For each t ∈ [ 0 , 1 ], let T t : L 2 ( R ) → L 2 ( R ) be the operator that shifts everything to the right by t, i.e. T t f ( x ) = f ( x − t )for all f ∈ L 2 ( R ).Then t ↦ T t defines a map γ : [ 0 , 1 ] → B ( L 2 ( R ) ) .Note that γ is not continuous.Question: Is γ Bochner-integrable, i.e. does the integral ∫ [ 0 , 1 ] T t d tconverge in B ( L 2 ( R ) )?Thoughts: Since each operator Tt has norm, the question is whether γ is strongly measurable. In particular, there is a question of whether there exists a subset of J ⊆ [ 0 , 1 ] of measure 1 such that γ ( J ) is separable. Certainly γ ( [ 0 , 1 ] ) is not separable, and I suspect that no such subset J exists, but I'm not sure how to show this.

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For each   t  ∈  [  0  ,  1  ], let       T    t    :      L    2    (      R    )  →      L    2    (      R    ) be the operator that shifts everything to the right by   t, i.e.      T    t    f  (  x  )  =  f  (  x  −  t  )for all   f  ∈      L    2    (      R    ).Then   t  ↦      T    t   defines a map  γ  :  [  0  ,  1  ]  →      B    (      L    2    (      R    )  )  .Note that   γ is not continuous.Question: Is   γ Bochner-integrable, i.e. does the integral      ∫          [      0      ,      1      ]            T    t      d  tconverge in       B    (      L    2    (      R    )  )?Thoughts: Since each operator Tt has norm, the question is whether   γ is strongly measurable. In particular, there is a question of whether there exists a subset of   J  ⊆  [  0  ,  1  ] of measure 1 such that   γ  (  J  ) is separable. Certainly   γ  (  [  0  ,  1  ]  ) is not separable, and I suspect that no such subset   J exists, but I&#039;m not sure how to show this.

Punktatsp · Accepted Answer

This question suggested me the following formal manipulation. The semigroup       T    t   can be written, formally, as      T    t    =      e          t              ∂                  ∂          x                      =      ∑          n      =      0        ∞              t      n              n      !                  d      n              d              x        n              .Integrating in   t  ∈  [  0  ,  1  ], again formally, we obtain      ∫    0    1        T    t      d  t  =      ∑          n      =      0        ∞        1          n      !      (      n      +      1      )                  d      n              d              x        n              .It seems to me that the comments are suggesting that       T    t   is NOT integrable. Therefore this last expression should make no sense. But I cannot see why.Not an answer, but i hope it helps

For each t ∈ [ 0 , 1 ] , let T t </msub>

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