By Theo Bühler

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We now turn to the main result on consequences of the Δ2 condition. 9 Let (M, μ) be a measure space with a nonatomic component and F a weak Young function. Then the following are equivalent: (i) F obeys the Δ2 condition. 7, is equivalent to) norm convergence. , L∞ is dense in L(F ) ). (iv) YF = L(F ) (v) YF is a vector space. 21) We will show (i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (v) ⇒ (vi) ⇒ (i). 22) (we use μ(M ) = 1 here). 22), QF (2k g) ≤ 1 so g F ≤ 2−k . This shows if QF (gn ) → 0, then gn F → 0. 40 Convexity (ii) ⇒ (iii) Suppose (ii) holds and let f ∈ L(F ) (M, dμ).

Conversely, if limx↓0 F (x)/x = y0 > 0, then F (x) ≥ xy0 so G(y0 ) = 0. 32) is part of the deﬁnition of Young functions – it is needed for the conjugate function to be a weak Young function. 17 Let F be a Young function and G its convex conjugate. Then G is a Young function. 45, G is a convex function which is steep and nonnegative. 16, G(x) = 0 if and only if x = 0. Since F is even, G is even. 16 implies that limx↓0 G(x)/x = 0. 18 If F (x) = xp , 1 < p < ∞, the conjugate function is not y q (where p−1 + q −1 = 1) but G(y) = p−q /p q −1 y q by a simple calculation.

Iii) ⇒ (iv) (iv) ⇒ (i) is trivial. Pick x0 so F (x0 ) ≥ ε. For x ≥ x0 , F (kx) ≤ B F (x) + ε ≤ (B + 1)F (x) Now pick so k ≥ 2. Then for x ≥ x0 , F (2x) ≤ F (k x) ≤ (B + 1)F (k (i) ⇒ (v) −1 x) ≤ · · · ≤ (B + 1) F (x) Since (D− F )(x) is monotone, if (i) holds and x ≥ x0 , 2x x(D− F )(x) ≤ x D− F (y) dy ≤ F (2x) ≤ C F (x) Orlicz spaces 37 so sup x≥x 0 x(D− F )(x) ≤C F (x) Since sup 1≤x≤x 0 x(D− F )(x) <∞ F (x) (v) holds. (v) ⇒ (iv) Suppose (v) holds. Let the sup be A and let x ≥ 1 and k > 1. 18) Pick k so A(log k) ≤ 12 .