# A question in real analysis which I am unable to prove

This question was asked in my real analysis mid term and I was totally stumped by it .

So , I am asking for help here .

Let $$g: [0,1/2]\to\mathbb{R}$$ be a continuous function. Define $$g_{n} : [0,1/2]\to \mathbb{R}$$ by $$g=g_{1}$$ and $$g_{n+1}(t)=\int_{0}^{t}g_{n}(s) ds$$ for all n$$\geq$$1 .

Then show that $$\lim_{n \rightarrow \infty} n! \cdot g_{n}(t)=0$$ for all $$t \in [0,1/2]$$.

I could only think of the fact that we can change limit and integration if function is uniformly convergent . But that cant be used here . So , I cannot provide anything in this question as attempt but thats due to the fact that I dont have any clue .

Kindly guide .

Let $$M$$ be an upper bound on $$|g(x)|$$, which exists by the compactness of $$[0, 1/2]$$.
Claim: for all $$t \in [0, 1/2]$$, $$|g_{n}(t)| \leq M \frac{t^{n - 1}}{(n - 1)!}$$. Proof: induction on $$n$$.
Base case: $$n = 1$$. Then we see that $$|g_1(t)| = |g(t)| \leq M = M \frac{t^{0}}{0!}$$.
Inductive step: suppose true for $$n$$. Then $$|g_{n + 1}(t)| = |\int\limits_0^t g_n(s) ds| \leq \int\limits_0^t |g_n(s)| ds \leq \int\limits_0^t M \frac{s^{n - 1}}{(n - 1)!} ds = M \frac{t^n}{n!}$$.
Then we see that $$|n! g_n(t)| \leq n! M \frac{t^{n - 1}}{(n - 1)!} = n M t^{n - 1}$$, which clearly goes to zero as $$n$$ goes to $$\infty$$ since $$|t| < 1$$.