Here $g:[0,1]\to \mathbb{R}$ is a continuous function and so is $f: [0,1]\to \mathbb{R}$. $R(f)$ is a little bit more complicated: \begin{equation} R(f)(x):=\int_0^x k(x,y)f(y)dy \end{equation} for a continuous $k: [0,1]\times [0,1]\to \mathbb{R}$ (we take the standard metric on $[0,1]^2\subset \mathbb{R}^2$ i.e. the Euclidean Metric/Distance Function). I have managed to prove that $R(f): [0,1]\to \mathbb{R}$ is continuous.
My initial idea was to solve it through Banach's fixed point theorem; for that I need to prove, however, that $R(f)$ is Lipschitz-continuous with a Lipschitz-constant $L<1$. I have not been able to prove that this far.
If $R$ turns out to have a L.C. $L<1$, then I will be able to utilize a variant of the Banach's fixed point theorem to prove the assertion. I greatly appreciate your help!
First, let's proceed formally to find an expression for $f$.
The equation is rewritten $(\text{id} - R)(f) = g$, whose solution is : $$f = (\text{id} - R)(g) = \sum_{n=0}^\infty R^n(g)$$
If we had $\|R \|<1$ this would be over, but this is not the case. To show that the series actually converges uniformly, notice that:
\begin{align} R^n(g) (x_0) &= \int_0^{x_0} \text{d}x_1 \ldots \int_0^{x_{n-1}} \text{d}x_n \prod_{i=0}^{n-1} k(x_i,x_{i+1}) g(x_n) \\ |R^n(g) (x_0)|& \leqslant (\|k \|_\infty)^n\|g\|_\infty \int_0^{x_0} \text{d}x_1 \ldots \int_0^{x_{n-1}} \text{d}x_n 1 \\ & \leqslant(\|k \|_\infty)^n\|g\|_\infty \text{vol}\left(\{(x_i) |0\leq x_n \leq x_{n-1}\leq \ldots \leq x_0\}\right)\\ & \leqslant \frac{x^n(\|k \|_\infty)^n}{n!}\|g\|_\infty \end{align}
Therefore, $\| R^n \|\leqslant \frac{(\|k\|_\infty)^n}{n!}$ and the series $\sum R^n(g)$ converges in $\mathcal C^0([0,1])$