Sequence of antiderivatives of a continuous function Let f : <mrow class="MJX-TeXAtom-ORD

ttyme411gl 2022-07-07 Answered
Sequence of antiderivatives of a continuous function
Let 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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Answers (1)

Franco Cohen
Answered 2022-07-08 Author has 8 answers
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 < 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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