Order of antiderivatives of Schwartz functionsFor a Schwartz function f ∈ S ( R ) it is known that ∃ F ∈ S ( R ) : F ′ = f ⟺ ∫ − ∞ + ∞ f ( x ) d x = 0in this case F ( x ) = ∫ x + ∞ f ( t ) d t, see related question here.Question: Does there exist f ∈ S ( R ) with f ( − k ) ∈ S ( R ) for all k ≥ 0 (my guess would be no)I.e. for a given f we can test if it has an antiderivate F in the Schwartz space by simply checking if its mean is 0. Then we can do the same with F. Does this process always stop at some point?My work: It seems that all functions of the type f ( x ) = p ( x ) e − α x 2 with p polynomial only have antiderivatives of finite order in S ( R ). (I think I can prove this). As one can construct an orthonormal basis of S ( R ) with such functions, e.g. the Hermite functions, I would guess that there is no function with antiderivates of arbitrary order in S ( R ).Remark: If there is such an f, then ( x k ⋆ f ) = 0 for all k ≥ 0; then via Fourier transform, this would imply that in the sense of distributions: 0 = F [ x k ⋆ f ] = F [ x k ] F [ f ] = ( i 2 π ) k δ ( k ) ( w ) f ^ ( w )

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Order of antiderivatives of Schwartz functionsFor a Schwartz function   f  ∈      S    (      R    ) it is known that  ∃  F  ∈      S    (      R    )  :      F    ′    =  f    ⟺        ∫          −      ∞              +      ∞        f  (  x  )  d  x  =  0in this case   F  (  x  )  =      ∫          x              +      ∞        f  (  t  )  d  t, see related question here.Question: Does there exist   f  ∈      S    (      R    ) with       f          (      −      k      )        ∈      S    (      R    ) for all   k  ≥  0 (my guess would be no)I.e. for a given f we can test if it has an antiderivate F in the Schwartz space by simply checking if its mean is 0. Then we can do the same with F. Does this process always stop at some point?My work: It seems that all functions of the type   f  (  x  )  =  p  (  x  )      e          −      α              x        2             with p polynomial only have antiderivatives of finite order in       S    (      R    ). (I think I can prove this). As one can construct an orthonormal basis of       S    (      R    ) with such functions, e.g. the Hermite functions, I would guess that there is no function with antiderivates of arbitrary order in       S    (      R    ).Remark: If there is such an f, then   (      x    k    ⋆  f  )  =  0 for all   k  ≥  0; then via Fourier transform, this would imply that in the sense of distributions:  0  =      F    [      x    k    ⋆  f  ]  =      F    [      x    k    ]      F    [  f  ]  =            (            i          2      π                          )              k        δ          (      k      )        (  w  )            f      ^        (  w  )

Savion Stanton · Accepted Answer

Explanation:You have pretty much answered your own question already: looking at the Fourier transforms, we want a   g  ∈      S   such that   g  (  ξ  )      ξ          −      k        ∈      S   also for all   k  ≥  1. This is easy: our only potential problem occurs at   ξ  =  0, so it suffices to take a   g  ∈      S   with   0  ∉      supp    (  g  )

Order of antiderivatives of Schwartz functions For a Schwartz function f ∈

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New Questions in Calculus 1