I have to determine the set which satisfies { p ∈ [ 0 , ∞ ] : f ∈ L p ( λ ) } for a function f : R → R defined by f ( x ) = { 1 x if x ∈ ( 0 , 1 ] 0 if x ∈ R ∖ ( 0 , 1 ] I have looked at three different cases, i.e. p = 0 , p = ( 0 , ∞ ) and p = ∞ where the last case is troubling me.If p = ∞ we have to check that f ∈ L ∞ . So assume that this is the case. Then by definition there ∃ R &gt; 0 such that | f | &lt; R. We have that | f | &lt; R ⇔ | 1 x | = 1 x &lt; R ⇔ 1 R &lt; x ⇔ x ∈ ( 1 R , 1 ] but how do I proceed from here?

Question

I have to determine the set which satisfies   {  p  ∈  [  0  ,  ∞  ]  :  f  ∈            L        p    (  λ  )  } for a function   f  :      R    →      R   defined by  f  (  x  )  =      {                                        1            x                                                          if                     x          ∈          (          0          ,          1          ]                                      0                                                if                     x          ∈                      R                    ∖          (          0          ,          1          ]                        I have looked at three different cases, i.e.   p  =  0  ,  p  =  (  0  ,  ∞  ) and   p  =  ∞ where the last case is troubling me.If   p  =  ∞ we have to check that   f  ∈            L        ∞  . So assume that this is the case. Then by definition there   ∃  R  &amp;gt;  0 such that       |    f      |    &amp;lt;  R. We have that      |    f      |    &amp;lt;  R  ⇔      |          1      x        |    =      1    x    &amp;lt;  R  ⇔      1    R    &amp;lt;  x  ⇔  x  ∈      (          1      R        ,    1    ]  but how do I proceed from here?

Adolfo Rich · Accepted Answer

lim          x      →              0        +              f  (  x  )  =  +  ∞  so   f  ∉      L    ∞  . There is no such that as       L    0  . For   p  ∈  (  0  ,  ∞  ) and   p  ≠  1, we see that:      ∫          R            |    f  (  x  )            |        p          d    x  =      ∫    0    1        1          x      p              d    x  =            [              1                  1          −          p                            x                  1          −          p                    ]                      0        +              1    =  1  −      lim          x      →              0        +                        x              1        −        p                    1      −      p      RHS is finite iff   p  &amp;lt;  1. Finally, for   p  =  1, we have:      ∫          R            |    f  (  x  )      |          d    x  =      ∫    0    1        1    x          d    x  =            [      ln      ⁡      (      x      )      ]                      0        +              1    =  −  ∞Thus, for   p  ∈  (  0  ,  ∞  ], we have:  f  ∈      L    p      ⟺    p  ∈  (  0  ,  1  )(Note that while       L    p   is well-defined for   0  &amp;lt;  p  &amp;lt;  1, they are not normed spaces as   ‖  ⋅      ‖    p   does not respect triangle inequality, and so are in general not studied)

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