ON WEIGHTED NORM INTEGRAL INEQUALITY ... - Kjm.pmf.kg.ac.rs

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168 Let n = 1, we have ∫ [T f] p wdµ = X 1 = ≤ ∫X 1 [∫ ∫X 1 [∫ K(x, y)f(y)dµ(y)] p wdµ K(x, y) 1 − p f(y)Φ 1 p ′ K(x, y) 1 p ′ Φ 1 p p dµ(y)] ′ wdµ [ ∫ ∫ ] ( K(x, y)f(y) ∫X p Φ − p p ′ dµ(y)) 1 p ( K(x, y)Φdµ(y)) 1 p p ′ wdµ 1 By Holder’s inequality ∫ [ = (T f p Φ 1−p )(T Φ) p−1 w ] dµ X 1 By a property of an inner product ∫ = f p vdµ X 1 Hence, valid for n = 1. Next, assume n = k > 1, we have k∑ ∫ [T f] p wdµ = [T f] p wdµ n=1 ∪ ∫ k n=1 Xn ∫ ∫ = [T f] p wdµ + [T f] p wdµ + . . . + [T f] p wdµ X 1 X 2 X ∫ ∫ ∫ k ≤ f p vdµ + f p vdµ + . . . + f p vdµ X 1 X 2 X k By Holder’s inequality and using the above argument for n = 1 ∫ = f p vdµ Then, for n = k + 1, we have k+1 ∑ ∫ ∪ k n=1 X n [T f] p wdµ = ∑ k ∫ n=1 X n [T f] p wdµ + ∫ X k+1 [T f] p wdµ n=1 X n k∑ ∫ ∫ ≤ f p vdµ + [T f] p wdµ n=1 X n X k+1 by assumption when n = k ≤ k∑ ∫ ∫ f p vdµ + X n f p vdµ X k+1 n=1 since µ(X k+1 ) < ∞ and by the proof of k = 1. On the other hand, let us assume that (3) holds for some v < ∞ µ−almost everywhere on X. By using the σ−finiteness of µ, we can find a positive function Φ such that ∫ X Φp vdµ < ∞ and (4) holds.
169 Finally, suppose (4) holds and let v denotes the weight in (5) then k+1 ∑ n=1 ∫ X n Φ p vdµ = = = = = ∫ ∫ ∫ Φ p vdµ + Φ p vdµ + . . . + Φ p vdµ X 1 X 2 X ∫ k+1 Φ [ p (Φ 1−p T ∗ (T Φ) p−1 w) ] ∫ dµ + Φ [ p (Φ 1−p T ∗ (T Φ) p−1 w) ] dµ X 1 X ∫ 2 + . . . + Φ [ p (Φ 1−p T ∗ (T Φ) p−1 w) ] dµ X ∫ k+1 [ [ ΦT ∗ (T Φ) p−1 w ]] ∫ [ dµ + [ ΦT ∗ (T Φ) p−1 w ]] dµ X 1 X ∫ 2 [ + . . . + [ ΦT ∗ (T Φ) p−1 w ]] dµ X ∫ k+1 [ (T Φ)(T Φ) p−1 w ] ∫ [ dµ + (T Φ)(T Φ) p−1 w ] dµ X 1 X ∫ 2 [ + . . . + (T Φ)(T Φ) p−1 w ] dµ X ∫ k+1 ∫ ∫ [(T Φ) p w] dµ + [(T Φ) p w] dµ + . . . + [(T Φ) p w] dµ X 1 X 2 X k+1 ∫ ⇒ [(T f) p w] dµ = X 1 +X 2 +...+X k+1 by (5). k+1 ∑ n=1 ∫ X n [(T Φ) p w] dµ < ∞ We conclude that the result in (3) is valid for all n = 1, 2, . . . and the proof is complete. ✷ Remark. If we set n = 1 with X 1 = X in (3), then our result yields Theorem 1 obtained by Kerman and Sawyer [6]. let Theorem 2.2. For X ∈ R and 1 < p < ∞. Let f(t) ≥ 0 and g(t) ≥ 0 and also h(t) = { f(t) if t > 0 g(t) if t < 0 and zero otherwise, also suppose T is a completely arbitrary integral operator, then ∫ R ∫ [T h](t) p dµ ≤ R ∫ f p v 1 dµ + If and only if there exists a positive function Φ on t with X R g p v 2 dµ (6) ∫ v 1 = [T Φ] p dµ < ∞ (7)
Page 1 and 2: 165 Kragujevac J. Math. 29 (2006) 1
Page 3: 167 2. STATEMENT OF RESULTS In this
Page 7 and 8: 171 Finally, suppose (7) and (8) ho
Page 9: 173 [11] K. Rauf, A Characterizatio

ON WEIGHTED NORM INTEGRAL INEQUALITY ... - Kjm.pmf.kg.ac.rs

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