Peetre theorem

Peetre theorem In mathematics, the (linear) Peetre theorem, named after Jaak Peetre, is a result of functional analysis that gives a characterisation of differential operators in terms of their effect on generalized function spaces, and without mentioning differentiation in explicit terms. The Peetre theorem is an example of a finite order theorem in which a function or a functor, defined in a very general way, can in fact be shown to be a polynomial because of some extraneous condition or symmetry imposed upon it.

This article treats two forms of the Peetre theorem. The first is the original version which, although quite useful in its own right, is actually too general for most applications.

Contents 1 The original Peetre theorem 1.1 Proof 2 A specialized application 3 Example: Laplacian 4 References The original Peetre theorem Let M be a smooth manifold and let E and F be two vector bundles on M. Let {displaystyle Gamma ^{infty }(E), {hbox{and}} Gamma ^{infty }(F)} be the spaces of smooth sections of E and F. An operator {displaystyle D:Gamma ^{infty }(E)rightarrow Gamma ^{infty }(F)} is a morphism of sheaves which is linear on sections such that the support of D is non-increasing: supp Ds ⊆ supp s for every smooth section s of E. The original Peetre theorem asserts that, for every point p in M, there is a neighborhood U of p and an integer k (depending on U) such that D is a differential operator of order k over U. This means that D factors through a linear mapping iD from the k-jet of sections of E into the space of smooth sections of F: {displaystyle D=i_{D}circ j^{k}} where {displaystyle j^{k}:Gamma ^{infty }Erightarrow J^{k}E} is the k-jet operator and {displaystyle i_{D}:J^{k}Erightarrow F} is a linear mapping of vector bundles.

Proof The problem is invariant under local diffeomorphism, so it is sufficient to prove it when M is an open set in Rn and E and F are trivial bundles. At this point, it relies primarily on two lemmas: Lemma 1. If the hypotheses of the theorem are satisfied, then for every x∈M and C > 0, there exists a neighborhood V of x and a positive integer k such that for any y∈V{x} and for any section s of E whose k-jet vanishes at y (jks(y)=0), we have |Ds(y)|0. Let ρ(x) denote a standard bump function for the unit ball at the origin: a smooth real-valued function which is equal to 1 on B1/2(0), which vanishes to infinite order on the boundary of the unit ball. Consider every other section s2k. At x2k, these satisfy j2ks2k(x2k)=0. Suppose that 2k is given. Then, since these functions are smooth and each satisfy j2k(s2k)(x2k)=0, it is possible to specify a smaller ball B′δ(x2k) such that the higher order derivatives obey the following estimate: {displaystyle sum _{|alpha |leq k} sup _{yin B'_{delta }(x_{2k})}|nabla ^{alpha }s_{k}(y)|leq {frac {1}{M_{k}}}left({frac {delta }{2}}right)^{k}} where {displaystyle M_{k}=sum _{|alpha |leq k}sup |nabla ^{alpha }rho |.} Now {displaystyle rho _{2k}(y):=rho left({frac {y-x_{2k}}{delta }}right)} is a standard bump function supported in B′δ(x2k), and the derivative of the product s2kρ2k is bounded in such a way that {displaystyle max _{|alpha |leq k} sup _{yin B'_{delta }(x_{2k})}|nabla ^{alpha }(rho _{2k}s_{2k})|leq 2^{-k}.} As a result, because the following series and all of the partial sums of its derivatives converge uniformly {displaystyle q(y)=sum _{k=1}^{infty }rho _{2k}(y)s_{2k}(y),} q(y) is a smooth function on all of V. We now observe that since s2k and {displaystyle rho } 2ks2k are equal in a neighborhood of x2k, {displaystyle lim _{krightarrow infty }|Dq(x_{2k})|geq C} So by continuity |Dq(x)|≥ C>0. On the other hand, {displaystyle lim _{krightarrow infty }Dq(x_{2k+1})=0} since Dq(x2k+1)=0 because q is identically zero in B2k+1 and D is support non-increasing. So Dq(x)=0. This is a contradiction.

We now prove Lemma 2.

First, let us dispense with the constant C from the first lemma. We show that, under the same hypotheses as Lemma 1, |Ds(y)|=0. Pick a y in V{x} so that jks(y)=0 but |Ds(y)|=g>0. Rescale s by a factor of 2C/g. Then if g is non-zero, by the linearity of D, |Ds(y)|=2C>C, which is impossible by Lemma 1. This proves the theorem in the punctured neighborhood V{x}. Now, we must continue the differential operator to the central point x in the punctured neighborhood. D is a linear differential operator with smooth coefficients. Furthermore, it sends germs of smooth functions to germs of smooth functions at x as well. Thus the coefficients of D are also smooth at x. A specialized application Let M be a compact smooth manifold (possibly with boundary), and E and F be finite dimensional vector bundles on M. Let {displaystyle Gamma ^{infty }(E)} be the collection of smooth sections of E. An operator {displaystyle D:Gamma ^{infty }(E)rightarrow Gamma ^{infty }(F)} is a smooth function (of Fréchet manifolds) which is linear on the fibres and respects the base point on M: {displaystyle pi circ D_{p}=p.} The Peetre theorem asserts that for each operator D, there exists an integer k such that D is a differential operator of order k. Specifically, we can decompose {displaystyle D=i_{D}circ j^{k}} where {displaystyle i_{D}} is a mapping from the jets of sections of E to the bundle F. See also intrinsic differential operators.

Example: Laplacian Consider the following operator: {displaystyle (Lf)(x_{0})=lim _{rto 0}{frac {2d}{r^{2}}}{frac {1}{|S_{r}|}}int _{S_{r}}(f(x)-f(x_{0}))dx} where {displaystyle fin C^{infty }(mathbb {R} ^{d})} and {displaystyle S_{r}} is the sphere centered at {displaystyle x_{0}} with radius {displaystyle r} . This is in fact the Laplacian. We show will show {displaystyle L} is a differential operator by Peetre's theorem. The main idea is that since {displaystyle Lf(x_{0})} is defined only in terms of {displaystyle f} 's behavior near {displaystyle x_{0}} , it is local in nature; in particular, if {displaystyle f} is locally zero, so is {displaystyle Lf} , and hence the support cannot grow.

The technical proof goes as follows.

Let {displaystyle M=mathbb {R} ^{d}} and {displaystyle E} and {displaystyle F} be the rank {displaystyle 1} trivial bundles.

Then {displaystyle Gamma ^{infty }(E)} and {displaystyle Gamma ^{infty }(F)} are simply the space {displaystyle C^{infty }(mathbb {R} ^{d})} of smooth functions on {displaystyle mathbb {R} ^{d}} . As a sheaf, {displaystyle {mathcal {F}}(U)} is the set of smooth functions on the open set {displaystyle U} and restriction is function restriction.

To see {displaystyle L} is indeed a morphism, we need to check {displaystyle (Lu)|V=L(u|V)} for open sets {displaystyle U} and {displaystyle V} such that {displaystyle Vsubseteq U} and {displaystyle uin C^{infty }(U)} . This is clear because for {displaystyle xin V} , both {displaystyle [(Lu)|V](x)} and {displaystyle [L(u|V)](x)} are simply {displaystyle lim _{rto 0}{frac {2d}{r^{2}}}{frac {1}{|S_{r}|}}int _{S_{r}}(u(y)-u(x))dy} , as the {displaystyle S_{r}} eventually sits inside both {displaystyle U} and {displaystyle V} anyway.

It is easy to check that {displaystyle L} is linear: {displaystyle L(f+g)=L(f)+L(g)} and {displaystyle L(af)=aL(f)} Finally, we check that {displaystyle L} is local in the sense that {displaystyle suppLfsubseteq suppf} . If {displaystyle x_{0}notin supp(f)} , then {displaystyle exists r>0} such that {displaystyle f=0} in the ball of radius {displaystyle r} centered at {displaystyle x_{0}} . Thus, for {displaystyle xin B(x_{0},r)} , {displaystyle int _{S_{r'}}(f(y)-f(x))dy=0} for {displaystyle r'

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