Given \(a< b\), there exists a smooth function \(h:\mathbb{R}\to \mathbb{R}\) such that \(h(t)\equiv 1\) for \(t\le a\), \(h(t)\in (0,1)\) for \(t\in (a,b)\) and \(h(t)\equiv 0\) for \(t\ge b\).

Such a function is called smooth transition function or cutoff function.

Proof

Consider

\begin{equation*} f(t)= \begin{cases} e^{-1/t}, & t>0 \\ 0, & t\le 0. \end{cases} \end{equation*}

This function is smooth [1, Lemma 2.20]. Then

\begin{equation*} h_{0,1}(t)=\frac{f(1-t)}{f(1-t)+f(t)}, \end{equation*}

satisfies the stated properties for \(b=1\) and \(a=0\). For other values set

\[ h_{a,b}(t)=h_{0,1}\bigl(\tfrac{t-a}{b-a}\bigr). \]

See also Link to heading

References Link to heading

  1. J. Lee, Introduction to Smooth Manifolds. New York ; London: Springer, 2013.