I'm shoring up my understanding of basic real analysis and encountered this problem.
Consider the operator $$K(x)(t) = \int_0^2 B(t,s)x(s) ds + g(t)$$
where $B$ and $g$ are continuous and $|B(t,s)| < 0.25$ for $t,s \in [0,2]$ acting in the Banach space $C[0,2]$ with norm $\|x\| = \sup_{t \in [0,2]}|x(t)|$. Show using Banach contraction principle that $K$ has a fixed point.
In the solution, it states that we first must show that $K$ is a contraction, and then we must find a closed set that $K$ maps to itself. Proving that it's a contraction is clearly necessary, but why is the second step necessary?
It is not directly clear to me that if $K : C[0, 2] \to A$, then $A \subseteq C[0, 2]$. In order to apply the Banach fixed point theorem you need to know that $K$ maps a complete metric space to itself.
If you can show that there is a closed $B \subseteq C[0, 2]$ such that $K(B) \subseteq B$ then you could apply the theorem. This is because any closed subset of a complete metric space is also a complete metric space. Note, $B$ may be all of $C[0, 2]$ as well.