> [!definition] > > Let $X$ be a [[n-Manifold|topological manifold]], $E, F \subset X$ be [[Closed Set|closed sets]] with $E \subset F$. A **smooth bump function** is a [[Space of Smooth Functions|smooth function]] $f: X \to \real$ with $f = 1$ on $E$ and $F = 0$ outside of $F$. # Construction on Real > [!definition] > > The function $f: \real \to \real$ with > $ > f(x) = \begin{cases} > e^{-1/t} &t > 0 \\ > 0 &t \le 0 > \end{cases} > $ > is smooth. > > *Proof*. Since $f$ is a composition of smooth functions outside of $0$, it's sufficient to show that $f$ is smooth at $0$. > > We claim that $\frac{d^k}{dt^k}f(t) = p_k(t)\frac{e^{-1/t}}{t^{2k}}$, where $p_k(t)$ is a polynomial of degree up to $k$. This holds for $k = 0$. Now suppose that this is true for some $k \ge 0$, then by the product rule, > $ > \begin{align*} > \frac{d^{k + 1}}{dt^{k + 1}}f(t) &= p_k'(t) \frac{e^{-1/t}}{t^{2k}} + p_k(t)\frac{e^{-1/t}}{t^{2(k + 1)}} - 2kp_k(t)\frac{e^{-1/t}}{t^{2k + 1}} \\ > &= (t^2p'_k(t) + p_k(t) - 2ktp_k(t))\frac{e^{-1/t}}{t^{2(k + 1)}} > \end{align*} > $ > From here, we can evaluate the limit > $ > \begin{align*} > \lim_{t \downto 0}\frac{\frac{d^{k}}{dt^k}f(t)}{t} &= \lim_{t \downto 0}\frac{d^{k + 1}}{dt^{k + 1}}f(t) \\ > &= p_{k + 1}(0) \cdot \lim_{t \downto 0}\frac{e^{-1/t}}{t^{2(k + 1)}} = 0 > \end{align*} > $ > so $\frac{d^k}{dt^k}f(t) = 0$ for all $k \in \nat$. > [!definition] The Cutoff Function > > Let $x, y \in \real$ with $x < y$, then there exists a smooth function $h: \real \to [0, 1]$ such that $h(t) = 1$ for all $t \le x$ and $h(t) = 0$ for all $t \ge y$. > > *Proof*. Let > $ > h(t) = \frac{f(y - t)}{f(y - t) + f(t - x)} > $ > then $h(t) = 0$ for all $t \ge y$, and $h(t) = 1$ for all $t \le x$. The denominator is never smooth because at least one of the two functions in the sum must be strictly positive. > [!definition] The Smooth Bump Function > > Let $x, y \in \real$, then there exists a smooth function $H: \real^n \to [0, 1]$ such that $H = 1$ on $B(0, x)$ and $H = 0$ outside of $B(0, y)$. > > *Proof*. Let $H = h(\abs{x})$, then $H$ is smooth everywhere except $0$ as a composition of smooth functions. However as $H = 1$ on $B(0, x)$, it is smooth there as well.