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Let $h:[0,1] \times [0,1] \times \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Assume that there is a constant $0<c<1$ such that

$|h(x,y,s)-h(s,y,t)| \leq c|s-t|$

for all $x,y \in [0,1]$ and $s,t \in \mathbb{R}$. Prove that there exists a unique continuous function $u: [0,1] \rightarrow \mathbb{R}$ such that

$u(x)=\lmoustache^{1}_{0} h(x,y,u(y))dy$.

The hint says define a mapping of the vector space $C[0,1]$ that is a strict contraction for the supremum norm. How the hint helps to prove? I really need help on this problem.

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Hint: The contraction you need is a mapping for which the $u$ you are looking for will be a fixed point. Try to find a mapping for which $u$ is a fixed point, so find some formula for which $u=F(u)$ hold.

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  • $\begingroup$ could you please explain a little bit more? I am not sure how this relates to the first part of problem $\endgroup$
    – user220055
    Feb 20, 2014 at 7:55
  • $\begingroup$ Hint number two: In your question, you already wrote one equation of the type $u=F(u)$. Also, when you have a contraction $F$, use Banach's fixed point theorem. $\endgroup$
    – 5xum
    Feb 20, 2014 at 7:58

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