Ellipticity Let a ∈ D

326 D. Kapanadze – B.-W. Schulze The function ψ satisfies the assumptions of Lemma 1, while M ht i ̺−δ belongs to S 0,0 n−1 × n−1 ; ✁ + , ✁ + by Lemma 2. This gives us M ht i ̺−δ f y, η ∈ S 0,0 n−1 × n−1 ; L 2 + , L 2 + Hence, setting again φ y, t = hyi ̺ hxi −̺ , we see that hxi ̺−δ hyi δ−̺ Op f hyi ̺ hxi −̺ = Op M ψ Op M ht i ̺−δ f Op M φ is a continuous operator L 2 n + → L 2 n + . In an analogous manner we can proceed in the remaining cases concerning the sign of ̺ and ̺ − δ. The compactness of ✁ for ord ✁ µ; δ then follows from the continuity of ✁ to spaces of better smoothness and weight and from corresponding compact embeddings of Sobolev spaces.

3.5. Ellipticity

The principal symbol structure of the preceding section gives rise to an adequate notion of ellip- ticity of pseudo-differential boundary value problems globally on the half-space. D EFINITION 7. An operator ✁ ∈ µ, d;δ cl n + ; N − , N + is called elliptic of order µ; δ if i σ ψ ✁ x, ξ for all x, ξ ∈ n + × n \ 0, σ e ✁ x, ξ for all x, ξ ∈ n + \ 0 × n , σ ψ, e ✁ x, ξ for all x, ξ ∈ n + \ 0 × n \ 0 are non-zero, ii σ ∂ ✁ y, η for all y, η ∈ n−1 × n−1 \ 0 , σ e ′ ✁ y, η for all y, η ∈ n−1 \ 0 × n−1 , σ ∂, e ′ ✁ y, η for all y, η ∈ n−1 \ 0 × n−1 \ 0 are isomorphisms ✁ + ⊕ N − −→ ✁ + ⊕ N + . R EMARK 16. Condition ii in the latter definition can equivalently be replaced by bijec- tivities in the sense H s + ⊕ N − −→ H s−µ + ⊕ N + for any s maxµ, d − 1 2 . D EFINITION 8. Given ✁ ∈ µ, d;δ cl n + ; N − , N + , an operator ∈ −µ,e;−δ cl n + ; N + , N − for some e ∈ ✁ is called a parametrix of ✁ if ✁ − ∈ −∞,d l ;−∞ n + ; N − , N − and ✁ − ∈ −∞,d r ;−∞ n + ; N + , N + for certain d l , d r ∈ ✁ . We shall see below that the ellipticity of an operator ✁ ∈ µ, d;δ cl n + ; N − , N + entails the existence of a parametrix. First we want to construct further examples of elliptic boundary value problems. Boundary value problems 327 The Dirichlet problem for c − 1, with the Laplace operator 1 = P n j =1 ∂ 2 ∂ x 2 j , and a constant c 0 is represented by the operator ✁ 1 = c − 1 r ′ : H s n + −→ H s−2 n + ⊕ H s− 1 2 n−1 . 52 For convenience we pass to ✁ 2 = c − 1 Qr ′ : H s n + −→ H s−2 n + ⊕ H s−2 n−1 53 where Q is an order reduction on the boundary that we take of the form Q = Op y hηi 3 2 , such that Q : H s n−1 −→ H s− 3 2 n−1 is an isomorphism for all s ∈ . Then we have ✁ 2 ∈ 2,1;0 cl n + ; 0, 1 . We want to show that 53 is an isomorphism for all s 3 2 and construct the inverse. We have for x = y, t , ξ = η, τ σ ψ ✁ 2 = |ξ | 2 , σ e ✁ 2 = c + |ξ | 2 , σ ψ, e ✁ 2 = |ξ | 2 , σ ∂ ✁ 2 = |η| 2 − ∂ 2 t |η| 3 2 r ′ , σ e ′ ✁ 2 = c + |η| 2 − ∂ 2 t hηi 3 2 r ′ , σ ∂, e ′ ✁ 2 = |η| 2 − ∂ 2 t |η| 3 2 r ′ . Hence ✁ 2 is elliptic in the sense of Definition 7. First we invert the operator family α − ∂ 2 t β r ′ : ✁ + −→ ✁ + ⊕ , 54 where α := c + |η| 2 1 2 , β := hηi 3 2 . Let us write l ± τ = α ± i τ ; then l − τ l + τ = α 2 + τ 2 and α 2 − ∂ 2 t = op + l − l + = op + l − op + l + the latter identity is true because l − is a minus function; op + l − : ✁ + → ✁ + is an isomorphism . Thus, to invert 54, it suffices to consider op + l + β r ′ : ✁ + −→ ✁ + ⊕ which is an isomorphism, because op + l + : ✁ + → ✁ + is surjective and βr ′ induces an isomorphism of ker op + l + = γ e −αt : γ ∈ to . Let us form the potential k = kα : → ✁ + , defined by kγ = γ β −1 e −αt , γ ∈ . Then op + l + β r ′ op + l −1 + k = 1 1 , because r ′ op + l −1 + = 0, and hence op + l + β r ′ −1 = op + l −1 + k . 328 D. Kapanadze – B.-W. Schulze Consider now aτ = α 2 + τ 2 . The operator in 54 can be written op + a β r ′ = op + l − 1 op + l + β r ′ and hence op + a β r ′ −1 = op + l −1 + k op + l −1 − 1 = op + l −1 + op + l −1 − k . Here op + l −1 + op + l −1 − = op + a −1 + g for a certain g ∈ Ŵ + . It follows altogether op + a β r ′ −1 = op + a −1 + g k , i.e., we calculated the inverse of 54. Inserting now the expression for α = αη, β = βη, we easily see that the ingredients of σ e ′ ✁ 2 −1 = op + a −1 η + gη kη 55 belong to −2,0;0 cl n−1 × n−1 ; 1, 0 they are, of course, independent of y, and it is clear that ✁ −1 2 = Op y σ e ′ ✁ 2 −1 which belongs to −2,0;0 cl n + ; 1, 0 . The method of calculating 55 gives us analogously σ ∂ ✁ 2 −1 and σ ∂, e ′ ✁ 2 −1 , and σ ✁ −1 2 = |ξ | −2 , c + |ξ | 2 −1 , |ξ | −2 ; σ ∂ ✁ 2 −1 , σ e ′ ✁ 2 −1 , σ ∂, e ′ ✁ 2 −1 . It is then obvious how to express ✁ −1 1 , namely ✁ −1 1 = ✁ −1 2 1 Q −1 . R EMARK 17. Similar arguments apply to the Neumann problem for c − 1 in the half- space, with r ′ ∂ t in place of r ′ . To get an element with unified orders we can pass to the boundary operator Rr ′ ∂ t for R = Op y hηi 1 2 . We see that c − 1 Rr ′ ∂ t ∈ 2,2;0 cl n + ; 0, 1 is also elliptic in the sense of Definition 7 and even invertible as an operator H s n + → H s−2 n + ⊕ H s−2 n−1 for s 3 2 . The inverse belongs to −2,0;0 cl n + ; 1, 0 . We shall construct in Section 5.3 below a general class of further examples of this kind. Boundary value problems 329 T HEOREM 7. For every N ∈ ✁ there exist elliptic elements ✁ + N ∈ 0,0;0 cl n + ; 0, N and ✁ − N ∈ 0,0;0 cl n + ; N , 0 that induce isomorphisms ✁ + N : H s n + −→ H s n + ⊕ H s n−1 , N , ✁ − N : H s n + ⊕ H s n−1 , N −→ H s n + for all s − 1 2 , where ✁ − N = ✁ + N −1 . Proof. Let us start from the above operator ✁ 2 and form ✁ = s −2 ✁ 2 e −s ∈ 0,0;0 cl n + ; 0, 1 56 for any fixed s 2, where e := R 1 ∈ B 1,0;0 cl n + is the order reducing element from Theorem 4 and := diag R 1 , R ′ for R ′ = Op y hηi ⊗ id ✁ N . Then, setting ✁ + 1 = ✁ , we can form ✁ + N inductively by ✁ + N = A + N T + N :=    A + N −1 T + N −1 1    A + 1 T + 1 =    A + N −1 A + 1 T + N −1 A + 1 T + 1    . Here, ✁ + 1 = A + 1 T + 1 . Moreover, from the above construction of ✁ −1 2 and Theorem 4 it follows that we may set ✁ − N := ✁ + N −1 .

3.6. Parametrices and Fredholm property