${p in [1, infty]: f in L^p}$ is an interval?











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I understood the proof but why does this imply that ${p in [1, infty]: f in L^p}$ is an interval?










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    enter image description here



    I understood the proof but why does this imply that ${p in [1, infty]: f in L^p}$ is an interval?










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      enter image description here



      I understood the proof but why does this imply that ${p in [1, infty]: f in L^p}$ is an interval?










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      enter image description here



      I understood the proof but why does this imply that ${p in [1, infty]: f in L^p}$ is an interval?







      real-analysis measure-theory






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      asked Nov 24 at 8:27









      conrad

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          The inequality implies that if $fin L^p$ and $fin L^q$ for $p<q$, then $fin L^r$ for all $p<r<q$. This implies that the set is an interval.






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            The inequality implies that if $fin L^p$ and $fin L^q$ for $p<q$, then $fin L^r$ for all $p<r<q$. This implies that the set is an interval.






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              up vote
              2
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              The inequality implies that if $fin L^p$ and $fin L^q$ for $p<q$, then $fin L^r$ for all $p<r<q$. This implies that the set is an interval.






              share|cite|improve this answer























                up vote
                2
                down vote










                up vote
                2
                down vote









                The inequality implies that if $fin L^p$ and $fin L^q$ for $p<q$, then $fin L^r$ for all $p<r<q$. This implies that the set is an interval.






                share|cite|improve this answer












                The inequality implies that if $fin L^p$ and $fin L^q$ for $p<q$, then $fin L^r$ for all $p<r<q$. This implies that the set is an interval.







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                answered Nov 24 at 11:37









                Seewoo Lee

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