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I was recently presented this within the context of topological spaces:

I am asked to show that there exists a unique continuous function $ f\colon \left[0,\frac{1}{2}\right] \rightarrow \Bbb R $ such that for all $ x \in \left[0,\frac{1}{2}\right] $ the following equality holds: $$ f(x) = \frac{x}{2}\sin f(x) +\sin\left( f\left(\frac{x}{2}\right)\right)+1. $$ I can do this because it is simply looking for a fixed point of the contracting operator $ T(f(x)) = \frac{x}{2}\sin(f(x))+\sin(f(\frac{x}{2}))+1. $ I can show it to be contracting for $ x \in [0,\frac{1}{2}] $ and the space of continuous functions $ C\left[0,\frac{1}{2}\right] $ is of course complete, so Banach's fixed point theorem holds.

Now the hard part: the exact same problem as before, only now the interval is $[0,1)$. How do I do it? I know that $C[0,1)$ is not complete and the operator itself is not a contraction, so Banach's fixed point theorem is out the window. Is there another solution?

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    $\begingroup$ @Omnomnomnom: I tried but how can I show it exists in C[0,1]? It's complete sure but the contraction? How can I show it is a contraction? The 1/2 there was OK now it is 1 which makes it harder to show contraction $\endgroup$ – kroner Aug 4 '15 at 12:17
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    $\begingroup$ I apologize I had a mistake with the parentheses now it makes more sense sorry $\endgroup$ – kroner Aug 4 '15 at 12:45
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Hint.

You can use Banach fixed point theorem for the operator $T$ restricted to the space $\mathcal C([0,1-\frac{1}{n}])$ for $n \ge 2$ as $T$ is a contraction on this space. For each $n$ you get a unique continuous function $f_n$ defined on $[0,1-\frac{1}{n}]$ solution of the Banach fixed point problem (the one defined by $T$ operator). By unicity, if $n < m$, $f_m$ extends $f_n$. And for all $x \in [0,1)$, you can find a solution of the Banach fixed point problem for $1-\frac{1}{n} > x$.

Finally you define a solution element $f$ of $\mathcal C([0,1))$ by $$f=\bigcup_{n \ge 2} f_n$$ which defines a function of $[0,1)$ as $f_n \subset f_m$ for $n \le m$.

The idea is similar to the one used to define maximal solutions of differential equations.

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    $\begingroup$ Thanks but I do not understand something: I understand your principle and reasoning but why do we necessarily have that this sequence converges, and why must it converge to a fixed point of the operator, obviously it is defined for all $ x \in [0,1) $, but why is it a fixed point and why does it converge to one? $\endgroup$ – kroner Aug 4 '15 at 13:10
  • $\begingroup$ Or rather, how do we define the solution on [0,1) based on the $ f_n $ $\endgroup$ – kroner Aug 4 '15 at 13:14
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    $\begingroup$ See the efit of my response for a precise definition of $f$. $\endgroup$ – mathcounterexamples.net Aug 4 '15 at 13:19
  • $\begingroup$ Got it thanks all clear now $\endgroup$ – kroner Aug 4 '15 at 13:20
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I am affraid it is not a contraction on $C([0,1/2])$ with standard norm. Take arbitrary $L\in(0,1)$, then for $f=0, ~g=L$ we have $||Tf-Tg||_\infty=\frac{5}{4}\sin{L}$ and $||f-g||_\infty=L$. But $h(L)=\frac{5}{4}\sin{L}-L^2>0$ for $L\in(0,1)$, therefore $||Tf-Tg||_\infty>L||f-g||_\infty$ (case $L=0$ is trivial).

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