I have to state if a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a contraction on the interval $I\subset \mathbb{R}$ and say if it admits fixed points.
The function is $f(x)=\left\{ \begin{array}{ll} sin\frac{1}{x}\:if\: x\neq 0\\ 0 \:\:\:\:\:\:\:\:if\:x=0 \end{array} \right.$
defined on $I = \mathbb{R}$.
How should I proceed?
The function is not a contraction on $\mathbb{R}$ : in fact
$$1=\left |f \left (\frac{2}{\pi} \right)-f(0) \right |>\left |\frac{2}{\pi}-0 \right |=\frac{2}{\pi}$$
Now, $\sin \left(\frac{1}{x}\right )-x$ is continuous in the interval $\left [\frac{1}{2\pi n + \frac{\pi}{2}}, \frac{1}{2\pi n-\frac{\pi}{2}}, \right ]$ for every $n \in \mathbb{N}$. But $$\sin \left (2\pi n + \frac{\pi}{2}\right )-\frac{1}{2\pi n + \frac{\pi}{2}}>0$$ And
$$\sin \left (2\pi n - \frac{\pi}{2}\right )-\frac{1}{2\pi n - \frac{\pi}{2}}<0$$
By the intermediate value theorem there exists $\xi \in \left [\frac{1}{2\pi n + \frac{\pi}{2}}, \frac{1}{2\pi n-\frac{\pi}{2}}, \right ]$ such that $\xi=\sin \left ( \frac{1}{\xi} \right )$ for every $n \in \mathbb{N}$; hence, there are infinitely many fixed points of $f$ in $\mathbb{R}$.