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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 .

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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 $|t| < 1$.

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