The Gaussian short-distance comparison · Part 2

In particular pairing independently transformed stencils with their inverse turns gives the same continuum two-point form for every pair. The relative endpoint positions are kept fixed. Bilinearity expresses the finite-torus covariance as

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c2
S
X
s,t
ordinarys,t(h; x, y; As, Bt).

Similarly expanding the two factors in the wave integral (PT18), then interchanging its finite sums and integral, expresses as the same double average of the infinite-lattice shell forms. The integrability already established justifies this interchange. The separated chart limit gives the same free pairing for every summand when tends to a nonzero physical vector. The return chart limit says that, at equal endpoints, each ordinary form minus its shell form tends to zero. Taking the finite double average proves (PT21) and the separated limit.

The reporting schedule meets both chart-limit hypotheses. With an integer side 1, its spacing and period are

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hm = L−m,
Nm = 4Lm2+m+1,
hmNm = 4Lm2+1.
(PT23)

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In particular Nmhq
m →∞for every fixed integer q, and 4 | Nm. Center the observed positions
at the quarter-period point. If their centered physical coordinates converge, their centered integer
coordinates are eventually less than Nm/4 in absolute value, since hmNm →∞.
Hence every
fixed stencil shift of their difference is represented without wrapping and has the required physical
limit. A sum of the two positions lies at twice the quarter-period center plus a bounded physical
displacement. Its lift has at least one coordinate of absolute value at least Nm/4. Its physical
distance is consequently at least hmNm/4 →∞.
This puts the inversion image outside every
bounded physical region, including when the two observed positions coincide. It supplies the return
limit used for the subtraction.

Finally apply (PT20) by induction on the number of occurrences. Covariance convergence (PT21) gives convergence of each moment to the recursion with zero within-insertion pairings and Maxwell pairings between distinct insertions. This is precisely the free Wick rule for reduced insertion polynomials. Finite polynomial expansion proves convergence of every moving row. The compactness and cell-integration argument in (PT15)–(PT16) then proves convergence of their profile integrals. This completes the Gaussian constant-coefficient step needed in the leading perturbative comparison.

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The reduction of the chart forms to scalar lattice kernels is also explicit. An elementary stencil
a is a finite signed sum of shifted edge readings, with coefficients at, offsets ut, a vector axis, and
an ordered derivative list Da. Its Fourier polynomial factors as a sign, a translation character, and
the product of the difference multipliers in Da. Thus, for two such stencils,
X
Y
atbsχp(z + ut ± vs) = σa,b,±χp(z + ua,± + ub,±)
qν(p).
(PT24)
t,s
ν∈DaDb

This follows by multiplying their two finite Fourier polynomials and using The prescribed offsets and signs are precisely those in the elementary stencil factorization, so no limiting replacement occurs in this identity. Multiplying by the torus inverse-Laplacian Fourier weight and summing over identifies their covariance with the corresponding differentiated scalar torus kernel, multiplied by and the same sign.

For the infinite-lattice expression, sum the sine and cosine channels first. The elementary identity cos() cos() + sin() sin() = cos(

shows that the sum over channels is zero unless both vector axes and colors match; when they match it is the cosine of the offset difference. Integrating against (2 gives exactly the infinite-cover scalar Green function at that difference. The bound cos1 and integrability of justify every such integral.

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A curvature coordinate is a difference of two elementary edge stencils.
Its two-point form
consequently has four terms, with signs +, −, −, +. Both its finite-torus covariance and its infinite-
cover subtraction have exactly this four-term expansion. At separated positions, (PT24) reduces
each torus term to the scalar differentiated-kernel limit. At inversion-image positions it reduces
each term to the scalar far-distance limit, which is zero.
At coincident positions subtract the
infinite-cover value before rescaling. The torus-to-cover return estimate makes this difference tend
to zero even after multiplication by any fixed inverse power of h, because Nmhq
m →∞for every
fixed q. These are respectively the separated, far-image, and return scalar estimates needed by the
chart argument.

Finally elementary coordinate readings span the finite-dimensional dual of every raw jet-response space: these responses are arrays of color components, vector-index pairs and derivative-index words, and coordinate projections span the dual of a finite array space. Expand each of two arbitrary dual vectors as a finite linear combination of these readings. Bilinearity reduces their covariance convergence to the elementary curvature calculations. This establishes the passage from elementary stencils to arbitrary chart functionals used in (PT22).

15.1.4 Scalar return estimate with its rescaling rate

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The scalar subtraction at a fixed integer displacement has a useful quantitative proof. Rescale the
Fourier cube to Q = [−1
2, 1
2]4, put d(x) = δ(2πx), and let

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fa(x) = d(x)−1 cos(2πa · x),
fa(0) = 0.

The value assigned at zero agrees with the zero-mode convention for the inverse Laplacian and does not affect its integral. The elementary lower bound for the lattice symbol implies

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|fa(x)| ≤∥x∥−2
(x ∈Q).

Let) be the cell representative in the torus Fourier quadrature. Outside cell boundaries, which have measure zero,, and, when. These imply, hence

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|fa(vN(x)) −fa(x)| ≤26∥x∥−2.

, and, then2. The numerator cosine and the symbol have bounded derivatives on a ball containing. Subtracting two quotients therefore gives

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|fa(y) −fa(x)| ≤Car−4∥y −x∥
(∥x∥, ∥y∥≥r),

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and with r/2 in place of r,

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|fa(vN(x)) −fa(x)| ≤AaN−1r−4
(∥x∥> r).

For example the quotient subtraction is

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(cosa(y) −cosa(x))d(y)−1 + (d(x) −d(y))d(y)−1d(x)−1 cosa(x),

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which proves the asserted bound directly from the Lipschitz bounds and d−1 ≤r−2.
The singular part is integrable. In four dimensions the change of variables x = ru gives
Z
∥x∥<2r
∥x∥−2 dx = r2
Z
∥u∥<2
∥u∥−2 du.

Split the quadrature error at radius, use the bound 26 inside and the uniform estimate outside, and obtain Re (PT25)

Here the torus Fourier sum is exactly the integral of over: every representative cell has volume. The infinite-cover value is exactly the integral of over. Even ensures the chosen symmetric representatives tile this same cube. For any fixed integer0, choose. Along the schedule, eventually1 and 8 Multiplying (PT25) by gives

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h−d |Re GN(a) −G∞(a)| ≤Cahd+2 + Aa(Nh5d+4)−1 −→0.
(PT26)

This proves the return estimate with every fixed rescaling power that occurs in a finite jet calculation.

15.1.5 Recovering the whole separated scalar kernel from annuli

Annular convergence alone determines differences of cutoffs and can leave an additive ambiguity. The small-cutoffbound and the large-cutofftail bound below fix that ambiguity. The comparison chooses the small cutofffirst, then a finite number of dyadic annuli, and only then the lattice index. No uniform convergence in a moving cutoffis assumed.

The separated Fourier limit uses a cutoffcomparison that also fixes the otherwise undetermined constant between annular limits. Here is the complete comparison principle. Suppose cutofffunctions), a continuum cutoff), and a candidate limit satisfy, for every

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Am(2s) −Am(s) −→A(2s) −A(s).

Suppose there are constants0, independent of, such that eventually in, for each fixed

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|Am(s)| ≤Csp,
|Fm −Am(s)| ≤B/s,
|A(s)| ≤Esp,
|F −A(s)| ≤D/s,
(PT27)

where1. Then Indeed choose 01 so that (3, and next an integer so that ( 3. Finite telescoping of the annular convergence gives

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Am(2qs) −Am(s) −→A(2qs) −A(s).

For sufficiently large its error is less than3. Adding the two large-cutofferrors, this difference error, and the two small-cutoffsizes bounds by. The order of choices is, then, then ; no uniform convergence in the cutoffparameter is assumed.

Apply this principle to the differentiated torus Green kernel at a nonzero limiting physical displacement, with) its smooth Fourier cutoff,) the corresponding continuum cutoff, and +2. The cutoffkernel is bounded by. Its complement is bounded by, uniformly

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when the physical displacement is bounded away from zero. The Fourier mesh condition required
for this estimate,
(2π/N)(|D| + 4) ≤hs,
hs ≤1
2,

holds eventually for each fixed, sinceand0. The annular Riemann sum converges to the continuum annular integral. Thus (PT27) proves the separated scalar limit used in (PT24). The continuum candidate in this argument is the ordinary derivative), with . One can identify it while retaining an integrable heat regulator 0: inverse Fourier transformation of the regulated differentiated inverse-square symbol gives

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∂DG(z) −∂D

G(z)e−π2∥z∥2/b
(z ̸= 0).

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The Fourier convention in this formula has character e2πik·z. To remove heat take b = v−2, v →∞.
The second term equals
v|D|+2∂D

Ge−π2∥·∥2
(vz).

Every derivative of the fixed function on the right decays faster than any inverse power at infinity, so this term tends to zero for= 0. At a fixed Fourier cutoff, dominated convergence also removes heat under the integral. The annular tail bound survives this limit. Summing its dyadic estimates ) yields at most 2, supplying the continuum half of (PT27).

For a physical displacement whose norm tends to infinity, the same comparison principle applies with) = 0. The annular summationby-parts estimate is

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2π
∥h lift(x)∥
|D|+3
.
| annulush,s,D(x)| ≤CD
s

For each fixed this tends to zero. The small-cutoffsize and tail estimates remain valid once the physical distance is at least one. This proves the far-image scalar limit. The uniform low-frequency bound in (PT27) can be proved without oscillation. In normalized momentum coordinates), put. The lattice denominator is

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3
X
ν=0
q2
ν sinc(tqν/2)2.
dt(q) =

On2, the inequality sinc( for1 gives

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dt(q) ≥(2/π)2∥q∥2.

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Each normalized difference numerator has modulus at most |qν| ≤1, because |eiu −1| ≤|u|. The
normalized cutoffsymbol for any derivative list is therefore bounded by (π/2)2∥q∥−2 on its unit-ball
support, independently of the length of that list.
For centered integer momentum lifts, let Bm contain those whose largest coordinate in absolute
value is at most m. Injectivity of the lift gives |Bm| ≤(2m + 1)4. On Bm \ B⌊m/2⌋, the inverse
squared Euclidean norm is at most 4/m2. Strong induction now gives
X

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p∈Bm
∥lift(p)∥−2 ≤432m2,

where the zero summand is assigned zero. Indeed the inner contribution is at most 432( = , and the outer contribution is at most (3; the case = 0 is zero.

Write. Rescaling this last inequality bounds the inverse-square sum on . The physical normalization of the cutoffsum is

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(Nh)−4s|D|−2.

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Since Ts = 2π/(Nh), multiplication gives

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| pieceh,s,D(x)| ≤432(π/2)2
(2π)4
s|D|+2.
(PT28)

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This is the first estimate in (PT27), uniformly also at coincident positions.
For the oscillatory estimate, repeated discrete integration by parts is exact on the finite torus.
If ∆−eν denotes a momentum difference, translation of the summation index gives
X

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p
χp(x)∆v
−eνf(p) = (χeν(x) −1)v X
p
χp(x)f(p).
(PT29)

The normalized annular symbol is smooth, supported where2, and all its derivatives through any prescribed finite order are bounded uniformly for 01. This follows from its cutoffaway from the denominator’s zero and the smooth sinc extension at = 0. A momentum difference has step size in. Repeated use of the fundamental theorem of calculus bounds its-th difference by1, the difference vanishes for 3. There are at most 8 momentum representatives in this enlarged support: each integer coordinate has absolute value at most and

These statements remain valid across a representative boundary. The periodic uncut symbol agrees on the two lifts; the annular cutoffis zero on both whenever a shift crosses that boundary, provided2. For the complementary high-frequency symbol, which has no compact support in, one instead uses its uniform derivative bound on the rescaled Fourier box when2. All representatives then contribute, but, so the same bound results. Combining these facts with (PT29) gives

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|χeν(x) −1|v
≤CD,vT v−4.
X
p
χp(x)f(p)

Choose a coordinate whose physical displacement has absolute value at least half its Euclidean norm. For centered lifts, the chord inequality for the circle implies

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|χeν(x) −1| ≥∥h lift(x)∥/(Nh).

Set + 3 and restore the physical normalization. Using) gives

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2π
∥h lift(x)∥
v
.
(PT30)
| annulush,s,D(x)| ≤CD,v
(2π)4s

The same bound holds for the high-frequency complement when 1

To bound the entire tail choose the first integer for which Then Decompose the tail into the annuli at and the final high-frequency complement. At a physical distance at least 0, (PT30) bounds the annular sum by 2 and the complement by. Hence jointpiece

This proves the torus tail estimate used in (PT27) and shows explicitly why an absolute bound on the uncut infinite-volume derivative sum is unnecessary.

The annular quadrature limit used above follows from a fixed-support dominated-convergence argument. Fix 0 and write, an integer for the reporting schedule. Physical Fourier sample points have spacing . The normalized annular integrand is jointly continuous in its lattice parameter, its physical displacement, and its Fourier variable, including It vanishes outside a fixed ball of radius 2) and is uniformly bounded there when 0 Choose an integer containing that support. Eventually 2, because this inequality is 1. Thus every integer Fourier address in the supporting box is its own centered torus

representative. The annular torus sum is exactly the unrestricted grid quadrature over this box. Let) be the grid representative of is the annular integrand, define the step function)). Its integral is the quadrature, since each cell has volume . The step functions converge pointwise to the continuum annular integrand by joint continuity, 0, convergence of the physical displacement, and. They are bounded by a common constant times the indicator of the ball of radius + 5: a cell whose representative lies in the support is within distance four of it. Dominated convergence proves convergence of their integrals. The exact phase identity underlying this representation is

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e2πi⟨hm lift(xm),lift(p)/cm⟩= χp(xm).

The normalized radial variable is also exactly the lattice radial variable, since. This proves the annular Riemann-sum statement with moving displacements. For clarity the continuum oscillatory bound can be obtained uniformly in the heat parameter. Let) be the normalized continuum annular symbol and

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HD,a(q) = e−a∥q∥2MD(q),
a ≥0.

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Its support is contained in the compact annulus 1/2 ≤∥q∥≤3. For each fixed derivative order
j, the derivatives through order j are bounded by a common constant times the indicator of this
annulus, uniformly in a ≥0. To prove the heat-factor part, put y = √a q. Its i-th derivative is
ai/2 times the i-th derivative of e−∥y∥2. When ∥y∥≤1, the lower bound ∥q∥≥1/2 implies √a ≤2,
so compact derivative bounds suffice. When ∥y∥≥1, use √a ≤2∥y∥and the rapid decay of the
Gaussian derivatives. Leibniz’s rule and bounded derivatives of MD give the stated common bound.
Integrating by parts in the continuum Fourier integral now gives, for every integer j ≥0,

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∥x∥j
Z
e2πik·xHD,a(k) dk
 ≤CD,j,
a ≥0.

For example select a coordinate with2, integrate times in that coordinate, and bound the resulting derivative in There are no boundary terms because the integrand is smooth and compactly supported. Rescaling in the continuum annular integral contributes ; taking + 3 leaves a factor. The rescaled heat parameter remains nonnegative, so the bound is uniform during heat removal. This proves the continuum annular bound and the summable dyadic tail asserted after (PT27).