In mathematical analysis, a bump function (also called a test function) is a function <math>f : \Reals^n \to \Reals</math> on a Euclidean space <math>\Reals^n</math> which is both smooth (in the sense of having continuous derivatives of all orders) and compactly supported. The set of all bump functions with domain <math>\Reals^n</math> forms a vector space, denoted <math>\mathrm{C}^\infty_0(\Reals^n)</math> or <math>\mathrm{C}^\infty_\mathrm{c}(\Reals^n).</math> The dual space of this space endowed with a suitable topology is the space of distributions.
ExamplesEdit
The function <math>\Psi : \mathbb{R} \to \mathbb{R}</math> given by <math display="block">\Psi(x) = \begin{cases} \exp\left( \frac{1}{x^2 -1}\right), & \text{ if } |x| < 1, \\ 0, & \text{ if } |x| \geq 1, \end{cases}</math>
is an example of a bump function in one dimension. Note that the support of this function is the closed interval <math> [-1,1]</math>. In fact, by definition of support, we have that <math> \operatorname{supp}(\Psi):=\overline{\{x\in \mathbb{R}:\Psi(x)\neq 0\}} =\overline{(-1,1)}</math>, where the closure is taken with respect the Euclidean topology of the real line. The proof of smoothness follows along the same lines as for the related function discussed in the Non-analytic smooth function article. This function can be interpreted as the Gaussian function <math>\exp\left(-y^2\right)</math> scaled to fit into the unit disc: the substitution <math>y^2 = {1} / {\left(1 - x^2\right)}</math> corresponds to sending <math>x = \pm 1</math> to <math>y = \infty.</math>
A simple example of a (square) bump function in <math>n</math> variables is obtained by taking the product of <math>n</math> copies of the above bump function in one variable, so <math display="block">\Phi(x_1, x_2, \dots, x_n) = \Psi(x_1) \Psi(x_2) \cdots \Psi(x_n).</math>
A radially symmetric bump function in <math>n</math> variables can be formed by taking the function <math>\Psi_n : \Reals^n \to \Reals</math> defined by <math>\Psi_n(\mathbf{x})=\Psi(|\mathbf{x}|)</math>. This function is supported on the unit ball centered at the origin.
For another example, take an <math>h</math> that is positive on <math>(c, d)</math> and zero elsewhere, for example
- <math>h(x) = \begin{cases}
\exp\left(-\frac{1}{(x-c)(d-x)}\right),& c < x < d \\ 0,& \mathrm{otherwise} \end{cases}</math>.
Smooth transition functions
Consider the function
- <math>f(x)=\begin{cases}e^{-\frac{1}{x}}&\text{if }x>0,\\ 0&\text{if }x\le0,\end{cases}</math>
defined for every real number x.
The function
- <math>g(x)=\frac{f(x)}{f(x)+f(1-x)},\qquad x\in\mathbb{R},</math>
has a strictly positive denominator everywhere on the real line, hence g is also smooth. Furthermore, g(x) = 0 for x ≤ 0 and g(x) = 1 for x ≥ 1, hence it provides a smooth transition from the level 0 to the level 1 in the unit interval [0, 1]. To have the smooth transition in the real interval [a, b] with a < b, consider the function
- <math>\mathbb{R}\ni x\mapsto g\Bigl(\frac{x-a}{b-a}\Bigr).</math>
For real numbers Template:Math, the smooth function
- <math>\mathbb{R}\ni x\mapsto g\Bigl(\frac{x-a}{b-a}\Bigr)\,g\Bigl(\frac{d-x}{d-c}\Bigr)</math>
equals 1 on the closed interval [b, c] and vanishes outside the open interval (a, d), hence it can serve as a bump function.
Caution must be taken since, as example, taking <math>\{a =-1\} < \{b = c =0\} < \{d=1\}</math>, leads to:
- <math>q(x)=\frac{1}{1+e^{\frac{1-2|x|}{x^2-|x|}}}</math>
which is not an infinitely differentiable function (so, is not "smooth"), so the constraints Template:Math must be strictly fulfilled.
Some interesting facts about the function:
- <math>q(x,a)=\frac{1}{1+e^{\frac{a(1-2|x|)}{x^2-|x|}}}</math>
Are that <math>q\left(x,\frac{\sqrt{3}}{2}\right)</math> make smooth transition curves with "almost" constant slope edges (a bump function with true straight slopes is portrayed this Another example).
A proper example of a smooth Bump function would be:
- <math>u(x)=\begin{cases} 1,\text{if } x=0, \\ 0, \text{if } |x|\geq 1, \\ \frac{1}{1+e^{\frac{1-2|x|}{x^2-|x|}}}, \text{otherwise}, \end{cases}</math>
A proper example of a smooth transition function will be:
- <math>w(x)=\begin{cases}\frac{1}{1+e^{\frac{2x-1}{x^2-x}}}&\text{if }0<x<1,\\ 0&\text{if } x\leq 0,\\ 1&\text{if } x\geq 1,\end{cases}</math>
where could be noticed that it can be represented also through Hyperbolic functions:
- <math>\frac{1}{1+e^{\frac{2x-1}{x^2-x}}} = \frac{1}{2}\left( 1-\tanh\left(\frac{2x-1}{2(x^2-x)} \right) \right)</math>
Existence of bump functionsEdit
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 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>Template:Sfn 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.Template:Sfn 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>Template:Sfn 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 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>Template:Sfn 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>Template:Sfn Moreover, for any non-negative integers <math>p_1, \ldots, p_n \in \Z,</math>Template:Sfn <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>th 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>Template:Sfn 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>
Properties and usesEdit
While bump functions are smooth, the identity theorem prohibits their being analytic unless they vanish identically. Bump functions are often used as mollifiers, as smooth cutoff functions, and to form smooth partitions of unity. They are the most common class of test functions used in analysis. The space of bump functions is closed under many operations. For instance, the sum, product, or convolution of two bump functions is again a bump function, and any differential operator with smooth coefficients, when applied to a bump function, will produce another bump function.
If the boundaries of the Bump function domain is <math>\partial x,</math> to fulfill the requirement of "smoothness", it has to preserve the continuity of all its derivatives, which leads to the following requirement at the boundaries of its domain: <math display="block">\lim_{x \to \partial x^\pm} \frac{d^n}{dx^n} f(x) = 0,\,\text { for all } n \geq 0, \,n \in \Z</math>
The Fourier transform of a bump function is a (real) analytic function, and it can be extended to the whole complex plane: hence it cannot be compactly supported unless it is zero, since the only entire analytic bump function is the zero function (see Paley–Wiener theorem and Liouville's theorem). Because the bump function is infinitely differentiable, its Fourier transform must decay faster than any finite power of <math>1/k</math> for a large angular frequency <math>|k|.</math><ref>K. O. Mead and L. M. Delves, "On the convergence rate of generalized Fourier expansions," IMA J. Appl. Math., vol. 12, pp. 247–259 (1973) {{#invoke:doi|main}}.</ref> The Fourier transform of the particular bump function <math display="block">\Psi(x) = e^{-1/(1-x^2)} \mathbf{1}_{\{|x|<1\}}</math> from above can be analyzed by a saddle-point method, and decays asymptotically as <math display="block">|k|^{-3/4} e^{-\sqrt{|k|}}</math> for large <math>|k|.</math><ref>Steven G. Johnson, Saddle-point integration of C∞ "bump" functions, arXiv:1508.04376 (2015).</ref>
See alsoEdit
CitationsEdit
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