Bump function

Remarks

Example: let $\psi(x) = e^{-1/(1 - x^2)}$ for $\abs{x} < 1$ and $\psi(x) = 0$ for $\abs{x} \ge 1$. Let $g(t) = e^{-1/t}$ for $t > 0$ and $g(t) = 0$ for $t \le 0$, which is smooth. Then $\psi(x) = g(1 - x^2)$ is a composition of smooth functions, so it is smooth. Its support is $[-1, 1]$, which is compact.

Non-example: the Gaussian $e^{-x^2}$ is smooth, but it is nonzero at every point, so its support $\R$ is not compact. The tent function $\max(0, 1 - \abs{x})$ has compact support $[-1, 1]$, but it is not differentiable at $0$ and $\pm 1$.

A bump function that is not identically zero is not analytic on $\R^n$, since an analytic function that vanishes on a nonempty open set of $\R^n$ is zero. For each compact set $K$ and open set $U \supseteq K$ in $\R^n$, some bump function equals $1$ on $K$, takes values in $[0, 1]$, and has support in $U$.