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Using the Banach fixed-point theorem, how do I prove that there exists exactly one continuous function $f:[0,1]\to\Bbb{R}$ that satisfies $$f(x)=x+\frac{1}{2}\sin(f(x))$$ for all $x\in[0,1]$.

I'm currently doing some analysis problems for practice and I really don't know how to solve this. Thank you for any solutions or hints to this problem.

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    $\begingroup$ You should include more context. What does the Banach theorem say, and what would that look like in the situation at hand. $\endgroup$ – zhw. Oct 9 '20 at 23:26
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Let $X$ be the Banach space $C([0,1])$ (continuous functions from $[0,1]$ to $\mathbb{R}$) with the usual norm $$ \| f-g \| = \sup_{x \in [0,1]} |f(x)-g(x)|. $$ Define $T: X \to X$ by $$ (Tf)(x) = x + \frac12 \sin(f(x)). $$ If we show that $T$ is a contraction, the Banach fixed point theorem implies that there is exactly one fixed point of $T$, which is what you want to prove.

We can estimate \begin{align*} \| Tf - Tg \| &= \sup_{x \in [0,1]} \left| x+ \frac12 \sin(f(x)) - \left( x + \frac12 \sin (g(x)) \right) \right| = \frac12 \sup_{x \in [0,1]} |\sin(f(x)) - \sin(g(x))| \\ &\leq \frac12 \sup_{x \in [0,1]} |f(x) - g(x)| = \frac12 \| f-g \|. \end{align*} The inequality going from the first line to the second comes from the fact that $|\sin y - \sin z| \leq |y-z|$ for any real $y,z$ (this can be proved using the mean value theorem, for example).

Therefore $T$ is a contraction mapping as desired.

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    $\begingroup$ Thanks a lot. So if I understand correctly, looking at the definition of Banach fixed point theorem, i.e. for a contraction $T$: $T(x^*)=x^*$, where $x^*\in X$ is unique, here $f(x)$ is our $x^*$ so to speak and so it is unique. This really helped me understand this topic a bit better $\endgroup$ – user833927 Oct 9 '20 at 23:47

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