Editing Bump function (section)

==Existence of bump functions==
[[Image:Venn diagram of three sets.svg|thumb|An illustration of the sets in the construction.|100x100px]]

It is possible to construct bump functions "to specifications". Stated formally, if <math>K</math> is an arbitrary [[compact set]] in <math>n</math> dimensions and <math>U</math> is an [[open set]] containing <math>K,</math> there exists a bump function <math>\phi</math> which is <math>1</math> on <math>K</math> and <math>0</math> outside of <math>U.</math> Since <math>U</math> can be taken to be a very small neighborhood of <math>K,</math> this amounts to being able to construct a function that is <math>1</math> on <math>K</math> and falls off rapidly to <math>0</math> outside of <math>K,</math> while still being smooth.

'''Bump functions defined in terms of convolution'''

The construction proceeds as follows. One considers a compact neighborhood <math>V</math> of <math>K</math> contained in <math>U,</math> so <math>K \subseteq V^\circ\subseteq V \subseteq U.</math> The [[indicator function|characteristic function]] <math>\chi_V</math> of <math>V</math> will be equal to <math>1</math> on <math>V</math> and <math>0</math> outside of <math>V,</math> so in particular, it will be <math>1</math> on <math>K</math> and <math>0</math> outside of <math>U.</math> This function is not smooth however. The key idea is to smooth <math>\chi_V</math> a bit, by taking the [[convolution]] of <math>\chi_V</math> with a [[mollifier]]. The latter is just a bump function with a very small support and whose integral is <math>1.</math> Such a mollifier can be obtained, for example, by taking the bump function <math>\Phi</math> from the previous section and performing appropriate scalings.

'''Bump functions defined in terms of a function <math>c : \Reals \to [0, \infty)</math> with support <math>(-\infty, 0]</math>'''

An alternative construction that does not involve convolution is now detailed. 
It begins by constructing a smooth function <math>f : \Reals^n \to \Reals</math> that is positive on a given open subset <math>U \subseteq \Reals^n</math> and vanishes off of <math>U.</math>{{sfn|Nestruev|2020|pp=13-16}} This function's support is equal to the closure <math>\overline{U}</math> of <math>U</math> in <math>\Reals^n,</math> so if <math>\overline{U}</math> is compact, then <math>f</math> is a bump function. 

Start with any smooth function <math>c : \Reals \to \Reals</math> that vanishes on the negative reals and is positive on the positive reals (that is, <math>c = 0</math> on <math>(-\infty, 0)</math> and <math>c > 0</math> on <math>(0, \infty),</math> where continuity from the left necessitates <math>c(0) = 0</math>); an example of such a function is <math>c(x) := e^{-1/x}</math> for <math>x > 0</math> and <math>c(x) := 0</math> otherwise.{{sfn|Nestruev|2020|pp=13-16}} 
Fix an open subset <math>U</math> of <math>\Reals^n</math> and denote the usual [[Euclidean norm]] by <math>\|\cdot\|</math> (so <math>\Reals^n</math> is endowed with the usual [[Euclidean metric]]). 
The following construction defines a smooth function <math>f : \Reals^n \to \Reals</math> that is positive on <math>U</math> and vanishes outside of <math>U.</math>{{sfn|Nestruev|2020|pp=13-16}} So in particular, if <math>U</math> is relatively compact then this function <math>f</math> will be a bump function.

If <math>U = \Reals^n</math> then let <math>f = 1</math> while if <math>U = \varnothing</math> then let <math>f = 0</math>; so assume <math>U</math> is neither of these. Let <math>\left(U_k\right)_{k=1}^\infty</math> be an [[Cover (topology)|open cover]] of <math>U</math> by open balls where the open ball <math>U_k</math> has radius <math>r_k > 0</math> and center <math>a_k \in U.</math> Then the map <math>f_k : \Reals^n \to \Reals</math> defined by <math>f_k(x) = c\left(r_k^2 - \left\|x - a_k\right\|^2\right)</math> is a smooth function that is positive on <math>U_k</math> and vanishes off of <math>U_k.</math>{{sfn|Nestruev|2020|pp=13-16}} 
For every <math>k \in \mathbb{N},</math> let
<math display="block">M_k = \sup \left\{\left|\frac{\partial^p f_k}{\partial^{p_1} x_1 \cdots \partial^{p_n} x_n}(x)\right| ~:~ x \in \Reals^n \text{ and } p_1, \ldots, p_n \in \Z \text{ satisfy } 0 \leq p_i \leq k \text{ and } p = \sum_i p_i\right\},</math>
where this [[supremum]] is not equal to <math>+\infty</math> (so <math>M_k</math> is a non-negative real number) because <math>\left(\Reals^n \setminus U_k\right) \cup \overline{U_k} = \Reals^n,</math> the partial derivatives all vanish (equal <math>0</math>) at any <math>x</math> outside of <math>U_k,</math> while on the compact set <math>\overline{U_k},</math> the values of each of the (finitely many) partial derivatives are (uniformly) bounded above by some non-negative real number.<ref group="note">The partial derivatives <math>\frac{\partial^p f_k}{\partial^{p_1} x_1 \cdots \partial^{p_n} x_n} : \Reals^n \to \Reals</math> are continuous functions so the image of the compact subset <math>\overline{U_k}</math> is a compact subset of <math>\Reals.</math> The supremum is over all non-negative integers <math>0 \leq p = p_1 + \cdots + p_n \leq k</math> where because <math>k</math> and <math>n</math> are fixed, this supremum is taken over only finitely many partial derivatives, which is why <math>M_k < \infty.</math></ref> 
The series 
<math display="block">f ~:=~ \sum_{k=1}^{\infty} \frac{f_k}{2^k M_k}</math>
converges uniformly on <math>\Reals^n</math> to a smooth function <math>f : \Reals^n \to \Reals</math> that is positive on <math>U</math> and vanishes off of <math>U.</math>{{sfn|Nestruev|2020|pp=13-16}} 
Moreover, for any non-negative integers <math>p_1, \ldots, p_n \in \Z,</math>{{sfn|Nestruev|2020|pp=13-16}}
<math display="block">\frac{\partial^{p_1+\cdots+p_n}}{\partial^{p_1} x_1 \cdots \partial^{p_n} x_n} f ~=~ \sum_{k=1}^{\infty} \frac{1}{2^k M_k} \frac{\partial^{p_1+\cdots+p_n} f_k}{\partial^{p_1} x_1 \cdots \partial^{p_n} x_n}</math>
where this series also converges uniformly on <math>\Reals^n</math> (because whenever <math>k \geq p_1 + \cdots + p_n</math> then the <math>k</math><sup>th</sup> term's absolute value is <math>\leq \tfrac{M_k}{2^k M_k} = \tfrac{1}{2^k}</math>). This completes the construction. 

As a corollary, given two disjoint closed subsets <math>A, B</math> of <math>\Reals^n,</math> the above construction guarantees the existence of smooth non-negative functions <math>f_A, f_B : \Reals^n \to [0, \infty)</math> such that for any <math>x \in \Reals^n,</math> <math>f_A(x) = 0</math> if and only if <math>x \in A,</math> and similarly, <math>f_B(x) = 0</math> if and only if <math>x \in B,</math> then the function 
<math display="block">h ~:=~ \frac{f_A}{f_A + f_B} : \Reals^n \to [0, 1]</math> 
is smooth and for any <math>x \in \Reals^n,</math> <math>h(x) = 0</math> if and only if <math>x \in A,</math> <math>h(x) = 1</math> if and only if <math>x \in B,</math> and <math>0 < h(x) < 1</math> if and only if <math>x \not\in A \cup B.</math>{{sfn|Nestruev|2020|pp=13-16}} 
In particular, <math>h(x) \neq 0</math> if and only if <math>x \in \Reals^n \smallsetminus A,</math> so if in addition <math>U := \Reals^n \smallsetminus A</math> is relatively compact in <math>\Reals^n</math> (where <math>A \cap B = \varnothing</math> implies <math>B \subseteq U</math>) then <math>h</math> will be a smooth bump function with support in <math>\overline{U}.</math>