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Let $C$ be a non-empty, closed, convex set of a normed linear space $E$. If $F: C \rightarrow C$ is a contraction(i.e. $\Vert F(x)-F(y) \Vert \leq L\Vert x-y \Vert $ where $0 \leq L <1 \thinspace \forall \thinspace x,y \in C$)

I want to prove $F$ has a unique fixed point in $C$.

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Hint for the uniqueness. If $F(x)=x$ and $F(y)=y$ then $$\Vert x-y \Vert=\Vert F(x)-F(y) \Vert \leq L\Vert x-y \Vert.$$ Hint for the existence. If $E$ is complete then show that for $x_0\in C$, the recursive sequence $x_{n+1}=F(x_n)$ is a Cauchy sequence and therefore it is convergent.

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  • $\begingroup$ Here $E$ is not complete. $\endgroup$ – Manu Rohilla Jul 16 '17 at 9:51
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    $\begingroup$ @Manu Rohilla If $E$ is not complete then I think that the existence of the fixed point does not hold. $\endgroup$ – Robert Z Jul 16 '17 at 9:55
  • $\begingroup$ can you please give an example where fixed point theorem does not hold? $\endgroup$ – Manu Rohilla Jul 16 '17 at 9:58
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    $\begingroup$ @Manu Rohilla $F(x)=x^2$ is a contraction in $C=(0,1/2)$ with no fixed point. I'm trying to find a counterexample with $E=C$ a not-complete normed linear space. $\endgroup$ – Robert Z Jul 16 '17 at 10:09

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