ScalingStacks

Proof. [02X5]

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Proof.

In view of Definition 7.1 both formulae in the above statement are equivalent and so it is enough to prove the second one. In case u=0u=0, we have Δ⁡(u)={Δ}\Delta(u)=\{\Delta\} and formula (7.4) holds because

∫Δf(n)​(⟨0,x⟩)​d​voln=vol⁡(Δ)​f(n)​(0)=∑k≥0Ck​(Δ,0,Δ)​f(k)​(0),\int_{\Delta}f^{(n)}(\langle 0,x\rangle)\,\text{\rm d}\operatorname{vol}_{n}=\operatorname{vol}(\Delta)f^{(n)}(0)=\sum_{k\geq 0}C_{k}(\Delta,0,\Delta)f^{(k)}(0),

We prove (7.4) by induction on the dimension nn. In case n=0n=0, we have u=0u=0 and so the verification reduces to the above one. Hence, we assume n≥1n\geq 1 and u≠0u\neq 0. For short, we write d​x=d​x1∧⋯∧d​xn\,\text{\rm d}x=\,\text{\rm d}x_{1}\wedge\dots\wedge\,\text{\rm d}x_{n}. Choose any vector v∈ℝnv\in\mathbb{R}^{n} of norm 11 and such that ⟨u,v⟩≠0\langle u,v\rangle\not=0. Performing an orientation-preserving orthonormal change of variables, we may assume v=(1,0,…,0)v=(1,0,\dots,0). We have

f(n)​(⟨u,x⟩)​d​x=1⟨u,v⟩​d​(f(n−1)​(⟨u,x⟩)​d​x2∧⋯∧d​xn).f^{(n)}(\langle u,x\rangle)\,\text{\rm d}x=\frac{1}{\langle u,v\rangle}\,\text{\rm d}\left(f^{(n-1)}(\langle u,x\rangle)\,\text{\rm d}x_{2}\wedge\dots\wedge\,\text{\rm d}x_{n}\right).

With Stokes’ theorem, we obtain

(7.5) ∫Δf(n)​(⟨u,x⟩)​d​voln\displaystyle\int_{\Delta}f^{(n)}(\langle u,x\rangle)\,\text{\rm d}\operatorname{vol}_{n} =∫Δf(n)​(⟨u,x⟩)​d​x\displaystyle=\int_{\Delta}f^{(n)}(\langle u,x\rangle)\,\text{\rm d}x
=1⟨u,v⟩​∑F∫Ff(n−1)​(⟨u,x⟩)​d​x2∧⋯∧d​xn.\displaystyle=\frac{1}{\langle u,v\rangle}\sum_{F}\int_{F}f^{(n-1)}(\langle u,x\rangle)\,\text{\rm d}x_{2}\wedge\dots\wedge\,\text{\rm d}x_{n}.

where the sum is over the facets FF of Δ\Delta, and we equip each facet with the induced orientation.

For each facet FF of Δ\Delta, we let ιuF​(d​x)\iota_{u_{F}}(\,\text{\rm d}x) be the differential form of order n−1n-1 obtained by contracting d​x\,\text{\rm d}x with the vector uFu_{F}. The form d​x2∧⋯∧d​xn\,\text{\rm d}x_{2}\wedge\dots\wedge\,\text{\rm d}x_{n} is invariant under translations and its restriction to the linear hyperplane LFL_{F} coincides with ⟨uF,v⟩​ιuF​(d​x)\langle u_{F},v\rangle\iota_{u_{F}}(\,\text{\rm d}x). Therefore,

∫Ff(n−1)​(⟨u,x⟩)​d​x2∧⋯∧d​xn=⟨uF,v⟩​∫F−mFf(n−1)​(⟨u,x+mF⟩)​ιuF​(d​x).\int_{F}f^{(n-1)}(\langle u,x\rangle)\,\text{\rm d}x_{2}\wedge\dots\wedge\,\text{\rm d}x_{n}=\langle u_{F},v\rangle\int_{F-m_{F}}f^{(n-1)}(\langle u,x+m_{F}\rangle)\iota_{u_{F}}(\,\text{\rm d}x).

Let voln−1\operatorname{vol}_{n-1} denote the Lebesgue measure on LFL_{F}. We can verify that voln−1\operatorname{vol}_{n-1} coincides with the measure induced by integration of −ιuF​(d​x)-\iota_{u_{F}}(\,\text{\rm d}x) along LFL_{F}. Let g:ℝ→ℝg\colon\mathbb{R}\to\mathbb{R} be the function defined as g⁡(z)=f⁡(z+⟨u,mF⟩)g(z)=f(z+\langle u,m_{F}\rangle). Then f(n−1)​(⟨u,x+mF⟩)=g(n−1)​(⟨πF​(u),x⟩)f^{(n-1)}(\langle u,x+m_{F}\rangle)=g^{(n-1)}(\langle\pi_{F}(u),x\rangle) for all x∈LFx\in L_{F}. Hence,

∫F−mFf(n−1)(⟨u,x+mF⟩)ιuF(dx)=−∫F−mFg(n−1)(⟨πF(u),x⟩)dvoln−1.\int_{F-m_{F}}f^{(n-1)}(\langle u,x+m_{F}\rangle)\iota_{u_{F}}(\,\text{\rm d}x)=-\int_{F-m_{F}}g^{(n-1)}(\langle\pi_{F}(u),x\rangle)\,\text{\rm d}\operatorname{vol}_{n-1}.

Applying the inductive hypothesis to FF and the function gg we obtain

∫Fg(n−1)​(⟨πF​(u),x⟩)​d​voln−1\displaystyle\int_{F}g^{(n-1)}(\langle\pi_{F}(u),x\rangle)\,\text{\rm d}\operatorname{vol}_{n-1} =∑V′∈F⁡(πF​(u))∑k≥0Ck​(F,πF​(u),V′)​g(k)​(⟨πF​(u),V′⟩)\displaystyle=\sum_{V^{\prime}\in F(\pi_{F}(u))}\sum_{k\geq 0}C_{k}(F,\pi_{F}(u),V^{\prime})g^{(k)}(\langle\pi_{F}(u),V^{\prime}\rangle)
=∑V′∈F⁡(πF​(u))∑k≥0Ck​(F,πF​(u),V′)​f(k)​(⟨u,V′⟩).\displaystyle=\sum_{V^{\prime}\in F(\pi_{F}(u))}\sum_{k\geq 0}C_{k}(F,\pi_{F}(u),V^{\prime})f^{(k)}(\langle u,V^{\prime}\rangle).

Each aggregate V′∈F⁡(πF​(u))V^{\prime}\in F(\pi_{F}(u)) is contained in a unique V∈Δ⁡(u)V\in\Delta(u) and it coincides with V∩FV\cap F. Therefore, we can transform the right-hand side of the last equality in

∑V∈Δ⁡(u)∑k≥0Ck​(F,πF​(u),V∩F)​f(k)​(⟨u,V⟩),\sum_{V\in\Delta(u)}\sum_{k\geq 0}C_{k}(F,\pi_{F}(u),V\cap F)f^{(k)}(\langle u,V\rangle),

where, for simplicity, we have set Ck​(F,πF​(u),V∩F)=0C_{k}(F,\pi_{F}(u),V\cap F)=0 whenever V∩F=∅V\cap F=\emptyset. Plugging the resulting expression into (7.5) and exchanging the summations on VV and FF, we obtain that ∫Δf(n)​(⟨x,u⟩)​d​voln\int_{\Delta}f^{(n)}(\langle x,u\rangle)\,\text{\rm d}\operatorname{vol}_{n} is equal to

(7.6) ∑V∈Δ⁡(u)∑k≥0(−∑F⟨uF,v⟩⟨u,v⟩Ck(F,πF(u),V∩F)f(k)(⟨u,V⟩)).\sum_{V\in\Delta(u)}\sum_{k\geq 0}\bigg(-\sum_{F}\frac{\langle u_{F},v\rangle}{\langle u,v\rangle}C_{k}(F,\pi_{F}(u),V\cap F)f^{(k)}(\langle u,V\rangle)\bigg).

Specialising this identity to v=uv=u, we readily derive formula (7.4) from Definition 7.1 of the coefficients Ck​(Δ,u,V)C_{k}(\Delta,u,V).

For the last statement, observe that the values f(k)​(⟨u,V⟩)f^{(k)}(\langle u,V\rangle) can be arbitrarily chosen. Hence, the coefficients Ck​(Δ,u,V)C_{k}(\Delta,u,V) are uniquely determined from the linear system obtained from the identity (7.4) for enough functions ff. ∎

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