Let $E\subset \mathbb{R}^n$ open. Consider the initial value problem
$\left\{ \begin{array}{lcc} y' = f(t,y(t)) \\ y(t_0)=y_0 \end{array} \right.$
where $t_0=[a,b]$ and $f\in C(J\times E, \mathbb{R}^n) $. Suppose that the Picard iteration $\{g_n(t)\}_{n\in \mathbb{N}}$ is uniformly convergent to a function $g(t)$ in $J$.
Do $g(t)$ have to be a solution to the previous problem?
I think that it does, however I do not know what arguments I can give. I am suppose to know it through Cauchy-Lipstchitz's and Peano's theorems. \ Any help would be appreciated.
The sequence $g_n(t)$ is defined as $$g_{n+1}=y_0+\int_{t_0}^t f(s,g_{n}(s))\mathop{ds}$$ where $g_0(t)= y_0$. Taking the limit as $n\to \infty$ in the expression above yields $$\begin{align*} g(t)&= y_0 +\lim_{n\to \infty}\int_{t_0}^tf(s,g_n(s))\mathop{ds}\\ &=y_0 +\int_{t_0}^t f(s, \lim_{n\to \infty} g_n(s))\mathop{ds}\\ &= y_0 + \int_{t_0}^t f(s, g(s)) \mathop{ds} &= \end{align*}$$ where we have used the hypothesis that $g_n(t)$ converges uniformly to $g(t)$ to justify the exchange between the limit and the integral, and the fact that $f$ is continuous to justify that $\lim f(s, g_n)= f(s, \lim g_n)$. Now, to see that $g(t)$ is indeed a solution to the IVP just differentiate both sides on the equality above and see what happens when plugging $t_0$.