Proving integral transform using banach fixed-point theorem I'm currently working on the following problem:

Let $K: [0,1]^2 \to \mathbb{R}$ be continuous with $|K(x,y)| < 1$ for all $(x,y) \in [0,1]^2$. Prove the existence of a function $f \in C([0,1])$ s.t. 
  $$f(x) + \int_0^1 f(y) K(x,y) dy = e^{(x^2)}$$ 
  for all $x \in [0,1]$. Is $f$ also unique?

I'm given the hint that I should use the banach fixed point theorem; and additionally that I should use the compactness of $[0,1] \times [0,1]$ to show that $\max|K(x,y)| < 1$. Honestly, this just confuses me more, as the task already states that $|K(x,y)| < 1$, so I'm not sure how it'll help.
All in all I'm admittedly at a loss with this problem, and I'd appreciate it if somebody could give some input on this.
 A: What you want to do is start with an $f$, and iteratively construct functions that are fit the equation better.
Specifically, you let $f_1(x)\equiv e^{x^2}$ and then define $$f_{n+1}(x)=e^{x^2}-\int_0^1f_n(y)K(x,y)\,dy.$$
Finally, we want to show that $f_n$ converges uniformly towards a function, $f$, satisfies the properties in the question.
Note that $$|f_{n+1}(x)-f_n(x)|=\int_0^1(f_{n-1}-f_n)K(x,y)\,dy\le\int_0^1|f_n-f_{n-1}||K|\,dy<|f_n-f_{n-1}|$$
and thus by Banach fixed-point theorem, the functions converge uniformly towards a solution, $f$, to the equation. A straightforward application of uniform limit theorem tells us that $f\in C([0,1])$ (after verifying that every $f_n\in C([0,1])$ as well).
As for uniqueness, suppose $f,g$ satisfy the equation. Subtracting the two, we get that $$f(x)-g(x)+\int_0^1(f(y)-g(y))K(x,y)\,dy=0.$$ Let $h=f-g$, and $t\in[0,1]$ such that $|h(t)|$ is maximal (such a $t$ exists by compactness). If $|h(t)|>0$, then $$|h(t)|=\left|\int_0^1h(y)K(x,y)\,dy\right|\le\int_0^1|h(y)||K|\,dy<\int_0^1|h|\,dy,$$ where the strict inequality uses the fact that $|h(t)|>0.$ Therefore $h=0$ and thus $f=g$, so the equation does indeed have a unique solution.
