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]
dµ
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