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L(H). Hence Λs u ∈ H implies D s1 ( x s2 u) ∈ H, hence, x s2 u ∈ C(R3 ), by ordinary Sobolev imbedding. So we also get u continuous. ] Notice that the ψdo Λs = x s2 D s1 belongs to Opψcs and its inverse Λ−1 s = D −s1 x −s2 = Λ∗−s belongs to Opψc−s . By ψdo-calculus conclude that Λs AΛ−1 s ∈ same constants, since the Fourier transform F is unitary. One ﬁnds that (iii): A∧ θι u(η) = 1 ∨1 ∨2 (ξ − η, κ − ξ, ξ). 11) with partial integration and Schur’s lemma. (iv):We have b∨ 2 (1−∆ξ ) a(x,ξ)ψθ (x)ψι (y) }eixζ eiyκ dxdy.

4. 7 An md-elliptic operator A = a(x, D) (of order m), considered as a continuous map S → S is a Fredholm operator. It has closed rank; its null space ker A is ﬁnite dimensional, and it is a subspace of S. 12) (f, ϕj ) = 0 , j = 1, . . , N where ϕj ∈ S , j = 1, . . , N is a set of given ﬁxed vectors. , also, they are independent of s. 11) are valid. Proof. About ker A: Let Au = 0 for any u ∈ S . Apply the operator B to this equation for u + Ku = 0 where K is an integral operator with kernel in S(R6 ).

For the open interval |t| < 1 since it is needed in that form, later on. But it is evident that the statement holds as well if |t| < 1 is replaced by any other interval containing 0. 6. 24) u x u|x ). 31 It satisﬁes ˘ with g˘ = v|t = g˘v + h g 0 x g|x g ˘=( , h h x h|x ) . ˘ ∈ ψcm , u Notice that g˘ ∈ ψc0 , h ˘0 = (ux0 u|x ) ∈ ψcm . 2) (for x) just as above. The rest is just iteration again. D. 2 into the following form. 4 For a 4 × 4-matrix-valued symbol k(t, x, ξ) assume that all tderivatives ∂tj k(t, x, ξ) , j = 0, 1, .