# Repeated integration of a real valued bounded function.

I was reviewing the proof the remainder estimate for a Taylor series expansion and I came across something I can't find an intuitive explanation for: if you have a function f that's bounded on an interval $[a-s, a+s]$ and define $f_1(x) := \int_a^xf(t)\,\mathrm dt$ and define $f_n(x) := \int_a^x f_{n-1}(t)\,\mathrm dx$, then $$\lim_{n\to\infty}f_n(x) = 0.$$

Can anyone explain why or how this is the case on an intuitive level?

Also if I try this iteration using $f(x) = \cos(x)$, at each iteration I get either $\sin(x)$ or $\cos(x)$ and a part of its Taylor expansion with the next iteration resulting in a better estimate.

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Just deal with the case $f$ constant: you have $f_n(x)=\frac{x^n}{n!}$ which converges pointwise to $1$. – Davide Giraudo Oct 7 '12 at 19:42
@Davide: You mean to $0$? – joriki Oct 7 '12 at 20:42
@joriki: yes, it was a typo. – Davide Giraudo Oct 7 '12 at 20:43

You are given that your function is bounded on a small region about $a$. First notice that $f\le |f|\le M$ for some $M$. Then $f_1(x)=\int_a^xf(t)\, \mathrm dt\le\int_a^xM\, \mathrm dt = M(x-a).$ Inductively using the previously established bound shows that
$$f_n(x)\le M\frac{(x-a)^n}{n!}.$$ A similar bound holds below using the same style of approximation as above. This implies the limit go to 0 by the squeeze theorem.