Let $K:[a,b] \times [a,b] \to \mathbb{R}$ continuous with $|K(t,s)| \leq M, \forall s,t \in [a,b]$. Prove that $\forall f \in C([a,b])$ and $\forall \lambda $ st. $ |\lambda|< \frac{1}{M(b-a)}$, the integral equation $y(t) = f(t) + \lambda \int_{a}^{b} K(t,s)y(s)ds$ has a unique solution in $C([a,b])$.
Here's my attempt:
Consider the map $T: C([a,b]) \to C([a,b])$, $(T(y)(t)) = f(t) + \lambda \int_{a}^{b} K(t,s) y(s) ds$.
$\|T(f)-T(g)\| = \sup_{x \in [a,b]} \| T(f(x)) - T(g(x)) \| = \sup_{ s \in [a,b]} \| \lambda \int_{a}^{b} K(t,s) f(s) ds - \lambda \int_{a}^{b} K(t,s)g(s)ds \| < \frac{1}{(b-a)} \sup_{s \in [a,b]} \| \int_{a}^{b}f(s)ds - \int_{a}^{b}g(s)ds \| \leq \frac{1}{b-a} (b-a) \sup_{ s \in [a,b] } \|f(s)-g(s)\| = \| f-g \|$.
Then $T$ is a contraction map in the complete space $C([a,b])$, hence there is a unique fixed point $y(t)$ satisfying $T(y(t))=y(t) = f(t) + \lambda \int_{a}^{b} K(t,s) y(s) ds$, so the equation has a unique solution as required. $\square $
Please verify if my proof makes sense. Any help would be appreciated. Thanks.