Construct a bump function

Question

I am looking for a explicit expression of a $C^{\infty}$ function $h$ with $$h(x) = 1$$ for $x\in\mathbb{R}^{n}\setminus B_{1/4}$ and $$h(x) = 0$$ for $x\in B_{1/8}$. Can someone help ?

Answer 1

I will construct a function $f$ such that $f(x)=1$ for $x\in B_{1/8}$ and $f(x)=0$ for $x\notin B_{1/4}$. You may then take $h=1-f$. Let $$\varphi(t):=\begin{cases}Ce^{\frac{1}{(16|t|)^2-1}} & |t|< \frac{1}{16}\\ 0 & |t|\geq \frac{1}{16}\end{cases} $$ where $C$ is a normalization constant: $$C:=\frac{1}{\int_{B_{1/16}}e^{\frac{1}{(16|t|)^2-1}}dt}$$

This function is smooth and supported in $B_{1/16}$, and $\int_{\mathbb{R}^n}\varphi(t)dt=1$. Now integrate over $B_{3/16}$.

$$f(x):=\int_{B_{3/16}}\varphi(x-t)dt $$

1. If $x\notin B_{1/4}$, i.e. $|x|> \frac{1}{4}$, then $$|x-t|\geq |x|-|t| > \frac{1}{4}-\frac{3}{16}=\frac{1}{16}$$ and so $\varphi(x-t)=0$ for all $t\in B_{3/16}$, and $f(x)=0$.

2. If $x\in B_{1/8}$, then for all $t$ such that $\varphi(x-t)\neq 0$, i.e. for all $x-t\in B_{1/16}$, we have

   $$ t=x-(x-t)\in B_{1/8}-B_{1/16}\subset B_{1/8+1/16}\subset B_{3/16}$$ and therefore $$f(x)=\int_{\mathbb{R}^n}\varphi(x-t)dt=1 $$