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Let $g:[0,\frac1{2}]\to \mathbb{R}$ be continuous function. Let us define the sequence $g_n(t)$ as $g_1(t)=g$ and $$g_{n+1}(t)=\int_0^tg_n(s)ds$$ for $n\ge1$. Is it true that $\lim_{n\to\infty}n!g_n(t)=0\forall t\in[0,\frac1{2}]$

I think yes, and may be we have to use the Cauchy formula for repeated integrals. Is there any other way to prove this? Thanks beforehand.

vidyarthi
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  • See https://math.stackexchange.com/questions/1888375/uniformly-convergence-of-f-n1x-intx-0-f-ntdt or https://math.stackexchange.com/questions/109559/f-nx-is-defined-by-f-n1x-int-0xf-ntdt-prove-sum-1-i – Martin R Apr 24 '19 at 12:08

2 Answers2

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Hint:

Note that $g_{n+1}(t)\le t\max g_n\le\frac12\max g_n$.

ajotatxe
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Just apply Taylor to $g_{n+1}(t)$ and note that $|g(t)|\leq C$ on $[0,\frac{1}{2}]$, since $g$ is continuous:

$$g_{n+1}(t) = \underbrace{\sum_{k=0}^{n-1}\frac{1}{k!}g_{n+1}^{(k)}(0)}_{=0}t^k + \underbrace{g_{n+1}^{(n)}}_{=g}(\tau_n)\frac{t^{n}}{n!} $$ $$\Rightarrow n!\cdot|g_{n+1}(t)|\leq C\frac{1}{2^n}$$

trancelocation
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