By Bibhutibhushan Datta

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**Example text**

For any F ∈ Ap,0 , let I(F ; p, q) = D |F (z)|p−q |F (z)|q (1 − |z|2 )q dm(z). First of all, we recall two basic facts for f ∈ H, fr (z) = f (rz), r ∈ (0, 1) and p ∈ (0, ∞). The ﬁrst one is the Hardy–Stein identity which reads as: fr p Hp = 2π|f (0)|p + p2 D |fr (z)|p−2 |fr (z)|2 (− log |z|)dm(z). The second one is the following Littlewood–Paley inequalities: fr and p Hp |f (0)|p + I(fr ; p, p), p ∈ (0, 2] |f (0)|p + I(fr ; p, p) fr p Hp , p ∈ [2, ∞). Next, we check (i) and (ii). (i) For the ﬁrst inequality, let q ∈ (0, min{2, p}].

1 In so doing, let B(x, y) = 0 tx−1 (1 − t)y−1 dt. Then for x, y ∈ (0, ∞) and j, k ∈ N, B(j + 1 + x, y) − B(j + k + 1 + x, y) 1 tj+x = 0 k−1 ≈ k−1 tn (1 − t)y dt n=0 (n + j + 1)−y−1 ≈ (j + 1)−y − (k + j + 1)−y . 5. Dirichlet Double Integral without Derivative 43 Since qr (ζ) is a member of A2,β , it can be represented by the reproducing formula for this space. Consequently, I(k) 1 = 0 D 1 = 0 ≈ ≈ D |qr (ζ)|2 (1 − |ζ|2 )β dm(ζ)(1 − r2 )β rdr 1+β π ∞ 1 0 D j=0 ⎛ 1 0 ⎜ ⎝ ∞ j=0 D qr (η)(1 − |η|2 )β dm(η) (1 − η¯ζ)2+β 2 1 − |ζ|2 (1 − r2 )−1 β dm(ζ)rdr 2 q (η)¯ η j ζ j (1 D r − |η|2 )β dm(η) B(j + 1, β + 1) 1 − |ζ|2 β dm(ζ)rdr (1 − r2 )−1 ⎞ 2 j 2 2 β q (η)¯ η (1 − |η| )(1 − r ) dm(η) ⎟ D r ⎠ rdr.

Applying this inequality to f ◦ σw , we derive I(f ; w) = D = |f (z) − f (w)|2 |1 − z¯w|−4−2β (1 − |z|2 )β dm(z) (1 − |w|2 )−(2+β) (1 − |w|2 )−(2+β) D D |f ◦ σw (z) − f ◦ σw (0)|2 (1 − |z|2 )β dm(z) |(f ◦ σw ) (z)|2 (1 − |z|2 )β+2 dm(z) 2 D |f (z)| |1 − z¯w|−2β−4 (1 − |z|2 )β+2 dm(z). Integrating I(f ; w) over D with respect to (1 − |w|2 )β dm(w) and changing the order of integration, we further get f 2 D D D |f (z)|2 D (1 − |w|2 )β dm(w) (1 − |z|2 )β+2 dm(z) |1 − wz| ¯ 4+2β I(f ; w)(1 − |w|2 )β dm(w) thanks to β > −1.