I am trying to solve the following problem:Let ( X , S , μ ) be a measure space. Let 1 &lt; p &lt; ∞. Let f : X × X → R be such that, for every y ∈ X, the section f y is p-integrable and that ∫ X ‖ f y ‖ p d μ ( y ) &lt; + ∞ . Define, for x ∈ X, g ( x ) = ∫ X f ( x , y ) d μ ( y ) . Show that g ∈ L p ( μ ) and that ‖ g ‖ p ≤ ∫ X ‖ f y ‖ p d μ ( y ) . "So above is the problem. We need to show (after simplifying the inequality) that ( ∫ X | ∫ X f ( x , y ) d μ ( y ) | p d μ ( x ) ) 1 p ≤ ∫ X ( ∫ X | f ( x , y ) | p d μ ( x ) ) 1 p d μ ( y ) . Can you please give me some hints on how to do it? Thanks.

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I am trying to solve the following problem:Let   (  X  ,  S  ,  μ  ) be a measure space. Let   1  &amp;lt;  p  &amp;lt;  ∞. Let   f  :  X  ×  X  →      R   be such that, for every   y  ∈  X, the section       f    y   is   p-integrable and that            ∫      X        ‖          f      y              ‖      p             d    μ    (    y    )    &amp;lt;    +    ∞    .  Define, for   x  ∈  X,  g  (  x  )  =            ∫      X        f    (    x    ,    y    )    d    μ    (    y    )    .  Show that   g  ∈      L    p    (  μ  ) and that  ‖  g      ‖    p    ≤            ∫      X        ‖          f      y              ‖      p             d    μ    (    y    )    .  &quot;So above is the problem. We need to show (after simplifying the inequality) that            (                        ∫          X                                      |                          ∫              X                        f            (            x            ,            y            )                         d            μ            (            y            )            |                    p                d        μ        (        x        )            )                      1        p              ≤            ∫      X                      (                  ∫          X                                      |            f            (            x            ,            y            )            |                    p                         d        μ        (        x        )        )                    1        p                   d    μ    (    y    )    .  Can you please give me some hints on how to do it? Thanks.

eurgylchnj · Accepted Answer

When   X is   σ-finite, this is just the Minkowski integral inequality with   Y  =  X, which you can prove using the &quot;duality&quot; representation of the   p-norm: If   (  E  ,      F    ,  μ  ) is   σ-finite, then for any measurable   p  ∈  [  1  ,  ∞  ],  ‖  f      ‖                  L        p              =  sup  {      ∫          E        f  g    d  μ  :  g  ≥  0  ,  ‖  g      ‖                  L        q              =  1  }  .The above identity for   ‖  f      ‖                  L        p             also holds when   (  E  ,      F    ,  μ  ) is not   σ-finite, provided   ‖  f      ‖                  L        p              &amp;lt;  ∞.To prove Minkowski&#039;s integral inequality, you can use the above identity and upper bound the integrals appearing in the supremum.

I am trying to solve the following problem: Let ( X , S , μ ) be a

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