If f ∈ L 2 ( a , b ), then ∫ a x f ( y ) d y ∈ L 2 ( a , b )?If f ∈ L 2 ( a , b ), then I want to show that the antiderivative F ( x ) := ∫ a x f ( y ) d y.is in L 2 (I guess this is true). If L 2 ( a , b ) would be closed under pointwise product, i.e. if f , g ∈ L 2 , then f ⋅ g ∈ L 2 , then this would follow easily, but I guess this is not true. So any hints how to show the closure under taking antiderivatives?

Question

If   f  ∈      L    2    (  a  ,  b  ), then       ∫    a    x    f  (  y  )  d  y  ∈      L    2    (  a  ,  b  )?If   f  ∈      L    2    (  a  ,  b  ), then I want to show that the antiderivative   F  (  x  )  :=      ∫    a    x    f  (  y  )  d  y.is in       L    2   (I guess this is true). If       L    2    (  a  ,  b  ) would be closed under pointwise product, i.e. if   f  ,  g  ∈      L    2  , then   f  ⋅  g  ∈      L    2  , then this would follow easily, but I guess this is not true. So any hints how to show the closure under taking antiderivatives?

nuvolor8 · Accepted Answer

Step 1Write   F  (  x  )  =      ∫    a    b        1          [      a      ,      x      ]        (  y  )  f  (  y  )    d  y. By the Cauchy-Schwarz inequality,       |    F  (  x  )      |    ≤  ‖      1          [      a      ,      x      ]            ‖    2    ⋅  ‖  f      ‖    2    =  (  x  −  a      )          1              /            2        ‖  f      ‖    2      (  a  &amp;lt;  x  &amp;lt;  b  )  .Step 2  ‖  F      ‖    2    ≤  ‖  (  x  −  a      )          1              /            2            ‖    2    ⋅  ‖  f      ‖    2    =            b      −      a              2        ‖  f      ‖    2    &amp;lt;  ∞  .

If f ∈ L 2 </msub> ( a , b ) , then <msubsup> &#

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