# There exists exactly one function satisfying $f(x)=x+\frac{1}{2}\sin(f(x))$ for all $x\in[0,1]$

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.

• You should include more context. What does the Banach theorem say, and what would that look like in the situation at hand. – zhw. Oct 9 '20 at 23:26

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.
• 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 – user833927 Oct 9 '20 at 23:47