I am stuck on this exercise. We are studying the boundary value problem \begin{equation} \begin{cases} u(x) - u''(x) = f(x),&x \in (0,1) \\ u(0) = u(1) = 0 \\ \end{cases} \label{prob} \end{equation}
which has the solution (in compact form) \begin{equation} u(x) = \int_{0}^{1}G(x,y)(f(y) - u(y))dy \label{sol} \end{equation} Where $G(x,y)$ is the Greens functions for this specific boundary value problem. We later define $\mathcal{H}$ as the mapping from $C([0,1])$ to $C_{0}([0,1])$ by \begin{equation} \mathcal{H}(v)(x) = \int_{0}^{1}G(x,y)(f(y) - v(y))dy \end{equation} and the sequence of solutions $\{u_n\}_{n = 0}^{\infty}$ by
\begin{equation} \begin{cases} u_0(x) = 0\\ u_{n+1}(x) = \mathcal{H}(u_n)(x),&n\geq 0 \end{cases} \end{equation}
From a previous exercise I have already shown that $$ ||\mathcal{H}(v)-\mathcal{H}(u)||_{\infty} \leq \frac{1}{8}||u - v||_{\infty} $$
I wish to show that $$u(x) = \lim\limits_{n\rightarrow\infty} u_n$$ exists and that $u(x)$ solves the problem.
Can I somehow use this result to show that the sequence converges and thus showing that the boundary value problem has a unique solution, or do I have to delve into the world of Cauchy analysis?
To see that $u_n$ converges in $C([0,1])$, it suffices to check that it is a Cauchy sequence for the $\infty$-norm since $C([0,1])$ is complete for that norm. For this, the thing you showed in the previous exercise is enough. Indeed, that inequality implies that \begin{align} \|u_{n+1} - u_n\| = \|\mathcal{H}(u_n) - \mathcal{H}(u_{n-1})\| \leq 8^{-1} \|u_n - u_{n-1}\| \leq \dots \leq 8^{-(n-1)} \|u_2 - u_1\| \end{align} and as a result that for $m > n$ \begin{align} \|u_m - u_n\| \leq \sum_{j = n}^{m-1} \|u_{j+1} - u_j\| \leq \|u_2 - u_1\| \sum_{j = n}^\infty 8^{-(j-1)} \to 0 \end{align} as $n \to \infty$ so that $u_n$ is a Cauchy sequence as desired. Hence there exists $u \in C_0([0,1])$ such that $\|u_n - u\| \to 0$.
Now we know that $$u_{n+1} = \int_0^1 G(\cdot,y) (f(y) - u_n(y)) dy$$ and so by sending $n \to \infty$ we get that $$u = \lim_{n \to \infty} \int_0^1 G(\cdot, y)(f(y)-u_n(y))dy = \int_0^1 G(\cdot,y)(f(y)-u(y))dy$$ where we can pass the limit through the integral since convergence for the $\infty$-norm is uniform convergence. This is exactly that $u$ solves the integral form of ODE we started with, as desired.