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 &lt; ∞, 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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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  &amp;lt;  ∞, 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&#039;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.

sniokd · Accepted Answer

We have by Holder&#039;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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