Further calculations and boundary examples for the short-block comparison

This appendix accompanies the main short-block argument. It expands three derivations that are easy to compress into a sentence, then uses concrete examples to explain the roles of the objects involved. Apart from explicitly identified external inputs, these are direct calculations for H Lemmas 7.1–7.2. The boundary examples refute overly strong intermediate claims, not H's short-block comparison.

A. How two-coordinate endpoints fill the exponent range

H, p. 4 · H, p. 54 · H, p. 55 · Return to the absolute estimate

Consider a nonnegative quadrilinear integral $\mathcal A(f_0,f_1,f_2,f_3)$. It may be a progression integral on a bounded normalized interval, or the integral on local counting measure times a torus from Section 7 of the main text. Suppose there is a uniform endpoint bound whenever at most two slots use $L^1$ and the rest use $L^\infty$. The endpoints with zero or one $L^1$ slot follow from the two-$L^1$ endpoints and bounded total local mass.

Write $r_j=1/p_j$. The polytope

has precisely the zero-one vectors with at most two ones as its vertices. To see this directly, if at least two coordinates are strictly between zero and one, they can be varied in opposite directions while keeping their sum fixed, so the point is not a vertex. If exactly one coordinate is nonintegral and the sum is not two, that coordinate can vary on its own. If the sum is two, the other integral coordinates force the remaining coordinate to be integral as well. Thus $r=\sum_\ell\lambda_\ell v^{(\ell)}$, where $\lambda_\ell\ge0$, $\sum_\ell\lambda_\ell=1$, and each $v^{(\ell)}$ has at most two ones.

First suppose all $p_j<\infty$. Since

Hölder applied to the underlying nonnegative integral, followed by the appropriate endpoint bound in each factor, gives

For slots with $p_j=\infty$, simply factor out the supremum norm. No multilinear interpolation theorem with unspecified constant dependence is needed.

The only additional fact at the discrete endpoint is injectivity of the integer coordinate map. Its image may have finite index, but enlarging a nonnegative sum to all coordinate pairs incurs no factor of that index. On the torus, the same integer matrix may be many-to-one, but it is a surjective group homomorphism. Its pushforward of normalized Haar measure is exactly Haar measure. Again, there is no factor growing with the number of lattice points.

B. From pointwise moments to moments of local square norms

H, p. 53 · H, p. 55 · H, p. 56 · H, p. 57 · Precise short-block hypotheses

Let $(S,\lambda)$ be any measure space and let $a\ge2$. Minkowski applied to the nonnegative function $|F|^2$ gives

Hence $\|\|F^\sigma\|_{L^2(S)}\|_{L_\sigma^a}\le C_a\|Z\|_{L^2(S)}$.

For the first sampling error, take $a=4$, apply (A2) in each slot, and then use four-factor Hölder on the probability space. For the heavy-slot tail, $\|G-G^A\|_1\le A^{-1}\|F\|_2^2$, so its fourth random moment requires the eighth moment of the original $L^2$ size. Thus take $a=8$. The ordinary-slot Fourier fourth-moment bound uses $a=4$. Finally, the mixed norm in the heavy slot uses $a=8/p_*\in(2,4)$. The pointwise eighth-moment bound and monotonicity of probability norms therefore suffice throughout.

When a stack is randomized by one or two independent Rademacher families, the moment constants can be dimension-independent. For real coefficients, first use

Exponential Markov with the optimal $t$ gives a two-sided Gaussian tail bound. Integrating $\mathbb E|X|^a=a\int_0^\infty u^{a-1}\mathbb P(|X|>u)\,\mathrm du$ then yields

For complex coefficients, split into real and imaginary parts, changing only $C_a$. If $X=\sum_{m,n}\varepsilon_m\varepsilon'_na_{mn}$, first apply (A3) in $\varepsilon$, then the finite-sum version of (A2) in $\varepsilon'$, to obtain

Lower moments follow from the second-moment bound. This proves that the at most two independent Rademacher randomizations mentioned in the original paper have the required moment bounds. The four slots need not be independent. Independence is used only for the sign families within each specific randomization.

C. One mass calculation for both sampling errors

H, p. 54 · H, p. 55 · H, p. 56 · Return to continuous inputs

Let $B_j$ be the fixed lift interval for slot $j$. For an internal cell $I$, let $S_{j,I}$ contain all slow indices that can meet the support of the original kernel, its slow-center version, or the narrow-pulse kernel. Each index corresponds to a complete window $\delta(n+B_j)$. The original support margin is at least $c_0r_I$, while every shift has size at most a fixed multiple of $\delta$. Since $s<t$ and $\delta/r_I\le\beta/2$, choosing $\beta$ sufficiently small places all these windows inside $I$.

One explicit choice of the mass is

If $B_j$ in a slot is larger than the actual support, keep the corresponding zero extension. Every $B_j$ has fixed length, so each real point is counted only a bounded number of times in (A5). Containment of the complete windows in $I$ gives

The fast Haar normalization and slow counting normalization satisfy

The first sampling difference has the additional factor $\delta/r_I$. The four $L^2$ norms in (A1) account for the remaining normalization. Equations (A2) and (A7) therefore give (24) in the main text.

For the second sampling difference, the fast functions have become $U_{j,\gamma}F_j^\sigma$, with clipping in the heavy slot. But for each $n$ the fast $L^2$ norm does not increase, for any $\gamma$. The right-hand side of (A7) remains an upper bound. Integration over the pulse-offset polygon adds only a fixed area factor. Even when the kernel difference depends on the offset, it is still bounded by $C\delta/r_I$. Applying (A1) and (A2) again gives exactly the same estimate (24).

The normalization constants can also be written out. With $T_0=16$, the first sampled form is

The factor $T_0^2$ exactly cancels the denominators from the two normalized Haar measures. On returning to narrow pulses, first fix the kernel at the center and then integrate over the fast variables. This gives

Since $a_1=(a_3+2a_0)/3$, the offset polygon has area $\kappa_c=4c^2/3$. Hence $\operatorname{Samp}_I(G)=(T_0^2/\kappa_c)\operatorname{BurstCenter}_I(G)$. This fixed factor is independent of $K,N,q$. Multiplying the second centering error by it preserves the preceding bound.

It would be incorrect to say that Fourier unitarity preserves support: it does not preserve support in the fast variable. What is preserved is the set of slow indices and the fast square norm at each index. The mass $M_{j,I}$ uses the complete windows before transformation.

D. Concrete exponents and a complete example

H, p. 53 · H, p. 54 · H, p. 55 · H, p. 56 · H, p. 57 · The rectangle partition

Take $q=5/2$. The exponents in the main text are

If there is a heavy slot, its fast exponent is three and all other fast exponents are four. Every random exponent is four. The three exponents actually used for the maximal operator are, respectively,

All are strictly greater than one, but no single increasing ordering of exponents works in every slot. The actual order is always real variable, random variable, fast variable.

Take $R=[0,1]$, with internal cells at depths zero and one and all four terminal leaves at depth two. Let $z_j(y)=1_R(y)\mathrm e(y)$, with no additional sign randomization. The normalized square norm on every leaf is exactly one. Choose any summation subfamily of these internal cells and any original kernels.

Every contributing leaf-depth quadruple is $(2,2,2,2)$. There is therefore just one rectangle, in the nonunique-minimum case, with $t=2$. We may designate $a=0,b=1$. All four masks equal $R$, and there is no heavy slot. Take $\delta=\beta/4$. The slow indices are finite in number. Formula (20) retains the actual sampled values on the fast intervals for slots zero and one, and uses the prescribed lift intervals in the other slots. For each kernel replacement, the depth factors are $\delta/r_0=\beta/4$ and $\delta/r_1=\beta/2$, summing to $3\beta/4$.

Shared quadratic randomization preserves the fast four-factor integral along each slow progression. Since each leaf has length exactly $r_t$, ordinary slots satisfy $e_j(n)\le C_\beta$. All four slots use the averaged fourth-moment estimate directly, without clipping. After returning to narrow-pulse inputs, the corresponding ordinary-slot exponents in (A8) allow the mixed-norm maximal estimate. Each $|E_j|=1$, so each mixed-norm contribution is bounded by a fixed constant. In this example we therefore obtain

without the position power or the logarithmic number of length groups from the general bound.

The form in this example need not vanish. On a sufficiently small open set where all four coordinates lie in the interior of $I$, choose a compactly supported nonnegative Gevrey function $\phi$, and take the kernel to be a fixed normalized multiple of $\prod_{j=0}^3(\partial_v-j\partial_x)\phi(x,v)$. Each marginal integral vanishes because of its corresponding directional derivative. The product of the four input phases is $\mathrm e(4x+6v)$, and integration by parts gives the four frequency factors $6,2,-2,-6$, all nonzero. Taking the support of $\phi$ sufficiently small ensures that the corresponding Fourier integral is nonzero. After fixed-scale normalization, the kernel belongs to a fixed admissible class. The point is not to compute its numerical value, but to show that the full short-block argument can operate on a nonzero form.

Now consider a depth quadruple that genuinely singles out a heavy slot: $(1,2,5,6)$. In the standard dyadic subdivision of $[0,8)$, the smallest common interval containing the minimum one and the second-smallest value two is $[0,4)$. Its lower half is $[0,2)$ and its upper half is $[2,4)$. Hence $a=0,b=1,t=2$, and only slot zero can be heavy. The other two slots are restricted to depths greater than three. This combinatorial example explains the rectangle masks. It does not assert that every abstract leaf-depth quadruple is realized by a contributing real progression in an arbitrary prescribed tree.

E. Eight limits of the argument

H, p. 52 · H, p. 53 · H, p. 54 · H, p. 55 · H, p. 56 · H, p. 57

E1. Leafwise square averages do not control higher averages

H, p. 53

On a leaf of length one, take $f_\varepsilon=\varepsilon^{-1/2}1_{[0,\varepsilon]}$. Then $\|f_\varepsilon\|_2=1$, whereas for fixed $q>2$ we have $\|f_\varepsilon\|_q=\varepsilon^{1/q-1/2}\to\infty$. Thus the original inputs cannot be passed directly to $C_N(q)$ before sampling and randomization.

E2. Interchanging mixed norms can introduce a dimension factor

H, p. 52 · H, p. 53

Let $I_1,\ldots,I_m$ be disjoint unit intervals on the real line, equip $X=\{1,\ldots,m\}$ with counting measure, and set $f(y,i)=1_{I_i}(y)$. Then

When $p\ne b$, no two-sided comparison with constants independent of $m$ can interchange these norms. This does not refute Minkowski's inequality in its correct exponent direction. It refutes an interchange made without checking that direction.

E3. The outer exponent cannot be taken at the maximal endpoint

H, p. 52 · H, p. 53

Take $f=1_{[0,1]}$. For $y>1$ we have $Mf(y)\ge1/y$, hence $\int Mf=\infty$. Thus $M_q$ is not strongly bounded on $L^q$. Setting $p=q$ in (12) already fails without any inner variables. This example does not claim to rule out every other inner endpoint by the same argument.

E4. A heavy slot cannot be treated as ordinary

H, p. 55 · H, p. 56

Inside a heavy leaf of length $\ell$, take a small interval of length $\delta$, denoted $J$, and set $z=(\ell/\delta)^{1/2}1_J$. Its normalized square norm on the whole leaf is one. But a fast window covering $J$ can have $e(n)^2$ of order $\ell/\delta$. This has no uniform bound when the heavy leaf is much longer than the sampling scale. The local total-mass bound (35) still holds, so the remedy is to clip just this slot.

E5. Four independent quadratic phases do not preserve the integral

H, p. 55

On the fast torus, take the four pure Fourier modes $F_j(w)=\mathrm e(c_jw/16)$. The frequency vector $m=c=(-1,3,-3,1)$ satisfies both progression constraints, so the original fast integral is one. With a shared $\gamma$, the additional total phase is $\mathrm e(\gamma\sum_jc_j)=1$. If each slot instead uses an independent $\gamma_j$, the averaged phase is $\prod_j\mathbb E_{\gamma_j}\mathrm e(\gamma_jc_j)=0$. The original integral has been erased, so this independent randomization is not an equivalent transformation.

E6. A fixed $\gamma$ does not give the averaged fourth-moment gain

H, p. 55

Take $\gamma=0$. Then $U_{j,0}$ is the identity. For tall narrow spikes on the torus, the $L^4/L^2$ ratio can be arbitrarily large. The average in (31) cannot be removed.

E7. Disjointness of two masks does not remove the remaining position factor

H, p. 57

Take $J$ designated masks of equal measure in each family, with $|E_{a,\iota}|=|E_{b,\iota}|=|R|/J$. The left-hand side of (51) in the main text is exactly

With only these two disjointness conditions and $p_*>2$, the factor $J^{1-2/p_*}$ is therefore necessary for this summation step. This does not prove optimality of H's final $K$ exponent. It explains why that factor cannot simply be discarded at this step.

E8. Finiteness is not uniform boundedness, and a vanishing exponent is not zero

H, p. 4 · H, p. 57

The sequence $a_N=N$ is finite for every finite $N$ but has no uniform upper bound. The same quantifier distinction applies to the bound $C_N(q)<\infty$ available here. Moreover, for any fixed $q>2$, the quantity $(K+2)^{11(q-2)/(q-1)}$ still grows with $K$. It can be compared with a separate negative-power gain. The statement that the exponent tends to zero as $q\downarrow2$ does not mean that the $K$ loss has already disappeared.

F. What still requires later arguments

H, p. 53 · H, p. 57 · Parameters and conclusion scope

The starting point is a fixed deterministic stack normalized on every leaf, together with its random scalarization. This chapter does not construct fixed stacks from sides that vary within a block, construct complete helper side bases, prove the square energy on the linear side, or establish a global estimate for the coupled near-band sums at power one. Choosing one common $q$, fixing its constants, and making the coefficient of $C_N(q)$ absorbable still require the later parameter choices in H.

In the later proof, each type of side input must still be checked against (13)–(14) of the main text: does the common stopping rule give normalization on leaves, does randomization act on the complete fixed stack, and is the original coupled sum preserved? The calculations in this appendix do not replace those later arguments.