Let $f,g:[0,1]\rightarrow \mathbb{R}$ be continuous functions. Define $$\displaystyle x_{n}(t)=f(t)+ \int_{0}^{t}x_{n-1}(s)ds\quad 0\leq t\leq 1,\;n=1,2,\cdots ,$$ where $x_{0}(t)=g(t),0\leq t\leq 1$ . How we can show that the sequence $(x_n)$ is uniformly convergent on $[0,1]$ and its limit is independent of $g$.
I have tried in any direction I can but I fail to succeed. Can you indicate the reference material that can help for this and related questions.
Thank you for your valuable time.
Let $$h(t)= f(t)+ \int_0^t f(s)ds + \sum\limits_{i \geq 1} \int_0^t \int_0^{x_1}...\int_0^{x_i} f(s)ds dx_i...dx_1$$
To show that $h$ is well-defined, notice that $$\int_0^t \int_0^{x_1} ... \int_0^{x_i} f(s)dsdx_i...dx_1 \leq ||f||_{\infty} \frac{t^{i+1}}{(i+1)!}$$
Now, you can show by induction that $$x_n(t)= f(t)+ \int_0^t f(s)ds + \sum\limits_{i =1}^{n-2} \int_0^t \int_0^{x_1}...\int_0^{x_i} f(s)ds dx_i...dx_1 + \int_0^t\int_0^{x_1}...\int_0^{x_{n-1}} g(s)dsdx_{n-1}...dx_1$$
So finally $$h(t)-x_n(t) \leq ||g||_{\infty} \frac{t^{n-1}}{(n+1)!} + ||f||_{\infty} \sum\limits_{i \geq n-1} \frac{t^{i+1}}{(i+1)!} $$
Hence $$||h-x_n||_{\infty} \leq \frac{||g||_{\infty}}{(n+1)!}+||f||_{\infty} \sum\limits_{i \geq n-1} \frac{1}{(i-1)!} $$
Consequently, $x_n \to h$ uniformly; notice also that $h$ doesn't depend on $g$.