Inmathematical analysis, abump function (also called atest function) is afunction${\displaystyle f:{\mathbb{R}}^n\to \mathbb{R}}$ on aEuclidean space${\displaystyle {\mathbb{R}}^n}$ which is bothsmooth (in the sense of havingcontinuousderivatives of all orders) andcompactly supported. Theset of all bump functions withdomain${\displaystyle {\mathbb{R}}^n}$ forms avector space, denoted${\displaystyle {\mathrm{C}}_0^{\mathrm{\infty }}({\mathbb{R}}^n)}$ or${\displaystyle {\mathrm{C}}_{\mathrm{c}}^{\mathrm{\infty }}({\mathbb{R}}^n).}$ Thedual space of this space endowed with a suitabletopology is the space ofdistributions.

The function${\displaystyle \mathrm{\Psi }:\mathbb{R}\to \mathbb{R}}$ given by

is an example of a bump function in one dimension. Note that the support of this function is the closed interval${\displaystyle [-1,1]}$. In fact, by definition ofsupport, we have that${\displaystyle \mathrm{supp}(\mathrm{\Psi }):=\overline{\{x\in \mathbb{R}:\mathrm{\Psi }(x)\ne 0\}}=\overline{(-1,1)}}$, 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 theNon-analytic smooth function article. This function can be interpreted as theGaussian function${\displaystyle \exp \left(-y^2\right)}$ scaled to fit into the unit disc: the substitution${\displaystyle y^2=1/\left(1-x^2\right)}$ corresponds to sending${\displaystyle x=\pm 1}$ to${\displaystyle y=\mathrm{\infty }.}$

A simple example of a (square) bump function in${\displaystyle n}$ variables is obtained by taking the product of${\displaystyle n}$ copies of the above bump function in one variable, so$${\displaystyle \mathrm{\Phi }(x_1,x_2,\dots ,x_n)=\mathrm{\Psi }(x_1)\mathrm{\Psi }(x_2)\cdots \mathrm{\Psi }(x_n).}$$

A radially symmetric bump function in${\displaystyle n}$ variables can be formed by taking the function${\displaystyle {\mathrm{\Psi }}_n:{\mathbb{R}}^n\to \mathbb{R}}$ defined by${\displaystyle {\mathrm{\Psi }}_n(\mathbf{x})=\mathrm{\Psi }(|\mathbf{x}|)}$. This function is supported on the unit ball centered at the origin.

For another example, take an${\displaystyle h}$ that is positive on${\displaystyle (c,d)}$ and zero elsewhere, for example

.

Smooth transition functions

Consider the function

defined for everyreal numberx.

The function

${\displaystyle g(x)=\frac{f(x)}{f(x)+f(1-x)},x\in \mathbb{R},}$

has a strictly positive denominator everywhere on the real line, henceg is also smooth. Furthermore,g(x) = 0 forx ≤ 0 andg(x) = 1 forx ≥ 1, hence it provides a smooth transition from the level 0 to the level 1 in theunit interval [0, 1]. To have the smooth transition in the real interval [a,b] witha < b, consider the function

${\displaystyle \mathbb{R}\ni x\mapsto g(\frac{x-a}{b-a}).}$

For real numbersa <b <c <d, the smooth function

${\displaystyle \mathbb{R}\ni x\mapsto g(\frac{x-a}{b-a})g(\frac{d-x}{d-c})}$

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, leads to:

${\displaystyle q(x)=\frac{1}{1+e^{\frac{1-2|x|}{x^2-|x|}}}}$

which is not an infinitelydifferentiable function (so, is not "smooth"), so the constraintsa <b <c <d must be strictly fulfilled.

Some interesting facts about the function:

${\displaystyle q(x,a)=\frac{1}{1+e^{\frac{a(1-2|x|)}{x^2-|x|}}}}$

Are that${\displaystyle q\left(x,\frac{\sqrt{3}}{2}\right)}$ make smooth transition curves with "almost" constant slope edges (a bump function with true straight slopes is portrayed thisAnother example).

A proper example of a smooth Bump function would be:

${\displaystyle u(x)=\{\begin{array}{l}1,\text{if }x=0,\\ 0,\text{if }|x|\ge 1,\\ \frac{1}{1+e^{\frac{1-2|x|}{x^2-|x|}}},\text{otherwise},\end{array}}$

A proper example of a smooth transition function will be:

where could be noticed that it can be represented also throughHyperbolic functions:

${\displaystyle \frac{1}{1+e^{\frac{2x-1}{x^2-x}}}=\frac{1}{2}\left(1-\tanh \left(\frac{2x-1}{2(x^2-x)}\right)\right)}$

It is possible to construct bump functions "to specifications". Stated formally, if${\displaystyle K}$ is an arbitrarycompact set in${\displaystyle n}$ dimensions and${\displaystyle U}$ is anopen set containing${\displaystyle K,}$ there exists a bump function${\displaystyle \varphi }$ which is${\displaystyle 1}$ on${\displaystyle K}$ and${\displaystyle 0}$ outside of${\displaystyle U.}$ Since${\displaystyle U}$ can be taken to be a very small neighborhood of${\displaystyle K,}$ this amounts to being able to construct a function that is${\displaystyle 1}$ on${\displaystyle K}$ and falls off rapidly to${\displaystyle 0}$ outside of${\displaystyle K,}$ while still being smooth.

Bump functions defined in terms of convolution

The construction proceeds as follows. One considers a compact neighborhood${\displaystyle V}$ of${\displaystyle K}$ contained in${\displaystyle U,}$ so${\displaystyle K\subseteq V^{\circ }\subseteq V\subseteq U.}$ Thecharacteristic function${\displaystyle {\chi }_V}$ of${\displaystyle V}$ will be equal to${\displaystyle 1}$ on${\displaystyle V}$ and${\displaystyle 0}$ outside of${\displaystyle V,}$ so in particular, it will be${\displaystyle 1}$ on${\displaystyle K}$ and${\displaystyle 0}$ outside of${\displaystyle U.}$ This function is not smooth however. The key idea is to smooth${\displaystyle {\chi }_V}$ a bit, by taking theconvolution of${\displaystyle {\chi }_V}$ with amollifier. The latter is just a bump function with a very small support and whose integral is${\displaystyle 1.}$ Such a mollifier can be obtained, for example, by taking the bump function${\displaystyle \mathrm{\Phi }}$ from the previous section and performing appropriate scalings.

Bump functions defined in terms of a function${\displaystyle c:\mathbb{R}\to [0,\mathrm{\infty })}$ with support${\displaystyle (-\mathrm{\infty },0]}$

An alternative construction that does not involve convolution is now detailed. It begins by constructing a smooth function${\displaystyle f:{\mathbb{R}}^n\to \mathbb{R}}$ that is positive on a given open subset${\displaystyle U\subseteq {\mathbb{R}}^n}$ and vanishes off of${\displaystyle U.}$^[1] This function's support is equal to the closure${\displaystyle \overline{U}}$ of${\displaystyle U}$ in${\displaystyle {\mathbb{R}}^n,}$ so if${\displaystyle \overline{U}}$ is compact, then${\displaystyle f}$ is a bump function.

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

If${\displaystyle U={\mathbb{R}}^n}$ then let${\displaystyle f=1}$ while if${\displaystyle U=\varnothing }$ then let${\displaystyle f=0}$; so assume${\displaystyle U}$ is neither of these. Let${\displaystyle {\left(U_k\right)}_{k=1}^{\mathrm{\infty }}}$ be anopen cover of${\displaystyle U}$ by open balls where the open ball${\displaystyle U_k}$ has radius${\displaystyle r_k>0}$ and center${\displaystyle a_k\in U.}$ Then the map${\displaystyle f_k:{\mathbb{R}}^n\to \mathbb{R}}$ defined by${\displaystyle f_k(x)=c\left(r_k^2-{\Vert x-a_k\Vert }^2\right)}$ is a smooth function that is positive on${\displaystyle U_k}$ and vanishes off of${\displaystyle U_k.}$^[1] For every${\displaystyle k\in \mathbb{N},}$ let$${\displaystyle M_k=sup\left\{\left|\frac{{\mathrm{\partial }}^p{f}_k}{{\mathrm{\partial }}^{p_1}{x}_1\cdots {\mathrm{\partial }}^{p_n}{x}_n}(x)\right|:x\in {\mathbb{R}}^n\text{ and }{p}_1,\dots ,p_n\in \mathbb{Z}\text{ satisfy }0\le p_i\le k\text{ and }p=\sum _i{p}_i\right\},}$$where thissupremum is not equal to${\displaystyle +\mathrm{\infty }}$ (so${\displaystyle M_k}$ is a non-negative real number) because${\displaystyle \left({\mathbb{R}}^n\setminus U_k\right)\cup \overline{U_k}={\mathbb{R}}^n,}$ the partial derivatives all vanish (equal${\displaystyle 0}$) at any${\displaystyle x}$ outside of${\displaystyle U_k,}$ while on the compact set${\displaystyle \overline{U_k},}$ the values of each of the (finitely many) partial derivatives are (uniformly) bounded above by some non-negative real number.^{[note 1]} The series$${\displaystyle f:=\sum _{k=1}^{\mathrm{\infty }}\frac{f_k}{2^k{M}_k}}$$converges uniformly on${\displaystyle {\mathbb{R}}^n}$ to a smooth function${\displaystyle f:{\mathbb{R}}^n\to \mathbb{R}}$ that is positive on${\displaystyle U}$ and vanishes off of${\displaystyle U.}$^[1] Moreover, for any non-negative integers${\displaystyle p_1,\dots ,p_n\in \mathbb{Z},}$^[1]$${\displaystyle \frac{{\mathrm{\partial }}^{p_1+\cdots +p_n}}{{\mathrm{\partial }}^{p_1}{x}_1\cdots {\mathrm{\partial }}^{p_n}{x}_n}f=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2^k{M}_k}\frac{{\mathrm{\partial }}^{p_1+\cdots +p_n}{f}_k}{{\mathrm{\partial }}^{p_1}{x}_1\cdots {\mathrm{\partial }}^{p_n}{x}_n}}$$where this series also converges uniformly on${\displaystyle {\mathbb{R}}^n}$ (because whenever${\displaystyle k\ge p_1+\cdots +p_n}$ then the${\displaystyle k}$^th term'sabsolute value is${\displaystyle \le \frac{M_k}{2^k{M}_k}=\frac{1}{2^k}}$). This completes the construction.

As a corollary, given two disjoint closed subsets${\displaystyle A,B}$ of${\displaystyle {\mathbb{R}}^n,}$ the above construction guarantees the existence of smooth non-negative functions${\displaystyle f_A,f_B:{\mathbb{R}}^n\to [0,\mathrm{\infty })}$ such that for any${\displaystyle x\in {\mathbb{R}}^n,}$${\displaystyle f_A(x)=0}$ if and only if${\displaystyle x\in A,}$ and similarly,${\displaystyle f_B(x)=0}$ if and only if${\displaystyle x\in B,}$ then the function$${\displaystyle h:=\frac{f_A}{f_A+f_B}:{\mathbb{R}}^n\to [0,1]}$$ is smooth and for any${\displaystyle x\in {\mathbb{R}}^n,}$${\displaystyle h(x)=0}$ if and only if${\displaystyle x\in A,}$${\displaystyle h(x)=1}$ if and only if${\displaystyle x\in B,}$ and if and only if${\displaystyle x\notin A\cup B.}$^[1] In particular,${\displaystyle h(x)\ne 0}$ if and only if${\displaystyle x\in {\mathbb{R}}^n\setminus A,}$ so if in addition${\displaystyle U:={\mathbb{R}}^n\setminus A}$ is relatively compact in${\displaystyle {\mathbb{R}}^n}$ (where${\displaystyle A\cap B=\varnothing }$ implies${\displaystyle B\subseteq U}$) then${\displaystyle h}$ will be a smooth bump function with support in${\displaystyle \overline{U}.}$

While bump functions are smooth, theidentity theorem prohibits them from beinganalytic unless theyvanish identically. Bump functions are often used asmollifiers, as smoothcutoff functions, and to form smoothpartitions of unity. They are the most common class oftest functions used in analysis. The space of bump functions is closed under many operations. For instance, the sum, product, orconvolution of two bump functions is again a bump function, and anydifferential operator with smooth coefficients, when applied to a bump function, will produce another bump function.

If the boundaries of the Bump function domain is${\displaystyle \mathrm{\partial }x,}$ 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:$${\displaystyle \underset{x\to \mathrm{\partial }{x}^{\pm }}{lim}\frac{d^n}{d{x}^n}f(x)=0,\text{ for all }n\ge 0,n\in \mathbb{Z}}$$

TheFourier transform of a bump function is a (real) analytic function, and it can be extended to the wholecomplex plane: hence it cannot be compactly supported unless it is zero, since the only entire analytic bump function is the zero function (seePaley–Wiener theorem andLiouville's theorem). Because the bump function is infinitely differentiable, its Fourier transform must decay faster than any finite power of${\displaystyle 1/k}$ for a large angular frequency${\displaystyle |k|.}$^[2] The Fourier transform of the particular bump functionfrom above can be analyzed by asaddle-point method, and decays asymptotically as$${\displaystyle |k{|}^{-3/4}{e}^{-\sqrt{|k|}}}$$for large${\displaystyle |k|.}$^[3]

The integral of the bump function${\displaystyle \mathrm{\Psi }(x)}$ is given by$${\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\mathrm{\Psi }(x)dx=e^{-1/2}\left(K_1\left(\frac{1}{2}\right)-K_0\left(\frac{1}{2}\right)\right)}$$where${\displaystyle K_1(x)}$ and${\displaystyle K_0(x)}$ are theModified Bessel functions of the second kind.^[4]

• Cutoff function – Integration kernels for smoothing out sharp featuresPages displaying short descriptions of redirect targets
• Laplacian of the indicator – Limit of sequence of smooth functions
• Non-analytic smooth function – Mathematical functions which are smooth but not analytic
• Schwartz space – Function space of all functions whose derivatives are rapidly decreasing

• Nestruev, Jet (10 September 2020).Smooth Manifolds and Observables.Graduate Texts in Mathematics. Vol. 220. Cham, Switzerland:Springer Nature.ISBN 978-3-030-45649-8.OCLC 1195920718.

• Smooth functions
• Schwartz distributions

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