 55 · A counterexample for independent phases

Set

Here $c_j$ are the fixed third-difference coefficients, not the constants specifying the kernel support margin. Define the action on Fourier coefficients in the fast variable by

The Fourier basis is $\mathrm e(mw/16)$. Equip $\gamma$ with normalized Haar measure of period three. Every $U_{j,\gamma}$ is unitary on $L^2(\mathbb T_0)$. The four slots must use the same $\gamma$.

First expand (22) for Fourier polynomials. The surviving frequencies satisfy exactly

This two-dimensional nullspace is spanned by $(c_j)_j$ and $(jc_j)_j$, so we can write $m_j=c_j(h+jh')$. Since $\sum_j j^\ell c_j=0$ for $\ell=0,1,2$, we have

The total phase multiplying every surviving Fourier term is therefore one. The fast integral is unchanged for each $\gamma$.

For an individual slot, however, averaging gives the additional fourth-moment bound

Expand the fourth power. Integration in $w$ equates the frequency sums on the two sides, and integration in $\gamma$ equates their sums of squares. Two numbers with the same sum and the same sum of squares form the same unordered pair. The two possible pairings give $2(\sum_m|\widehat F(m)|^2)^2-\sum_m|\widehat F(m)|^4$, proving (31).

These identities initially hold for Fourier polynomials. The fast progression integral is continuous in the four $L^2$ norms, and (31) applies to differences as well. Approximation in $L^2$ therefore gives both invariance of the integral and the fourth-moment estimate for general $L^2$ inputs. This does not assert that, for fixed $\gamma$, the operator $U_{j,\gamma}$ maps $L^2$ into $L^4$. The fourth-order gain holds only after averaging in $\gamma$.

9. Why ordinary slots need no clipping

H, p. 55

Let $e_j(n)$ be the fast $L^2$ size of the deterministic input corresponding to (20): use $\|z_j\|1_{E_j}$ in place of $z_j^\sigma1_{E_j}$ before taking the fast norm. The pointwise moment bound and Minkowski give

Lower orders follow from monotonicity of norms on a probability space. Sampling windows have length $O(\delta)$, and adjacent windows have bounded overlap. Hence

In an ordinary slot, every selected leaf has length at most $r_t$. Of the disjoint leaves meeting a window of length $O(\delta)$, all but at most two crossing its endpoints are contained in the window. Their full lengths sum to at most $C\delta+2r_t$. Apply (13) on these complete leaves and divide by $\delta$ to obtain

Since $\beta$ has already been fixed by the kernel class, this bound is independent of $K$.

Even for the heavy slot, the slow indices associated with an internal cell $I$ satisfy the local total-mass bound

This follows because the complete windows lie in $I$, together with (16). The heavy slot does not satisfy the bound (34) at each slow index. These two types of control must be distinguished.

10. Clipping only the heavy slot and estimating the full error

H, p. 56 · A spike in the heavy slot

Write $G_j^{\gamma,\sigma}=U_{j,\gamma}F_j^\sigma$. If there is a heavy slot $a$, take

with value zero when $G_a=0$. This radial clipping satisfies $|G_a^A|\le A$ and

On the product of local slow counting measure and fast Haar measure, unitarity gives

Taking the fourth norm in $(\gamma,\sigma)$ requires, on the right, the eighth moment of $F_a^\sigma$ measured in local $L^2$. Equation (14), Minkowski, and (35) bound this by $C/A$.

For an ordinary slot, the local $L^3$ norm has bounded fourth random moment. Indeed, the local measure has bounded total mass, so

The second step uses (31) and (32). The last step uses (34) followed by (35).

Apply the local absolute estimate with exponents $(1,3,3,3)$ to the heavy-slot tail, then four-factor Hölder to the shared random factors. The expected error for a single cell is at most $C|I|/A$. Each rectangle has at most $K$ depths, and the cell lengths at any depth sum to at most $|R|$. At a fixed interval length there are at most $C(K+1)$ positions. The total clipping error for that length group is therefore at most

This count is deliberately crude. The power ten is fixed in advance and does not vary with $q$. When a heavy slot is present, we continue to denote its clipped function by $G_a$.

11. Returning to continuous inputs by narrow pulses

H, p. 56 · Pulse area and normalization

Now, for fixed $\gamma,\sigma,u,v_1$, allowing $u,v_1$ to range over the entire fast torus, define

where $0<c<1/20$ is fixed. The function actually inserted in slot $j$ is $g_j(y,w_j)$. Each slow index corresponds to a narrow pulse, and neighboring pulses are disjoint.

If a real progression meets all four pulses, write $y_j=\delta(n_j+a_j)$ with $|a_j|<c$. The progression relation requires

The right-hand side has absolute value at most $2jc\le6c<1$. The left-hand side is an integer, so it must be zero. Thus the slow centers still form an exact integer arithmetic progression.

For fixed slow centers, the admissible $(a_0,a_1)$ form the fixed polygon

The two middle coordinates are convex combinations of the endpoint coordinates, so this polygon equals $\{|a_0|<c,|-2a_0+3a_1|<c\}$ and has area exactly $\kappa_c=4c^2/3>0$. The continuous change of variables has Jacobian $\delta^2$. After replacing the continuous kernel by its value at the slow center, the integral is exactly $\kappa_c\delta^2$ times the corresponding four-factor product at that lattice point. Integrating over the fast torus recovers the sampled form (22), up to a fixed normalization factor.

This kernel replacement also has error $O(\delta/r_s)$. Although Fourier evolution changes the fast support, it introduces no new slow indices. The original complete windows associated with those indices still lie in $I$. For each slow index, unitarity and clipping ensure that the transformed fast $L^2$ norm is no greater than that of the original $F_j^\sigma$. The same $M_{j,I}$ and the same absolute $L^2$ argument from Section 7 therefore apply. Equations (25)–(27) control the second sampling error as well.

We now have the original continuous kernels $w_I$ and the original summation cells, so (3) applies legitimately. Only the error comparisons use sampled kernels. All four marginal cancellations of the original kernel remain unchanged. The main term for the current rectangle is consequently bounded by a fixed multiple of

There are only finitely many slow indices. For almost every parameter, $g_j$ is a finite sum of pulses of finite height, hence a compactly supported $L^q$ function. The fact that the original $z_j$ lie only in $L^2$ does not prevent us from applying $C_N(q)$ here.

12. The order and exponents of the three Hölder steps

H, p. 56 · H, p. 57 · A concrete set of open-range exponents

Set

These four exponents satisfy

Indeed, $p_*-q=q-2>0$, and for the other two slots,

If there is a heavy slot, take $h_a=p_*$. In every ordinary slot take $h_j=4$. Then $h_j>q$ in every slot, the random exponent satisfies $4>q$, and $\sum_j1/h_j\le2$.

In (42), first fix $y,\gamma,\sigma$ and apply the exponent range from Section 7 to the fast progression integral:

Next apply four-factor Hölder in the shared variables $(\gamma,\sigma)$, and finally Hölder in $y$ with exponents $p_j$. This gives

The order of these three steps cannot be reversed. In particular, the random fourth norm has not been moved outside the real-variable norm.

Now (12), with exactly the same norm order, gives

All extrapolation exponents lie strictly in the open range. The constant $C_q$ may depend on $q$, but no dependence on $K,N$ or the number of signs has been introduced.

13. Computing the mixed norms in ordinary and heavy slots

H, p. 57

For ordinary slots, $h_j=4$. By (31) and (32), for every $n$,

The pulse supports are disjoint, and each pulse has length $2c\delta$. Therefore

The second step uses $p_j>2$ and the ordinary-slot bound (34), without interchanging mixed norms.

In the heavy slot use the clipped $G_a$. For each $\gamma,\sigma,n$,

To take the fourth random norm on the right, we need $F_a^\sigma$ to have a moment of order $8/p_*$ for its local $L^2$ norm. Since $2<p_*<4$, we have $2<8/p_*<4$, so (32) applies. Thus

Only now take the norm in $y$ with exponent $p_*$:

If there is no heavy slot, we may harmlessly multiply the upper bound by $A^{1-2/p_*}\ge1$ to keep a uniform formula.

14. How the two designated masks leave a vanishing exponent

H, p. 57 · Why the position factor cannot be dropped

Insert (47) and (49) into (46). The main term for the current rectangle is at most

The masks in the other two slots lie in $R$ and contribute at most $|R|^{1-2/p_*}$. At a fixed interval length and slot assignment, (17) and three-factor Hölder give

The third Hölder exponent is $(1-2/p_*)^{-1}>1$. We need only that each of the two designated mask families is disjoint. We cannot assume disjointness for all four families.

Using $J\le C(K+1)$ and $A=(K+2)^{10}$, the main term for this length group is at most

By (27), each sampling error contributes $C|R|$. By (39), clipping contributes $C|R|$. Sum over $C(1+\log(K+2))$ length groups and a fixed number of slot assignments to obtain (15). Restoring $s_j$ completes the proof.

15. What is fixed first, and what may then grow

H, p. 6 · H, p. 53 · H, p. 57 · What still requires later proof

The comparison holds for every fixed $2<q<3$. We need not yet know which $q$ will ultimately allow absorption. The parameter order is as follows. First fix the kernel class, the torus length, the value of $\beta$ required by the support margin, the pulse width $c$, the clipping power ten, and the additional power one from summing positions. These produce the absolute power eleven. Then choose $q$ and fix all exponents in (43) and all mixed-norm constants. Also fix in advance a uniform eighth-moment bound in (14). Every short-block constant may depend on this moment bound and the original kernel class. Only afterward do we allow arbitrary finite $K,N$, finite trees, and arbitrary random stacks satisfying the fixed hypotheses.

In particular,

so $A^{\vartheta(q)}$ contributes the power $10\vartheta(q)$, and the position sum contributes another $\vartheta(q)$. We cannot combine them into $(K+2)^{C_q(q-2)}$ with unspecified constant dependence and then claim that the exponent tends to zero. What we actually have is an absolute power.