Let $g:[0,\frac{1}{2}]\to \mathbb{R}$ be a continuous function.Define $g_n:[0,\frac{1}{2}] \to \mathbb{R}$ by $g_1=g $ and $g_{n+1}(t)=\int_0^t g_n(s)ds\, \forall n\ge 1$.Show that $\lim_{n\to \infty}n!g_n(t)=0 \, \forall t\in [0,\frac{1}{2}]$
i think since $g(t)$ is continuous on $t\in [0,\frac{1}{2}]$ so there exists $M$ such that $\left|g(t)\right|\le M$ $\, \forall t\in [0,\frac{1}{2}]$ . Now $g_2(t)=\int_0^t g(t)dt \le \int_0^t Mdt=Mt$ again $g_3(t)=\int_0^tg_2(t)dt\le \int_0^t Mtdt=M\frac{t^2}{2}$, similarly $g_{n+1}(t)\le M\frac{t^n}{n!}$ so now $$0< \left |n!g_n(t) \right| \le \left|Mnt^n \right |=Mnt^n $$ Now taking limit both side limit we get$$\lim_{n\to \infty}n!g_n(t)=0$$Since $t\in [0,\frac{1}{2}]$
Is there any problem or mistake ???
