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 ???

  • 3
    $\begingroup$ 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! $\endgroup$ Mar 27, 2019 at 21:21
  • 1
    $\begingroup$ thanks ........ $\endgroup$
    – RAM_3R
    Mar 27, 2019 at 21:37


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.