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Consider the mapping $T: C[0,1]\to C[0,1]$ by

$$ Tf(x)=\exp\!{\left[x-\int_0^{1}f(y)e^{-y}dy\right]} $$

I need to show that has unique fixed point but does not have contraction.

What I have been thinking is doing the following, If I can show that the composition of $n$ times $T$ has unique fixed point then $T$ has fixed point, but I dont know if it its correct. For the contraction I think the metric $d(f,g)=\max\limits_{0\leq x\leq 1}|f(x)-g(x)|$ might work but dont know.

Any help I will appreciate.

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    $\begingroup$ $T f(x) = K_f e^x$, hence the only fixed points are of the form $K e^x$, for a constant $K$ that fulfills $\exp\left(\frac{1-e^2}{2e^2}K\right)=K$ $\endgroup$ – Jack D'Aurizio May 2 '17 at 18:57
  • $\begingroup$ There is only one $K$ that fulfills such constraint, so there is a single fixed point. I find it difficult to prove that something is not a contraction without knowing which metric are we considering. $\endgroup$ – Jack D'Aurizio May 2 '17 at 18:59
  • $\begingroup$ I think that metric works and if I consider $g$ to be identically 0 and $f$ to be -exp(x), probably will get some contradiction. $\endgroup$ – Thetexan May 2 '17 at 19:02

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