I have some problems, help: Let ( X , <mrow class="MJX-TeXAtom-ORD"> <mi

Crystal Wheeler 2022-07-13 Answered
I have some problems, help:
Let ( X , A 1 , ν ) be a probability space. Let A : L 1 L 1 be a contraction such that A : L L is also a contraction. Suppose A preserves positivity.
In a proof of why A : L p L p is a contraction for every 1 p < , my lecture notes state that for all f 0, we have
A f p = sup g 0 , g q = 1 X ( A f g ) ,
where 1 p + 1 q = 1.
Why does the above equality hold? The strangest part to me is the fact that the LHS has a power of 1 p outside of the integral X | A f | p , but in the RHS the integral is not being raised to any power. I've tried to raise both sides to p but I could not do much with this. If it helps, this came up in the study of conditional expectations.
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Answers (1)

sniokd
Answered 2022-07-14 Author has 22 answers
We have by Holder's that | X ( A f g ) | A f p g q = A f p for all g L q with g q = 1. The question then becomes whether or not g can be chosen to obtain equality.
Note that 1 / p + 1 / q = 1 implies that 1 + p / q = p and 1 + q / p = q. Hence, choose g = ( A f ) p / q A f p p / q , which is nonnegative because f is and A preserves nonnegativity. Moreover, g q = 1.
But then A f g = ( A f ) 1 + p / q A f p p / q = ( A f ) p A f p p / q . Integrating, shows that
X A f g = A f p p A f p p / q = A f p .

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