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

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

170

and

or equivalently,

and


v 2 =

X

[T ∗ Φ] p dµ < ∞ (8)

v 1 = Φ 1−p T ∗ (T Φ) p−1 < ∞ (9)

v 2 = Φ 1−p T (T ∗ Φ) p−1 < ∞ (10)

Proof. Suppose T f = ∫ ∞

0 f(t)dt and T ∗ is the dual of T .

Let

Then,

Therefore,

I =

=


R


R

= ∫ x

I = T h(t)

h(t)dt

[f(t) + g(t)] dt

0 f(t)dt + ∫ ∞

x g(t)dt


∫ [∫ x ∫ ∞ p

[T h](t) p dµ = f(t)dt + g(t)dt]


R

R 0

x

∫ [∫ x

∫ ∞ ]

≤ f(t) p dt + g(t) p dt dµ

R 0

x

By Minkowski’s inequality



= (T f) p dµ + (T ∗ g) p dµ

R

R

∫ [

≤ (T f p Φ 1−p )(T Φ) p−1] ∫ [

dµ + (T ∗ g p Φ 1−p )(T ∗ Φ) p−1] dµ

∫R

[ R

= f p Φ 1−p T ∗ (T Φ) p−1] ∫ [

dµ + g p Φ 1−p T ∗∗ (T ∗ Φ) p−1] dµ

By definition of inner product



= f p v 1 dµ +

Conversely, we can assume that (6) holds for some v < ∞ µ−almost everywhere.

R

R

R

g p v 2 dµ

By using the σ−finiteness of µ, we can obtain a positive function Φ such

that ∫ X Φp 1(v 1 )dµ < ∞ and ∫ X Φp 2(v 2 )dµ < ∞, then (7) and (8) holds.

R

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