1

I want a family of smooth cutoff functions that satisfy the following:

  • $\psi_k:[0, \infty)\to [0, 1]$ (for all positive integers $k>4$)
  • $\psi_k([0, k])= 0 $
  • $\psi_k'(2k)>k$
  • $\psi_{k+1}(t)\leq \psi_k(t)$, for every $t$ in the domain.

Any help?

Micael
  • 198
  • Do you know how to construct a function for the first three properties? There is nothing particularly novel in these requirements. –  Jan 05 '16 at 06:41
  • No, @Normal, that's why I'm asking for some help. Can you show me how can I construct a function satisfying the first three properties? – Micael Jan 05 '16 at 10:12

1 Answers1

0

Suppose that you are given a $f\in C^\infty(\mathbb R,[0,1])$ function such that $f(\mathbb{R}_)={0}$ , $f([1,\infty))={1}$, and $f^\prime(\frac12)>1$. You can find one here. Then, take $$ \psi_{k}=f\left(k\left((x-2k+\frac1{2k}\right)\right). $$ you find that, $\psi_k\in C^\infty(\mathbb R,[0,1])$, $\psi_k(x)=0$ for all $x\leq 2k-\frac1{2k}$, $\psi_k(x)=1$ for all $x>2k+\frac1{2k}$, and $\psi^\prime(2k)=kf^\prime(\frac12)>k.$

Since $\psi_k=0$ for $x\leq{2k-1}$, and $\psi_k=1$ for $x\geq{2k+1}$, $\psi_{k+1}=0$ on $[0,2k+1]$, therefore $\psi_{k+1}\leq \psi_k$.

username
  • 198