Atomic decompositions on manifolds with bounded geometry

Abstract

We give the atomic decomposition of Triebel–Lizorkin and Besov scales on Riemannian manifolds of bounded geometry. In order to do it we modify the moment condition for atoms. 1 Preliminaries A fundamental technique in harmonic analysis is to represent a function or distribution as a linear combination of function of an elementary form. One of the example is the atomic decomposition. The decomposition appeared to be useful for characterizations of various function spaces as well as for the investigation of continuity properties of pseudodifferential operators. The decomposition comes from the theory of homogeneous Hardy spaces Hp, but it works for inhomogeneous function spaces as well. Frazier and Jawerth have got the atomic decomposition theorem for the Triebel-Lizorkin and Besov scales [2, 3]. A smooth (s,p)–atom centered in the ball B(xo, r) is a test function a satisfying the following three condition: supp a ⊂ B(xo, 2r) (1) sup x∈X | (Da)(x) | ≤ r n p , for any α, | α |≤ L, (2) ∫ Rn xa(x) = 0 for any β , | β |≤M, (3) where L,M are positive constant depending on s, p. The inhomogeneous spaces of F s p,q − Bs p,q type are invariant with respect to large class of diffeomorphism, some assumptions about derivatives are needed. But, if we look at the above condition of atoms we see immediately that the first two condition are invariant with respect to the diffeomorphism after the suitable normalization of the constants. But the third condition called the moment condition is not invariant. In the paper we give a new definition of the moment condition, which is invariant with respect to diffeomorphisms . In consequence we get atomic decompositions of the Triebel-Lizorkin and Besov scales on Riemannian manifolds with bounded geometry. 1.1 Function spaces For convenience we recall the definition of the Triebel-Lizorkin and Besov scales on Rn. Let η ∈ S(Rn) (the Schwartz space) be a non-negative function with η(x) = 1 if |x| ≤ 1 and η(x) = 0 if |x| ≥ 3 2 . We put η0 = η, η1 = η( x 2 )− η(x) ηk = η1(2x), k = 2, 3, . . . Then supp ηj ⊂ {x : 2j−1 ≤| x |≤ 32j−1}, j = 1, 2, . . . and ∑∞ j=0 ηj(x) = 1.