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Dimension threshold for the infinite-lattice massless Gaussian Green integral

Statement

Assume the Axiom of Countable Choice ${A C}_{ω}$ , meaning every sequence $(X_{n})_{n \in N}$ of nonempty sets has a function $f$ with $f (n) \in X_{n}$ for every $n$ . Let integer $d \geq 1$ , parameter $κ > 0$ , and $k = (k_{1}, \dots, k_{d}) \in [- π, π]^{d}$ , and put

$D_{d} (k) : = 4 \sum_{j = 1}^{d} \sin^{2} (k_{j} / 2) .$

Here $λ_{d}$ is $d$ -dimensional Lebesgue measure restricted to the cube. For a measurable function $h : [- π, π]^{d} \to R$ , write $h^{+} = \max {h, 0}$ and $h^{-} = \max {- h, 0}$ ; its signed Lebesgue integral is $\int h^{+} d λ_{d} - \int h^{-} d λ_{d}$ when at most one of these two integrals is infinite, and $h$ is integrable when $\int ∣ h ∣ d λ_{d} < + \infty$ . For $x = (x_{1}, \dots, x_{d}) \in Z^{d}$ , define $k \cdot x : = \sum_{j = 1}^{d} k_{j} x_{j}$ and $∣ k ∣_{2}^{2} : = \sum_{j = 1}^{d} k_{j}^{2}$ . Define the massless infinite-lattice Green integral when it is finite by

$G_{d, κ} (x) : = (2 π)^{- d} \int_{[- π, π]^{d}} \frac{\cos (k \cdot x)}{κ D_{d} (k)} d λ_{d} (k),$

where the integrand is assigned value $0$ at $k = 0$ , a choice that does not affect the integral because a singleton has Lebesgue measure zero. If $d > 2$ , this integral is absolutely finite for every fixed $x \in Z^{d}$ . If $1 \leq d \leq 2$ , the nonnegative diagonal integral with $x = 0$ is $+ \infty$ . Thus the dimension threshold concerns this free Gaussian Fourier integral; it does not assert an interacting critical exponent or a continuum scaling limit.

Statement references

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Finite periodic Gaussian lattice field with positive mass

Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$

The nonnegative Lebesgue integral

Integrable real and complex functions, and their integrals

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

Facts & Assumptions

Given: ${A C}_{ω}$ , an integer $d \geq 1$ , $κ > 0$ , $x \in Z^{d}$ , the cube $[- π, π]^{d}$ with Euclidean Lebesgue measure, and $D_{d}$ as in the Statement.

[F1]

For $∣ u ∣ \leq π / 2$ , $2 ∣ u ∣ / π \leq ∣ \sin u ∣ \leq ∣ u ∣$ (Chord and Lipschitz bounds for sine).

[F2]

Under ${A C}_{ω}$ , boxes are Lebesgue measurable with measure equal to the product of their side lengths, and Lebesgue measure is complete and countably additive (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F3]

The nonnegative Lebesgue integral is monotone and additive for finite sums; increasing nonnegative measurable functions pass to their limit under the integral (Monotonicity and nonnegative homogeneity of the nonnegative integral, Additivity of the nonnegative Lebesgue integral, Monotone convergence for the integral).

[F4]

A continuous real function is Borel measurable, and a geometric series with ratio in $(0, 1)$ has finite sum (A continuous map has Borel preimages of Borel sets, For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[F5]

$∣ \cos t ∣ \leq 1$ for every real $t$ (Signs, monotonicity intervals, and ranges of sine and cosine).

[F6]

The real signed Lebesgue integral and the criterion $\int ∣ h ∣ d λ_{d} < + \infty$ for finite integrability are as stated in the real-integral definition linked above; the nonnegative integrals used in the comparison are those of The nonnegative Lebesgue integral.

Proof

1.1givenF1F4algebra

For $k \neq 0$ , apply [F1] to $u = k_{j} / 2$ in each coordinate and square the bounds. Summing yields $\frac{4}{π^{2}} ∣ k ∣_{2}^{2} \leq D_{d} (k) \leq ∣ k ∣_{2}^{2},$ where $∣ k ∣_{2}^{2} = \sum_{j} k_{j}^{2}$ . The denominator is positive off the origin, so its reciprocal is continuous on the cube with the origin removed. Extending it by the stated value at the closed singleton ${0}$ makes it Borel measurable: each open-set preimage is the union of a Borel subset of that complement and, when applicable, the singleton. The cosine numerator is continuous, so all nonnegative comparison integrands below are measurable by [F4].

2.1givenF2F3F4step 1.1algebra

For $j \geq 0$ , let $Q_{j} = [- 2^{- j}, 2^{- j}]^{d} ∖ [- 2^{- (j + 1)}, 2^{- (j + 1)}]^{d}$ . The sets are pairwise disjoint, are contained in $[- 1, 1]^{d}$ , and by [F2] have measure $(2^{d} - 1) 2^{- j d}$ . On $Q_{j}$ , $∣ k ∣_{2} \geq ∥ k ∥_{\infty} > 2^{- (j + 1)}$ , so step 1.1 gives $(κ D_{d} (k))^{- 1} \leq (π^{2} / (4 κ)) 2^{2 j + 2}$ . Hence $\int_{Q_{j}} (κ D_{d} (k))^{- 1} d λ_{d} (k) \leq \frac{π^{2}}{κ} (2^{d} - 1) 2^{(2 - d) j} .$ For $d > 2$ , these upper bounds sum to a finite geometric series by [F4]. On the remaining part of $[- π, π]^{d}$ outside $[- 1, 1]^{d}$ , $∣ k ∣_{2} \geq 1$ , so step 1.1 bounds $(κ D_{d})^{- 1}$ by $π^{2} / (4 κ)$ ; that part has finite measure by [F2]. Countable additivity and monotone convergence in [F3] therefore show that $(κ D_{d})^{- 1}$ is integrable on the whole cube.

2.2givenF3step 1.1algebra

For $1 \leq d \leq 2$ , step 1.1 gives $(κ D_{d} (k))^{- 1} \geq (κ ∣ k ∣_{2}^{2})^{- 1}$ on every $Q_{j}$ . There $∣ k ∣_{2} \leq \sqrt{d} 2^{- j}$ , and hence $\int_{Q_{j}} (κ D_{d} (k))^{- 1} d λ_{d} (k) \geq \frac{1}{κ d} (2^{d} - 1) 2^{(2 - d) j} .$ The right side is bounded below by the positive constant $(2^{d} - 1) / (κ d)$ for all $j \geq 0$ when $d \leq 2$ . Since the $Q_{j}$ are disjoint, finite additivity in [F3] shows that the integral over their first $M + 1$ layers is at least $(M + 1) (2^{d} - 1) / (κ d)$ ; letting $M \to \infty$ and using monotonicity gives $\int (κ D_{d})^{- 1} = + \infty$ .

3.1givenF5F6step 2.1step 2.2algebra∎

If $d > 2$ , then $∣ \cos (k \cdot x) ∣ (κ D_{d} (k))^{- 1} \leq (κ D_{d} (k))^{- 1}$ away from the origin by [F5], and step 2.1 proves $\int ∣ h_{x} ∣ d λ_{d} < + \infty$ ; hence the signed integral in the Statement is finite by [F6], and multiplication by the finite constant $(2 π)^{- d}$ preserves it. If $x = 0$ and $d \leq 2$ , the cosine is $1$ , so step 2.2 gives the stated infinite nonnegative diagonal integral.

Dimension threshold for the infinite-lattice massless Gaussian Green integral

Statement

Statement references

Facts & Assumptions

Proof

Depends on

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