Let $f$ be a continuous function in $[0,1]$. Define, for $n \in \mathbb N$ $$f_0 = f \text{ and } f_{n+1}(x) = \begin{cases} \frac{1}{x}\int_0^xf_n(t)\text{d}t\text{ if } x\ne0 \\ f_n(0) \text{ if } x = 0 \end{cases} $$
Study the pointwise and uniform convergence of $(f_n)$.
I studied the case in which $f$ is a polynomial : we have $$f_0(x) = \sum_{i=0}^da_ix^i$$ then $$\begin{equation*} \begin{split} f_0(x) & = \frac{1}{x}\int_0^xf_0(t)\text{d}t \\ & = \frac{1}{x} \sum_{i=0}^d \int_0^xx^i\text{d}t \\ & = \sum_{i=0}^d\frac{a_i}{i+1}x^i \\ \end{split} \end{equation*}$$ By induction we show $$f_n(x) = a_0 + \sum_{i=1}^d\frac{a_i}{(i+1)^n}x^i$$ So $\displaystyle{\forall x \in [0, 1], \lim_{n \rightarrow \inf} f_n(x) = a_0 = f(0)}$. This shows the pointwise convergence of $(f_n)$ toward $x \mapsto f(0)$.
I would need help for the general case.
Essential expanding the brilliant hint given by Daniel:
Given $\epsilon>0$, find a polynomial $p$ such that $\sup_{[0,1]}|f-p|<\epsilon$.
Define $p_{0}=p$ and \begin{align*} p_{n+1}(x)=\dfrac{1}{x}\int_{0}^{x}p_{n}(t)dt,~~~~x\in(0,1],~~~~p_{n+1}(0)=p_{n}(0). \end{align*} One has \begin{align*} |f_{n+1}(x)-p_{n+1}(x)|&\leq\dfrac{1}{x}\int_{0}^{x}|f_{n}(x_{1})-p_{n}(x_{1})|dx_{1}\\ &\leq\dfrac{1}{x}\int_{0}^{x}\dfrac{1}{x_{1}}\int_{0}^{x_{1}}|f_{n-1}(x_{2})-p_{n-1}(x_{2})|dx_{2}dx_{1}\\ &\leq\cdots\\ &\leq\dfrac{1}{x}\int_{0}^{x}\dfrac{1}{x_{1}}\int_{0}^{x_{2}}\cdots\dfrac{1}{x_{n}}\int_{0}^{x_{n}}|f(x_{n+1})-p(x_{n+1})|dx_{n+1}\cdots dx_{2}dx_{1}\\ &\leq\sup_{[0,1]}|f-p|\\ &<\epsilon. \end{align*} We also have \begin{align*} |f_{n+1}(x)-f(0)|&\leq|f_{n+1}(x)-p_{n+1}(x)|+|p_{n+1}(x)-p(0)|+|f(0)-p(0)|\\ &<2\epsilon+|p_{n+1}(x)-p(0)|. \end{align*} With such, the sequence converges uniformly to $x \mapsto f(0)$.