Holomorphic Functions

C. Fernández, A. Galbis, M.C. Gómez-ColladoPROOF.Given σ ∈ {−1, 1} N , ɛ > 0 and ϕ ∈ D(R N ), one has that ˇf ∗ ϕ ∈ D Lp (R N ) and〈(S N (ψ) ∗ f)(· + iσɛ), ϕ〉 = 〈S N (ψ)(· + iσɛ), ˇf ∗ ϕ〉.Since S N (ψ) ∈ H1N ⊂ Hp N, ′ p′ being the conjugate number of p, it suffices to apply Proposition 1 forω(t) = log(1 + t) and Proposition 2. The following lemma permit us to get a similar result for p = ∞.Lemma 4 Let K be a compact set in R and let ψ ∈ D(K) be given. Then S 1 (ψ) ∈ H(C \ K) and, forsome positive constants A and C,|S 1 (ψ)(x ± iɛ)| ≤ C|x| 2whenever |x| ≥ A and 0 < ɛ < 1.PROOF. It is clear from the definition of S 1 . Proposition 4 Given Γ ∈ D(R N ) and f ∈ L ∞ (R N ) we have S N (Γ) ∗ f ∈ H N ∞ and T (S N (Γ) ∗ f) =Γ ∗ f.PROOF. We already know that S N (Γ) ∗ f ∈ H∞. N To see that T (S N (Γ) ∗ f) = Γ ∗ f we proceedin two steps. First, let K be a compact set in R and let ϕ 1 , . . . , ϕ N ∈ D(K) be given. We considerΓ := ϕ 1 ⊗ · · · ⊗ ϕ N and we put f j := S 1 (ϕ j ) ∈ H1. 1 Then F := S N (Γ) is given by F (z 1 , . . . , z N ) =∏ Nj=1 f j(z j ). Let us check that T (F ∗ f) = Γ ∗ f. We observe that⎛∑⎝σ∈{−1,1} NN∏j=1σ j⎞⎠ F (x + iσɛ) =N∏(f j (x j + iɛ) − f j (x j − iɛ)) .From Lemma 4, we choose ˜K a compact subset in R, K ⊂ ˜K, and C > 0 such thatj=1|f j (x j ± iɛ)| ≤ C|x| 2whenever x /∈ ˜K and 0 < ɛ < 1. Let η ∈ D(R) be identically one on a neighborhood of ˜K. For eachψ ∈ D(R N ) and each 0 < ɛ < 1 we have,〈(∑(σ∈{−1,1} N j=1N∏σ j )F (x + iσɛ)) ∗ f, ψ〉N∏= 〈 (f j (x j + iɛ) − f j (x j − iɛ)),j=1= I 1 (ɛ) + I 2 (ɛ) + I 3 (ɛ),N∏(1 − η(x j ) + η(x j ))( ˇf ∗ ψ)〉j=1whereN∏N∏I 1 (ɛ) := 〈 (f j (x j + iɛ) − f j (x j − iɛ)), ( η(x j ))( ˇf ∗ ψ)〉,j=1j=1N∏N∏I 2 (ɛ) := 〈 (f j (x j + iɛ) − f j (x j − iɛ)), (1 − η(x j ))( ˇf ∗ ψ)〉,j=1j=1252