is an example of a bump function in one dimension. It is clear from the construction that this function has compact support, since a function of the real line has compact support if and only if it has bounded closed support. 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 scaled to fit into the unit disc: the substitution corresponds to sending to
A simple example of a bump function in variables is obtained by taking the product of copies of the above bump function in one variable, so
It is possible to construct bump functions "to specifications". Stated formally, if is an arbitrary compact set in dimensions and is an open set containing there exists a bump function which is on and outside of Since can be taken to be a very small neighborhood of this amounts to being able to construct a function that is on and falls off rapidly to outside of while still being smooth.
The construction proceeds as follows. One considers a compact neighborhood of contained in so The characteristic function of will be equal to on and outside of so in particular, it will be on and outside of This function is not smooth however. The key idea is to smooth a bit, by taking the convolution of with a mollifier. The latter is just a bump function with a very small support and whose integral is Such a mollifier can be obtained, for example, by taking the bump function from the previous section and performing appropriate scalings.
An alternative construction that does not involve convolution is now detailed.
Start with any smooth function that vanishes on the negative reals and is positive on the positive reals (that is, on and on where continuity from the left necessitates ); an example of such a function is for and otherwise.
Fix an open subset of and denote the usual Euclidean norm by (so is endowed with the usual Euclidean metric).
The following construction defines a smooth function that is positive on and vanishes outside of  So in particular, if is relatively compact then this function will be a bump function.
If then let while if then let ; so assume is neither of these. Let be an open cover of by open balls where the open ball has radius and center Then the map defined by is a smooth function that is positive on and vanishes off of 
For every let
which is a real number because the supremum vanishes at any outside of while on the compact set the values of the partial derivatives are bounded.[note 1]
converges uniformly on to a smooth function that is positive on and vanishes off of 
Moreover, for any non-negative integers 
where this series also converges uniformly on (because whenever then the th term's absolute value is ).
As a corollary, given two disjoint closed subsets of and smooth non-negative functions such that for any if and only if and similarly, if and only if then the function is smooth and for any if and only if if and only if and if and only if 
In particular, if and only if so if in addition is relatively compact in (where implies ) then will be a smooth bump function with support in
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 for a large angular frequency . The Fourier transform of the particular bump function
^The partial derivatives are continuous functions so the image of the compact subset is a compact subset of The supremum is over all non-negative integers where because and are fixed, this supremum is taken over only finitely many partial derivatives, which is why