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

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 !

• You have not specified what kind of function $K(x,y)$ is what $\|K\|$ stands for. – Kavi Rama Murthy Dec 26 '18 at 11:53
• You need to fix the notation! You're using "$K$" for two different things... – David C. Ullrich Dec 26 '18 at 14:30

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