3

How to construct a smooth function with compact support satisfying $$ f(x)+f(x^{-1})=1 $$

For example, let $$ g(x)=\left\{\begin{array}{ll} 0,&\mbox{if $x\leq 0$},\\ \frac{1}{1+e^{\frac{1}{x}-\frac{1}{1-x}}},&\mbox{if $0<x<1$},\\ 1,&\mbox{if $x\geq 1$}, \end{array} \right. $$ Then $$ G(x)=g\left(\frac{x-a}{b-a}\right)g\left(\frac{d-x}{d-c}\right) $$ is a smooth function which equals 1 on [b,c] and vanishes outside (a,d). However, I don't know how to modify it to satisfy $$ G(x)+G(x^{-1})=1 $$

Sun
  • 31
  • What have you tried? Can you do this without the constraint that $f(x)+f(x^{-1})=1$? – Milo Brandt Dec 10 '14 at 01:25
  • Yes, i can do this without the constraint $f(x)+f(x^{-1})=1$. – Sun Dec 10 '14 at 06:19
  • For example, let $$ g(x)=\left{\begin{array}{ll} 0,&\mbox{if $x\leq 0$},\ \frac{1}{1+e^{\frac{1}{x}-\frac{1}{1-x}}},&\mbox{if $0<x<1$},\ 1,&\mbox{if $x\geq 1$}. \end{array} \right. $$ Then $$G(x)=g\left(\frac{x-a}{b-a}\right)g\left(\frac{d-x}{d-c}\right)$$ is a smooth function which equals 1 on $[b,c]$ and vanishes outside $(a,d)$. – Sun Dec 10 '14 at 06:23

1 Answers1

3

Let $g : \Bbb{R} \to [0, 1]$ be a smooth function such that

  • $g(x) = 1$ for $x < -1$,
  • $g(x) = 0$ for $x > 1$,
  • $g(x) + g(-x) = 1$.

For example, we may define

$$ g(x) = \int_{x}^{\infty} C \exp\{-(1-t^{2})^{-1}\} \mathbf{1}_{(-1,1)}(t) \, dt, $$

where $C = \left( \int_{-1}^{1} \exp\{-(1-t^{2})\} \, dt \right)^{-1}$ is the normalization constant.

Now define

$$ f(x) = g(\log|x|) $$

with the convention that $f(0) = 1$. Then $f$ is a smooth function supported on $[-e, e]$ and satisfies all the desired properties.

Sangchul Lee
  • 167,468