By A. I. Fetísov

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4. The solution u(x, t) to (96) is (i) outgoing if and only if (Sf )(ω, s) = 0 for s > 0, all ω. (ii) incoming if and only if (Sf )(ω, s) = 0 for s < 0, all ω. 44 Proof. For (i) suppose (Sf )(ω, s) = 0 for s > 0. For |x| < t we have x, ω + t ≥ −|x| + t > 0 so by (97) u(x, t) = 0 so u is outgoing. Conversely, suppose u(x, t) = 0 for |x| < t. Let t0 > 0 be arbitrary and let ϕ(t) be a smooth function with compact support contained in (t0 , ∞). Then if |x| < t0 we have Sn−1 R = dω Sn−1 R (Sf )(ω, x, ω + t)ϕ(t) dt dω u(x, t)ϕ(t) dt = 0 = R (Sf )(ω, p)ϕ(p − x, ω ) dp .

Then H = C ((I − T )H) ⊕ Null space (I − T ) is an orthogonal decomposition, C denoting closure, and I the identity. Proof. If T g = g then since T ∗ = T ≤ 1 we have g 2 = (g, g) = (T g, g) = (g, T ∗ g) ≤ g T ∗g ≤ g 2 so all terms in the inequalities are equal. Hence g − T ∗g 2 = g 2 − (g, T ∗ g) − (T ∗ g, g) + T ∗g 2 =0 so T ∗ g = g. Thus I −T and I −T ∗ have the same null space. But (I −T ∗ )g = 0 is equivalent to (g, (I − T )H) = 0 so the lemma follows. 50 Definition. An operator T on a Hilbert space H is said to have property S if (109) fn ≤ 1, T fn −→ 1 implies (I − T )fn −→ 0 .

Proof. 6 we have (ΛS)∨ (f ) = (ΛS)(f ) = S(Λf) = S((Λf )∨ ) = cS(f ) . The other inversion formula then follows, using the lemma. In analogy with βA we define the “sphere” σA in Pn as σA = {ξ ∈ Pn : d(0, ξ) = A} . 4. 7. Suppose n is odd. Then if S ∈ E (Rn ) , supp(S) ∈ σR ⇒ supp(S) ⊂ SR (0) . To see this let > 0 and let f ∈ D(Rn ) have supp(f ) ⊂ BR− (0). Then supp f ∈ βR− and since Λ is a differential operator, supp(Λf ) ⊂ βR− . Hence cS(f ) = S((Λf)∨ ) = S(Λf) = 0 so supp(S) ∩ BR− (0) = ∅. Since > 0 is arbitrary, supp(S) ∩ BR (0) = ∅ .

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Acerca de la Demostración en Geometría by A. I. Fetísov
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