# How to evaluate the following limit rearding sequence of functions

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

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.

• How we have got $\vert g_1(t)\vert\leq Mt$. Would you please explain – J.Doe Apr 12 at 6:57
• @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. – Hayk Apr 12 at 7:07

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 $$t\in \left[0,\frac12\right]$$

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

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