Is there a fixed point theorem I could use to solve this problem?

Question

let $E = C([0,1]),\,\,$ $K : E \to E, \,\, (Kf)(x) = \int_0^1K(x,y)f(y)dy$

also $\|K\| \leq a < 1$

I want to prove that there for $g \in E$ there exists a unique $f_g \in E$ that satisfies the following equation :

$f_g + Kf_g = g$

which is equivalent to showing that $T : E \to E,\,\,T(f) = g-Kf$ has a fixed point.

with what I have in hands I feel like there must be some theorem I'm missing.

any help will be greatly appreciated !

Answer 1

You can apply the Contraction mapping, a.k.a. Banach's Fixed Point Theorem. Given $f,h\in C([0,1])$, $$ \|Tf-Th\|\le\int_0^1|K(x,y)|\,|f(y)-h(y)|\,dy\le\|K\|\,\|f-h\|<a\,\|f-h\|, $$ with $0<a<1$.

Answer 2

This is not a fixed point theorem, but it is well-known that if $T:E\to E$ is a bounded linear operator with $\|I-T\|<1,$ then $T$ has a bounded inverse $$ T^{-1}=\sum_{k=0}^\infty (I-T)^k. $$ In your case, since $\|I-(I+K)\|<1$, we have $$ f_g = (I+K)^{-1}g,\quad \|f_g\|\leq \|(I+K)^{-1}\|\|g\|. $$