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.

  • $\begingroup$ 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 $x=n$? $\endgroup$ – MasterYoda Aug 24 '18 at 0:41
  • $\begingroup$ @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. $\endgroup$ – user586256 Aug 25 '18 at 9:54
  • $\begingroup$ 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 $x\in \mathbb{R}^+.$ $\endgroup$ – user586256 Aug 25 '18 at 10:14

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