“…Proposition 17 DtN − , −DtN + , V −1 ∈ E(1, −Arg, H 1/2 ( )).Proof Let φ ∈ H 1/2 ( ) and define u as the solution tou − s 2 u = 0, in − , γ u = φ, on , so that ∂ ν u = DtN − (s)φ.Then A − s)u, u = DtN − (s)φ, φ and hence by(7),(5) and the trace theoremRe e −ıArg s DtN − (s)φ, φ =…”