SOME REMARKS ON THE FEFFERMAN-STEIN INEQUALITY 1 ...

6 ANDREI K. LERNERWe have∫R n (m λ f)g dx =≤∫ ∞0∫{m λ f>α}∫ ∞ ∫c λ,nwhich proves the lemma.0{f>α}g dxdα ≤∫ ∞0∫{Mχ {f>α} (x)≥λ}Mg dxdα = c λ,n∫R n f(Mg)dx,3. Proofs of Theorems 1.1 and 1.2g dxdαProof of Theorem 1.1. Let us show first that (i) ⇒ (ii). We can supposethat f, g ≥ 0. Also, it is enough to assume, for instance, thatg is compactly supported, and hence g ∈ L 1 (R n ). The general casewill follow by the standard limiting argument. If g ∈ L 1 (R n ), thenM ⋆ g ∈ L 1 (R n ), and thus M ⋆ g ∈ S 0 (R n ). Therefore, by Theorem 2.1,∥M ⋆ g∥ X ≤ c∥(M ⋆ g) # ∥ X ≤ c∥Mg∥ X .From this, applying Theorem 2.1 again, we get∫∫Mf(x)g(x)dx ≤ c Mf(x)g(x)dxR n R∫n= c f(x)M ⋆ g(x)dxR n≤ c∥f∥ X ′∥M ⋆ g∥ X≤ c∥f∥ X ′∥Mg∥ X .We prove now (ii) ⇒ (i). Using Theorem 2.2 and (2.5), we get∫∫|f(x)φ(x)|dx ≤ c M # λ f(x)Mφ(x)dxR n R n≤From this, by (2.1), we obtain (i).c∥φ∥ X ′∥MM # λ f∥ X≤ c∥φ∥ X ′∥f # ∥ X .Remark 3.1. The proof of Theorem 2.2 in [15] shows that actually onecan replace Mg by the dyadic maximal function M ∆ g, namely, we have∫∫|f(x)g(x)|dx ≤ c M # λ f(x)M ∆ g(x)dx.R n R nTherefore, taking into account the proof of Theorem 1.1, in order toverify the Fefferman-Stein property of X, it is enough to check that∫(M ∆ f)|g|dx ≤ c∥f∥ X ′∥Mg∥ X .R n□□