Sequence of antiderivatives of a continuous functionLet f : R → R be a continuous function and ( f n ) n ≥ 0 a sequence of functions such that f 0 = f and f n + 1 is an antiderivative (primitive) of f n for each n ≥ 0, with the property that for each x ∈ R there exists n ∈ N such that f n ( x ) = 0. Prove that f is identically 0.Honestly, I do not have a viable starting point; but the problem looks very nice. Obviously, f n ( n ) = f and one can use the general formula for the solutions of this differential equation of order n, but...

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Sequence of antiderivatives of a continuous functionLet   f  :      R    →      R   be a continuous function and   (      f    n        )          n      ≥      0       a sequence of functions such that       f    0    =  f and       f          n      +      1       is an antiderivative (primitive) of       f    n   for each   n  ≥  0, with the property that for each   x  ∈      R   there exists   n  ∈      N   such that       f    n    (  x  )  =  0. Prove that f is identically 0.Honestly, I do not have a viable starting point; but the problem looks very nice. Obviously,       f    n          (      n      )        =  f and one can use the general formula for the solutions of this differential equation of order n, but...

Franco Cohen · Accepted Answer

Explanation:Assume that   f  (      x    0    )  ≠  0 for some       x    0  . Then   f  (  x  )  ≠  0 for all x in an interval [a, b] with   a  &amp;lt;  b. Now   [  a  ,  b  ]  =      ⋃          n      ∈              N              {  x  ∈  [  a  ,  b  ]  ∣      f    n    (  x  )  =  0  } is the countable union of closed sets. The Baire category theorem implies that one of these sets has non-empty interior, i.e.       f    n    (  x  )  =  0 for some n and all x in an open interval   I  ⊂  [  a  ,  b  ]. But then       f    n          (      n      )        =  f implies that f is zero is on I, a contradiction.

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