Β· 21,706 chars Β· 40 deps Β· depth 26 Reason: First publication: local finiteness of the lattice sum, the resulting local representation as a finite sum of translates, and the regularity, mass bound, representation and joint measurability that follow from it.
At any point only finitely many lattice translates of the kernel are nonzero, and the same finite set works on a whole ball, so the periodised kernel is locally a finite sum of translates. Regularity, the mass bound, the integral representation and joint measurability all follow from that finite sum together with the tiling of space by the cell.
Proof
Each result cited below is universally quantified over the data appearing in its own statement, and is applied here to the data named in the statement of this lemma. We use throughout that dEβ(a,b)=β₯aβbβ₯ and that β£aiββ£β€β₯aβ₯ for every iβ[n], by claims 2 and 4 of Elementary Properties of the Euclidean Norm on Rn, and the triangle inequality of claim 6 of that lemma. We also use the following domination criterion: if f is measurable and β£fβ£β€g pointwise with g measurable, nonnegative and of finite integral, then β«β£fβ£β€β«g<β by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, so f is integrable by Integrable Function and the Lebesgue Integral, a measurable map being integrable exactly when the integral of its absolute value is finite.
Step 0. Bounded sets of lattice vectors are finite.
Consequently: (F) if Ξ is a nonnegative real number and TβZn is such that β£miββ£β€Ξ for every mβT and every iβ[n], then T is finite. Indeed by The Archimedean Property of the Real Numbers there is a natural number N with Ξβ€N, so TβGNβ, and subsets of finite sets are finite by claim 3 of Basic Properties of Finite Sets.
As this holds for j=1 and for j=2, the sums over F1β and over F2β agree. Thus Ξ¨(x,y) is well defined.
Step 3. Proof of claim 3.
Periodicity in the second variable. Fix x and let kβZn. For every m, Ο(xβ(y+k)βm)=Ο(xβyβ(k+m)), so mβS(x,y+k) exactly when k+mβS(x,y). Let F be finite with S(x,y)βF and put Fβ²={mβZn:k+mβF}, the image of F under mβ²β¦mβ²βk and hence finite by claim 4 of Basic Properties of Finite Sets; then S(x,y+k)βFβ². The map mβ¦k+m is a bijection of Fβ² onto F, so composing an enumeration of Fβ² with it gives an enumeration of F, and the corresponding terms coincide; by A Sum over a Finite Index Set Does Not Depend on the Enumeration, which says a sum over a finite index set may be computed from any enumeration, the two sums are equal. Hence Ξ¨(x,y+k)=Ξ¨(x,y), so yβ¦Ξ¨(x,y) is Zn-periodic.
For mβT0β, with a witness y, we have β₯xβy0ββmβ₯β€β₯xβyβmβ₯+β₯yβy0ββ₯β€R+1, so β£miββ£β€β₯xβy0ββ₯+R+1 for every i, exactly as in step 2; by (F) the set T0β is finite, hence T is finite by Peeling an Element off a Finite Set, and Unions of Finite Sets, and T is nonempty, so sums indexed by T are defined. For every y with β₯yβy0ββ₯β€1 we have S(x,y)βT, so by claim 2,
Continuity of Ξ¨(x,β ) at y0β follows, since the two maps agree on the open set B containing y0β and g is continuous at y0β. As y0β was arbitrary, Ξ¨(x,β ) is continuous on Rn.
For smoothness, suppose Ο is smooth, so that g is smooth. We first record a locality principle. Let UβRn be open and let Ο1β,Ο2β agree at every point of U. For y1ββU and iβ[n] there is, by Open Subset of a Metric Space, a positive real s with y1β+teiββU whenever β£tβ£<s; by Partial Derivative on a Euclidean Open Set the partial derivative at y1β is the limit as t tends to 0 of the quotients (Ο(y1β+teiβ)βΟ(y1β))/t, and for β£tβ£<s these quotients agree for Ο1β and Ο2β. Hence βiβΟ1β(y1β) exists exactly when βiβΟ2β(y1β) does, and then they are equal. So: if Ο1β and Ο2β agree on U then their first partial derivatives exist and agree at every point of U.
Now argue by induction on the natural number k for the following assertion P(k), quantified over all open sets UβRn and all pairs of maps: if Ο1β,Ο2β:RnβR agree at every point of U and Ο2β is of class Ck on U, then Ο1β is of class Ck on U, with the same iterated partial derivatives as Ο2β there. For k=1: the locality principle applied on U gives that each βiβΟ1β exists on U and equals βiβΟ2β there, and βiβΟ2β is continuous on U since Ο2β is of class C1; so Ο1β is of class C1 on U by C^k Maps on a Euclidean Open Set. For the step, suppose P(k), let U be open, and let Ο1β,Ο2β agree at every point of U with Ο2β of class Ck+1 on U. For each i the maps βiβΟ1β and βiβΟ2β exist and agree at every point of U, by the locality principle, and βiβΟ2β is of class Ck on U because Ο2β is of class Ck+1; so P(k), applied to U and to the pair βiβΟ1β, βiβΟ2β, gives that βiβΟ1β is of class Ck on U with the same iterated partial derivatives as βiβΟ2β. With the case k=1 this makes Ο1β of class Ck+1 on U, with the same iterated partial derivatives as Ο2β, by C^k Maps on a Euclidean Open Set. This proves P(k) for every natural number k. Since Ξ¨(x,β ) and g agree at every point of the open set B and g is smooth, hence of class Ck on B for every k, P(k) applied with U=B gives that Ξ¨(x,β ) is of class Ck on B for every natural number k, with the same iterated partial derivatives as g there.
Finally, C^k Maps on a Euclidean Open Set requires, for class Ck on Rn, the existence of the relevant iterated partial derivatives at every point of Rn and their continuity there, and continuity on Rn is continuity at each point. Each point y0β has an open ball B about it on which, by the induction just performed, those derivatives exist and are continuous; continuity at y0β relative to Rn follows from continuity relative to the open set B. Hence Ξ¨(x,β ) is of class Ck on Rn for every natural k, so smooth by Smooth Map on a Euclidean Open Set.
The symmetric statement. Fix y and put Ξ²mβ(x)=xβyβm, which satisfies Ξ²mβ(x)βΞ²mβ(xβ²)=xβxβ² and is therefore continuous, and smooth by the same argument. For kβZn we have Ο(x+kβyβm)=Ο(xβyβ(mβk)), so mβS(x+k,y) exactly when mβkβS(x,y), and the reindexing argument above applies with the bijection mβ¦mβk. The local representation and the continuity and smoothness arguments are unchanged with x in place of y. Hence xβ¦Ξ¨(x,y) lies in Cperβ, and in Cperββ when Ο is smooth.
By The Archimedean Property of the Real Numbers choose a natural number N with β₯xβ₯+R+1β€N. Let yβQ and mβS(x,y). Then 0β€yiβ<1 for every i, so β£yiββ£β€1, and as in step 2,
the last equality by claim 2 of Translation and Reflection Invariance of Lebesgue Measure on Rn, applied with a=x to the measurable nonnegative map β£Οβ£, whose reflection identity gives β«β£Ο(xβz)β£dΞ»nβ(z)=β«β£Οβ£dΞ»nβ. This proves the bound for Ξ¨xβ.
For the symmetric statement, fix y and choose N with β₯yβ₯+R+1β€N; the same computation applies with the roles of x and y exchanged, the relevant cells being Qβm for mβGNβ, which are again pairwise disjoint since Qβm=Q+(βm) and βm runs over lattice vectors.
where u~ is the periodic extension of u. The integrand vanishes off the union of the sets Q+m with mβGNβ: if z lies outside that union, let m be the unique lattice vector with zβmβQ, given by The Half-Open Unit Cell Tiles Euclidean Space Β§tiling; then mβ/GNβ, and if Ο(xβz)ξ =0 we would have mβS(x,zβm)βGNβ, a contradiction. Since the cells are pairwise disjoint, βmβGNββ1Q+mβ is the indicator of their union, so
Let gmβ:QβR be given by gmβ(y)=Ο(xβyβm)u(y), so that yβ¦h(y+m) is the extension of gmβ by the value 0 off Q, using u~=u on Q. Claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions identifies the integral of such a zero extension with the integral over Q against Ξ»Qβ, but is stated for maps into [0,β]; we therefore split gmβ into nonnegative parts. Put
By The Archimedean Property of the Real Numbers choose a natural number N with R+2β€N. For x,yβQ and mβS(x,y) we have β£xiββ£β€1 and β£yiββ£β€1, so as in step 4, β£miββ£β€β£xiββ£+β£yiββ£+Rβ€R+2β€N; hence S(x,y)βGNβ and