I'm reading about this in one of the articles that the teacher gave meSuppose that ( S 1 , μ 1 ) and ( S 2 , μ 2 ) are two σ-finite measure spaces and F : S 1 × S 2 → R is measurable. Then [ ∫ S 2 | ∫ S 1 F ( x , y ) μ 1 ( d x ) | p μ 2 ( d y ) ] 1 p ≤ ∫ S 1 ( ∫ S 2 | F ( x , y ) | p μ 2 ( d y ) ) 1 p μ 1 ( d x )with obvious modifications in the case p = ∞. If p &gt; 1, and both sides are finite, then equality holds only if | F ( x , y ) | = φ ( x ) ψ ( y ) a.e. for some non-negative measurable functions φ and ψ.If F is non-negative, then the function f : y ↦ ∫ S 1 F ( x , y ) μ 1 ( d x )and thus | f | p are measurable. For the LHS to be well-defined, the function | f | p should be measurable. In our case, F is not assumed to be non-negative. Could you elaborate on how to prove it?

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I&#039;m reading about this in one of the articles that the teacher gave meSuppose that       (          S              1              ,          μ              1              )   and       (          S              2              ,          μ              2              )   are two   σ-finite measure spaces and   F  :      S          1        ×      S          2        →      R   is measurable. Then            [              ∫                              S                          2                                                            |                      ∫                                          S                                  1                                                              F          (          x          ,          y          )                      μ                          1                                (                      d                    x          )          |                          p                            μ                  2                    (              d            y      )      ]                      1        p              ≤      ∫                  S                  1                                (              ∫                              S                          2                                                  |            F      (      x      ,      y      )                        |                          p                            μ                  2                    (              d            y      )      )                      1        p                  μ          1        (      d    x  )with obvious modifications in the case   p  =  ∞. If   p  &amp;gt;  1, and both sides are finite, then equality holds only if       |    F  (  x  ,  y  )      |    =  φ  (  x  )  ψ  (  y  ) a.e. for some non-negative measurable functions   φ and   ψ.If   F is non-negative, then the function  f  :  y  ↦      ∫                  S                  1                      F  (  x  ,  y  )      μ          1        (      d    x  )and thus       |    f            |        p   are measurable. For the LHS to be well-defined, the function       |    f            |        p   should be measurable. In our case,   F is not assumed to be non-negative. Could you elaborate on how to prove it?

Schetterai · Accepted Answer

You are correct that they left out some hypotheses. One hypothesis that works is   F  ≥  0. Another hypothesis that works is that the right hand side is finite. You can find a statement and proof of this result in &quot;Real Analysis&quot; by Folland.

I'm reading about this in one of the articles that the teacher gave me Suppose that (

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