Prove that there is a unique bounded continuous real-valued function f on $X=\{ t \in \mathbb{R} : t\geq 0 \}$ , equipped with the sup metric, such that $$ f(s) = 1+ \int_{0}^s e^{-t^2} f(st)dt \quad for\quad s\in X $$
I define a new function $ T :C([0, \infty) \rightarrow C([0, \infty) ) $ by $$ Tf(s)= 1+ \int_{0}^s e^{-t^2} f(st)dt$$ where $C([0, \infty)$ is the space of continuous functions on $ [0, \infty).$
If $T$ is a contraction then by Banach Fixed Point thm, it has a fixed point and it is the unique solution of the above integral equation.
First question is that how do I show that $T$ is a well defined function, meaning that it takes continuous function to continuous function.
Second question is that how do I show $T$ is a contraction ? To answer the second question, I tried to consider $$ |Tf(s) - Th(s)| \leq s d(f,h) $$ Taking the $\sup $ of both sides over $s$, RHS blows off. Does it mean that all I have done is useless?