TheoremBase

The cell integral of a periodic function is unchanged by every translation, so the integral of a shifted function does not depend on the shift; differentiating under the integral sign at zero gives the vanishing mean, and the product rule then gives integration by parts.

Proof

Each result cited below is universally quantified over the data appearing in its own statement, and is applied to the data named here. For a point x∈Rnx\in\mathbb{R}^{n}, an index i∈[n]i\in[n] and a real tt, write xβŠ•tx\oplus t for the point obtained from xx by replacing its iith coordinate by xi+tx_{i}+t; thus xβŠ•t=x+t ex\oplus t=x+t\,e, where ee is the point whose iith coordinate is 11 and whose other coordinates are 00. Note that βˆ₯(xβŠ•t)βˆ’xβˆ₯=∣t∣\lVert(x\oplus t)-x\rVert=|t| by claim 1 of Properties of the Euclidean Norm on Rn\mathbb{R}^n: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square.

Claim 1. Let g∈Cper1g\in C^{1}_{\mathrm{per}} and i∈[n]i\in[n]. A function of class C1C^{1} on Rn\mathbb{R}^{n} is continuous, by clause 2 of the calculus setting, so g∈Cperg\in C_{\mathrm{per}}. By Elementary Properties of Lattice-Periodic Functions (ii) we also have βˆ‚ig∈Cper\partial_{i}g\in C_{\mathrm{per}}, and by Elementary Properties of Lattice-Periodic Functions (iii) there are nonnegative reals M0M_{0} and M1M_{1} with ∣g(x)βˆ£β‰€M0|g(x)|\le M_{0} and βˆ£βˆ‚ig(x)βˆ£β‰€M1|\partial_{i}g(x)|\le M_{1} for every x∈Rnx\in\mathbb{R}^{n}.

Work on the measure space (Rn,B(Rn),Ξ»n)(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}),\lambda_{n}), and let UU be the open interval with endpoints βˆ’1-1 and 11. Define F:UΓ—Rnβ†’RF:U\times\mathbb{R}^{n}\to\mathbb{R} by

F(t,x)=1Q(x) g(xβŠ•t).F(t,x)=\mathbf{1}_{Q}(x)\,g(x\oplus t).

We verify the three hypotheses of Differentiation under the Integral Sign.

(i) Fix t∈Ut\in U. The map x↦xβŠ•tx\mapsto x\oplus t satisfies dE(xβŠ•t,yβŠ•t)=βˆ₯xβˆ’yβˆ₯=dE(x,y)d_{E}(x\oplus t,y\oplus t)=\lVert x-y\rVert=d_{E}(x,y) by claim 2 of Properties of the Euclidean Norm on Rn\mathbb{R}^n: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, so x↦g(xβŠ•t)x\mapsto g(x\oplus t) is continuous on Rn\mathbb{R}^{n}: given x1x^{1} and a positive Ξ΅\varepsilon, the Ξ΄\delta furnished by continuity of gg at x1βŠ•tx^{1}\oplus t serves in Continuous Map Between Metric Spaces. It is therefore measurable with respect to B(Rn)\mathcal{B}(\mathbb{R}^{n}) and B(R)\mathcal{B}(\mathbb{R}) by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets together with claim 5 there, and multiplying by 1Q\mathbf{1}_{Q} preserves measurability exactly as in the proof of Continuous Periodic Functions are Power-Integrable and Dense on the Torus (ii). Moreover ∣F(t,β‹…)βˆ£β‰€M01Q|F(t,\cdot)|\le M_{0}\mathbf{1}_{Q}, whose integral is M0Ξ»n(Q)=M0M_{0}\lambda_{n}(Q)=M_{0} by The Integral of an Indicator Function is the Measure of the Set, The Half-Open Unit Cell Tiles Euclidean Space (i) and claim 1 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; so F(t,β‹…)F(t,\cdot) is integrable by the criterion of Integrable Functions and Their Lebesgue Integral.

(ii) Fix x∈Rnx\in\mathbb{R}^{n}. If xβˆ‰Qx\notin Q then F( ⋅ ,x)F(\,\cdot\,,x) is identically 00 on UU, hence differentiable at every point of UU with derivative 00, since all its difference quotients vanish. If x∈Qx\in Q then F(t,x)=g(xβŠ•t)F(t,x)=g(x\oplus t), and for t∈Ut\in U and a nonzero real hh small enough that t+h∈Ut+h\in U we have (xβŠ•t)βŠ•h=xβŠ•(t+h)(x\oplus t)\oplus h=x\oplus(t+h), so the difference quotient

F(t+h,x)βˆ’F(t,x)h=g((xβŠ•t)βŠ•h)βˆ’g(xβŠ•t)h\frac{F(t+h,x)-F(t,x)}{h}=\frac{g\bigl((x\oplus t)\oplus h\bigr)-g(x\oplus t)}{h}

is exactly the difference quotient whose limit as hh tends to 00 defines the partial derivative βˆ‚ig(xβŠ•t)\partial_{i}g(x\oplus t), which exists because gg is of class C1C^{1} on Rn\mathbb{R}^{n}. Since every point of UU is an interior point of UU, this limit is also the derivative of F( ⋅ ,x)F(\,\cdot\,,x) at tt. In both cases D1F(t,x)=1Q(x)β€‰βˆ‚ig(xβŠ•t)D_{1}F(t,x)=\mathbf{1}_{Q}(x)\,\partial_{i}g(x\oplus t).

(iii) ∣D1F(t,x)βˆ£β‰€M11Q(x)|D_{1}F(t,x)|\le M_{1}\mathbf{1}_{Q}(x) for all t∈Ut\in U and x∈Rnx\in\mathbb{R}^{n}, and M11QM_{1}\mathbf{1}_{Q} is integrable as in (i).

By Differentiation under the Integral Sign the function Ξ¦(t)=∫RnF(t,x) dΞ»n(x)\Phi(t)=\int_{\mathbb{R}^{n}}F(t,x)\,d\lambda_{n}(x) is differentiable at every point of UU with

Ξ¦β€²(t)=∫Rn1Q(x)β€‰βˆ‚ig(xβŠ•t) dΞ»n(x),soΞ¦β€²(0)=∫Rn1Qβ€‰βˆ‚ig dΞ»n.\Phi'(t)=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}(x)\,\partial_{i}g(x\oplus t)\,d\lambda_{n}(x),\qquad\text{so}\qquad\Phi'(0)=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,\partial_{i}g\,d\lambda_{n}.

Ξ¦\Phi is constant on UU. Fix t∈Ut\in U and put h=t eh=t\,e, so that xβŠ•t=x+hx\oplus t=x+h. The map 1Qg\mathbf{1}_{Q}g is integrable by Continuous Periodic Functions are Power-Integrable and Dense on the Torus (ii), so by The Half-Open Unit Cell Tiles Euclidean Space (v), applied to the measurable Zn\mathbb{Z}^{n}-periodic map gg, the map 1Q+hg\mathbf{1}_{Q+h}g is integrable with

∫Rn1Q+h g dΞ»n=∫Rn1Q g dΞ»n=Ξ¦(0).\int_{\mathbb{R}^{n}}\mathbf{1}_{Q+h}\,g\,d\lambda_{n}=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,g\,d\lambda_{n}=\Phi(0).

For every x∈Rnx\in\mathbb{R}^{n} we have x+h∈Q+hx+h\in Q+h exactly when x∈Qx\in Q, so

(1Q+h g)(x+h)=1Q(x) g(x+h)=F(t,x).(\mathbf{1}_{Q+h}\,g)(x+h)=\mathbf{1}_{Q}(x)\,g(x+h)=F(t,x).

Claim 3 of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ, applied with a=ha=h to the integrable map 1Q+hg\mathbf{1}_{Q+h}g, therefore gives

Ξ¦(t)=∫Rn(1Q+hg)(x+h) dΞ»n(x)=∫Rn1Q+h g dΞ»n=Ξ¦(0).\Phi(t)=\int_{\mathbb{R}^{n}}(\mathbf{1}_{Q+h}g)(x+h)\,d\lambda_{n}(x)=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q+h}\,g\,d\lambda_{n}=\Phi(0).

Since Ξ¦\Phi is constant on UU, every difference quotient of Ξ¦\Phi at 00 is 00, so Ξ¦β€²(0)=0\Phi'(0)=0 by Derivative of a Real Function on an Interval. Hence ∫Rn1Qβ€‰βˆ‚ig dΞ»n=0\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,\partial_{i}g\,d\lambda_{n}=0, and Continuous Periodic Functions are Power-Integrable and Dense on the Torus (ii), applied to βˆ‚ig∈Cper\partial_{i}g\in C_{\mathrm{per}}, turns this into

∫Tnβˆ‚ig dx=∫Rn1Qβ€‰βˆ‚ig dΞ»n=0.\int_{\mathbb{T}^{n}}\partial_{i}g\,dx=\int_{\mathbb{R}^{n}}\mathbf{1}_{Q}\,\partial_{i}g\,d\lambda_{n}=0 .

Claim 2. Let g,h∈Cper1g,h\in C^{1}_{\mathrm{per}} and i∈[n]i\in[n]. The product ghgh lies in Cper1C^{1}_{\mathrm{per}} by Elementary Properties of Lattice-Periodic Functions (i), and by claim 1 of Constants, Affine Functions, Sums, Products and Quotients of Cᡏ Functions, applied on the open set Rn\mathbb{R}^{n},

βˆ‚i(gh)=(βˆ‚ig) h+g (βˆ‚ih)\partial_{i}(gh)=(\partial_{i}g)\,h+g\,(\partial_{i}h)

as maps on Rn\mathbb{R}^{n}. By Elementary Properties of Lattice-Periodic Functions (ii) the maps βˆ‚ig\partial_{i}g and βˆ‚ih\partial_{i}h lie in CperC_{\mathrm{per}}, so by Elementary Properties of Lattice-Periodic Functions (i) the products (βˆ‚ig)h(\partial_{i}g)h and g(βˆ‚ih)g(\partial_{i}h) lie in CperC_{\mathrm{per}}, and their restrictions to QQ lie in L1(Tn)\mathcal{L}^{1}(\mathbb{T}^{n}) by Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i). Restricting the displayed identity to QQ and integrating, claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

∫Tnβˆ‚i(gh) dx=∫Tn(βˆ‚ig) h dx+∫Tng (βˆ‚ih) dx.\int_{\mathbb{T}^{n}}\partial_{i}(gh)\,dx=\int_{\mathbb{T}^{n}}(\partial_{i}g)\,h\,dx+\int_{\mathbb{T}^{n}}g\,(\partial_{i}h)\,dx .

The left-hand side is 00 by claim 1 applied to ghgh, so the two integrals on the right are negatives of each other, which is the assertion.

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