Prove by induction that $|f_n(x)|\leq k\frac{x^n}{n!}\leq \frac{k}{n!}$

Question

I'm working on sequences and series of function and got stuck on this question:
Let $f:[0,1]\to\mathbb{R}$ a continuous function and define a sequence of function $(f_n)$ by $$ f_0=f \text{ and } f_{n+1}=\int_0^xf_n(s)ds, \forall n\in\mathbb{N} $$

I want to prove by induction that $$|f_n(x)|\leq k\frac{x^n}{n!}\leq \frac{k}{n!}$$ where's $k>0$ that satisfied $|f(x)|\leq k$ for all $x\in[0,1]$

There's more thing to prove, like the series of that sequence is convergent, but this i know how to do with the Weierstrass M test and i need this induction to do it but i don't even know how to start.

Answer 1

Hint: $$\begin{align}|f_n(x)| &\le \int_0^x |f_{n-1}(s)| \, ds \\&\overset{\text{induction}}{\le} \frac{k}{(n-1)!} \int_0^x s^{n-1} \, ds\end{align}$$