Let $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $\lim_{n→∞} n!g_n(t) = 0,$

Question

Let $g : [0,\frac{ 1} {2} ] → \mathbb R$ be a continuous function. Define $g_n : [0,\frac{ 1} {2} ] → \mathbb R$ by $g_1 = g$ and $g_{n+1}(t) = \int_0^t g_n(s) ds,$ for all $n ≥ 1.$ Show that $\lim_{n→∞} n!g_n(t) = 0,$ for all $t ∈ [0,\frac{1}{2}].$

My attempt:- If $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}<1$, then it converges to zero. For any $t\in [0,1/2]$ Consider the sequence $a_n(t)=n!g_n(t)$ $\lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty}(n+1)\frac{g_{n+1}(t)}{g_{n}(t)}=\lim_{n\to \infty}(n+1)\frac{\int_0^t g_n(s) ds}{g_{n}(t)}.$ But I am not able to simplify further. Please help me.

Answer 1

For a fixed $t$, define $$ M_n(t)=\sup_{0\leqslant t'\leqslant t}\left\lvert g_n(t')\right\rvert. $$ Then one can prove that for all $x\in [0,1/2]$, $M_{n+1}(t)\leqslant M_n(t)/2$.