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

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.

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