Pointwise limit of a sequence of functions defined recursively as integrals.

Question

Problem: Let $g$ be a continuous function from $[0,\dfrac{1}{2}]$ to $\mathbb{R}$. Consider the sequence of functions $\{g_n\}$ defined on $[0,\dfrac{1}{2}]$ to $\mathbb{R}$ with $g_1=g$ and $$g_{n+1}(t)=\int_{0}^{t} g_n(s)ds$$ for all $t\in [0,\dfrac{1}{2}]$. Show that $ \lim_{n \to \infty} n!g_{n}(t) =0$. for all $t \in [0,\dfrac{1}{2}]$.

I tried some examples of such functions but I don't know how to prove it in general. Any hint please regarding how to start a proof? Thank you.

Answer 1

The function $g$ is bounded, say $|g(t)|\leq M$ for all $t\in [0,1/2]$. This means that $$ |g_{2}(t)|= \left|\int_0^t g_1(s)\,ds\right|\leq \int_0^t|g_1(s)|ds\leq \int_0^t M\,ds= tM $$ This gives us: $$ |g_3(t)| = \left|\int_0^tg_2(s)\,ds\right|\leq \int_0^t|g_2(s)|ds\leq\int_0^t sM\,ds = \frac12t^2M $$ Now generalize the pattern and finish with a proof, say by induction.