Is there a 'distributional antiderivative'?Suppose I have a distribution F which I know on the derivative of an arbitrary test function. That is, I know F ( ϕ ′ ) for all ϕ ∈ D ( ℝ ). I also know that F ( ϕ ′ ) = − F ′ ( ϕ ). I want, however, to know F ( ϕ ). Is there a 'distributional antiderivative' that immediately tells me what this should be? If not, is it possible for such F′ that are induced by simpler, specific functions? (e.g. polynomials)For Φ ( x ) := ∫ ϕ ( x ) d x, simply putting F ( ϕ ) = − F ′ ( Φ ) seems a little dangerous.

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Is there a &#039;distributional antiderivative&#039;?Suppose I have a distribution F which I know on the derivative of an arbitrary test function. That is, I know   F  (      ϕ    ′    ) for all   ϕ  ∈      D    (      ℝ    ). I also know that   F  (      ϕ    ′    )  =  −      F    ′    (  ϕ  ). I want, however, to know   F  (  ϕ  ). Is there a &#039;distributional antiderivative&#039; that immediately tells me what this should be? If not, is it possible for such F′ that are induced by simpler, specific functions? (e.g. polynomials)For   Φ  (  x  )  :=  ∫  ϕ  (  x  )  d  x, simply putting   F  (  ϕ  )  =  −      F    ′    (  Φ  ) seems a little dangerous.

frogoogg31 · Accepted Answer

Step 1If there exists   ψ  ∈      C    ∞    (      R    ) such that   ϕ  =      ψ    ′   then we can just set   F  (  ϕ  )  =  −      F    ′    (  ψ  )  ..But this is in general not the case. It is true precisely when       ∫          ∞              ∞        ϕ  (  x  )    d  x  =  0.. Now fix   ρ  ∈      C    c    ∞    (      R    ) such that       ∫          ∞              ∞        ρ  (  x  )    d  x  =  1 and given   ϕ  ∈      C    c    ∞    (      R    ) set             ϕ      ~        =  ϕ  −  (      ∫          ∞              ∞        ϕ  (  x  )    d  x  )    ρ  .. Then       ∫          ∞              ∞                  ϕ      ~        (  x  )    d  x  =  0 so we can apply the first paragraph on             ϕ      ~        .. Now we note that   F      (    (          ∫              ∞                    ∞              ϕ    (    x    )        d    x    )    ρ    )    =  F  (  ρ  )  (      ∫          ∞              ∞        ϕ  (  x  )    d  x  ) is just some constant   F  (  ρ  ) times the constant distribution   ϕ  ↦      ∫          ∞              ∞        ϕ  (  x  )    d  x acting on   ϕ  .Step 2Thus, we can set   F  (  ϕ  )  =  F  (            ϕ      ~        )  +  F  (  ρ  )  (      ∫          ∞              ∞        ϕ  (  x  )    d  x  )  .

Is there a 'distributional antiderivative'? Suppose I have a distribution F which I know on the der

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New Questions in Calculus 1