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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 ???

RAM_3R
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    On your first line, it's $|g(t)|\leq M$, and the subsequent inequalities need this change too. Then at the end, it's better to not write the limits immediately in the inequalities, since technically you don't know if they exist, aka first write the inequalities and then use the "squeeze theorem". Otherwise, great job! – folouer of kaklas Mar 27 '19 at 21:21
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    thanks ........ – RAM_3R Mar 27 '19 at 21:37

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