# Problem in proving fixed point theorem

Let $f(x)=x^r, 1<r<\infty, x\in \mathbb{R^+}=[0,\infty) ~and ~n\in \mathbb{N}.$ Define $$\pi(x)= 1 ~~if~~ x\leq n ~and =0 ~if~ x< (n+1).$$ Then for any $x,y\in\mathbb{R},~~$ I want to prove that $$\|\pi(\|x\|_{L^r})x^r-\pi(\|y\|_{L^r})y^r\|_{L^1}\leq C\|x-y\|_{L^r},$$ where $C$ is constant and strictly less than $1.$

Can Any one help me to prove it? Any help will be appreciated.

• I might be wrong, but I think we run into a problem with the $L^r$- norm when $|x|>1$ for large $r$. Can you verify if the domain for $x$ should be $x\in\mathbb{R}$? Also, is there any information about the case that $x=n$? – MasterYoda Aug 24 '18 at 0:41
• @Master Yoda , I think the equality sign is with $n$ , which I have edited now. And here we have taken $L^1$ on left because we have $rth$ power inside the norm, which may manage the $L^r$ norm on the right side. – user586256 Aug 25 '18 at 9:54
• Or we can also take the cut off function as .> $\pi(x)=1 ~~if 0 \leq x\leq n,~~ =n+1-x ~~if n<y<n+1~~ and =0$ ~~otherwise for $x\in \mathbb{R}^+.$ – user586256 Aug 25 '18 at 10:14