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Let $g:[0,\frac{1}{2}]\rightarrow\mathbb{R}$ be a continuous function. Define $g_n:[0,\frac{1}{2}]\rightarrow\mathbb{R}$ by $g_1=g$ and $$g_{n+1}(t)=\int\limits_{0}^{t}g_n(s)ds$$ for all $n\geq1$. Show that $$\lim_{n\to\infty}n!g_n(t)=0$$ for all $t\in[0,\frac{1}{2} ]$.

How I can approach this problem. I have tried recursion on $g_n$ but it is not helping much

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Since $g$ is continuous on a compact set, it is bounded. Let $|g| \leq M$ on $[0,1/2]$, for some constant $M>0$. Then $|g_1(t)| \leq M t$, for $t\in[0,1/2]$, and $$ |g_2(t)| \leq \int\limits_0^t |g_1(x)| dx \leq M \int\limits_0^t xdx = \frac{1}{2}t^2M. $$ We now prove by induction on $n$ that $$ |g_n(t)| \leq \frac{1}{n!}M t^n, \text{ for all } t \in[0,1/2]. \tag{1} $$ The base case of $n=2$ was proved above. To pass from $n$ to $n+1$ we use $(1)$ and the definition of $g_{n+1}$ to get $$ |g_{n+1}(t)| \leq \int\limits_0^t |g_n(x)| dx \leq \frac{1}{n!} M \int\limits_{0}^t x^n dx = \frac{1}{(n+1)!} M t^{n+1}, \text{ for all } t \in [0,1/2]. $$ The claim in question now follows easily.

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  • $\begingroup$ How we have got $\vert g_1(t)\vert\leq Mt$. Would you please explain $\endgroup$ – J.Doe Apr 12 at 6:57
  • $\begingroup$ @J.Doe, that was a notational inconsistent with yours, I started with index $0$, while you start with $1$. If you let $g_0 = g$ and $g_1 = \int_0^t g(x) dx $, then $|g_1(t)| \leq \int\limits_0^t |g_0(x)| dx \leq M t$. So $g_n$ in my notation would be $g_{n+1}$ in yours. $\endgroup$ – Hayk Apr 12 at 7:07
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Using induction, you can prove that for $n \ge 2$, $$g_n(t) =\frac{1}{(n-2)!} \int_0^t \left(t-s\right)^{n-2} g(s) \mathrm d s$$

Indeed, $g_2 = \int_0^t g_1(s) \mathrm d s = \int_0^t (t-s)^{2-2}g(s) \mathrm d s$

By using the Leibniz integral rule : \begin{align} \frac{\mathrm d}{\mathrm d t} \left\{\frac{1}{n!} \int_0^t \left(t-s\right)^n g(s) \mathrm d s\right\} &= \frac1{n!}\left\{(t-t)^ng(t) + \int_0^t\frac {\partial}{\partial t} \left\{(t-s)^n g(s)\right\}\mathrm d s\right\}\\ &= \frac{1}{(n-1)!} \int_0^t (t-s)^{n-1} g(s) \mathrm d s \end{align}

The rest of the induction is trivial. Now $$\left|n!g_n(t)\right| = n(n-1)\left|\int_0^t (t-s)^{n-2} g(s) \mathrm d s \right| \le n(n-1)\left\|g\right\| \int_0^t (t-s)^{n-2} \mathrm d s = n\left\|g\right\| t^{n-1} \to 0$$ since $t\in \left[0,\frac12\right]$

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It's a direct consequence of Taylor.

You have (setting $g_0 = g$ "for convenience"):

  • $g_n^{(k)}(t) = g_{n-k}(t)$ and $g_n^{(n)}(t) = g(t)$ for $1\leq k \leq n$
  • $g_n^{(k)}(0) = 0$ for $0 \leq k < n$

Hence, for $n \in \mathbb{N}$ there is $\tau_n \in (0,t)$ such that

$$g_n(t) = \sum_{k=0}^{n-1}\frac{g_n^{(k)}(0)}{k!}t^k + \frac{g_n^{(n)}(\tau_n)}{n!}t^n = \frac{g(\tau_n)}{n!}t^n$$

It follows with $C = \max_{t\in [0,\frac{1}{2}]}|g(t)|$

$$|n!g_n(t) | =|g(\tau_n)t^n|\leq C\frac{1}{2^n}\stackrel{n \to \infty}{\longrightarrow} 0$$

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