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For a scalar-valued function $f$ on a topological space $X$, its support isThe closure matters. On the real line, $f(x)=\max (0,1-|x|)$ is nonzero exactly on $(-1,1)$, but its support is the closed interval $[-1,1]$. A point where $f$ vanishes can still belong to the support if nonzero values occur arbitrarily near it.

For the discrete topology, every subset is closed, so the support is simply the set of inputs with nonzero values. This is the convention used when representing a finitely supported count function. A function has compact support when the displayed closed set is compact.

For distributions $u$ on ${\mathbb{R}}^m$ and $v$ on ${\mathbb{R}}^n$, their tensor product acts on a joint test function by $⟨u\otimes v,\Phi ⟩=⟨u_x,⟨v_y,\Phi (x,y)⟩⟩$. The inner function is smooth with compact support in the projection of $\mathrm{supp}\Phi$, and finite order of a distribution estimates prove continuity. Reversing the pairings gives the same value: finite sums of product kernels are dense in the required smooth seminorms, and the two orders agree on product kernels. Its support is contained in the Cartesian product of the two supports. This provides a rigorous replacement for formal integration in separate distribution variables.

A test function is a smooth function with compact support. Weak and distributional equations are identities obtained by integrating against every test function.

The space of test functions has the locally convex inductive-limit topology of the spaces of smooth functions supported in a fixed compact set $K$. Each ${\mathcal{D}}_K$ uses seminorms $p_m(\phi )={\max }_{|\alpha |\le m}\Vert {\partial }^{\alpha }\phi {\Vert }_{\infty }$. Sequential convergence means a common compact support and uniform convergence of every derivative. A linear functional on this space is a distribution exactly when its restriction to each ${\mathcal{D}}_K$ satisfies a finite order of a distribution estimate; the order and constant may depend on $K$.

A vanishing moment is a cancellation condition: positive and negative parts of an analyzing function balance when weighted by a monomial. Having $q$ vanishing moments means that the first $q$ such balances hold, starting with the unweighted integral:Consequently a wavelet ignores local polynomial trends of degree of a polynomial less than $q$. A smooth feature is close to its Taylor polynomial, so only the small Taylor remainder contributes to its detail coefficient. A jump discontinuity cannot be approximated this way across its edge; its jump response of a localized analyzing function can remain large relative to details from a smooth region.^[ref]

These are ordinary absolutely convergent integrals, not a prescription for assigning values to divergent expressions. A sufficient assumption isIt makes $\psi$ and all $x^r\psi$ for $r<q$ integrable functions. Being square-integrable alone does not ensure these weighted integrals exist. For a square-integrable function with compact support, the Cauchy-Schwarz inequality guarantees every such weighted integral is finite. Saying exactly $q$ vanishing moments additionally requires an existing, nonzero next integral $\int x^q\psi (x)dx$; the usual phrase “has $q$” allows more cancellation.

Use the Fourier transform convention $\hat{\psi }(\omega )=\int \psi (x)e^{-i\omega x}dx$. If $\psi$ and $x^r\psi$ for $r<q$ are integrable functions, moment differentiation of the Fourier transform yields a $C^{q-1}$ Fourier transform andThe dominated convergence theorem justifies these derivatives. Under these assumptions, $q$ vanishing moments are equivalent to vanishing derivatives through order $q-1$ at zero. For $q>1$ this gives a Peano zero $\hat{\psi }(\omega )=o(|\omega {|}^{q-1})$; for $q=1$ it gives $\hat{\psi }(\omega )=o(1)$. It does not by itself justify a $q$th derivative or an $O(|\omega {|}^q)$ estimate.

If also $\int |x{|}^q|\psi (x)|dx<\infty$, subtract the degree-$(q-1)$ Taylor polynomial of $e^{-i\omega x}$ inside the integral. Its Taylor remainder gives the stronger boundFor an integrable function with compact support, every weighted integral exists and its Fourier transform extends to an entire function. The language “zero of order at least $q$” is then literal analytic function terminology. If the next integral is nonzero, the order is exactly $q$. The smooth-mask vanishing-moment criterion relates these cancellations to a refinement mask under its additional hypotheses.^[ref]

The variational derivative of a functional $H[u]$ is the coefficient of a test variation $v$ after integration by parts:The allowed variations and boundary conditions are part of the definition; compact support of $v$ in the interior removes boundary contributions without discarding physical boundary terms arbitrarily.

For example, $H[u]={\int }_a^b[u_x^2/2+V(u)]dx$ has first variationThus its interior variational derivative is $-u_{xx}+V^{\prime }(u)$. Fixed endpoint values impose $v(a)=v(b)=0$. If the endpoint values are free, stationarity also imposes the natural Neumann boundary conditions $u_x(a)=u_x(b)=0$. The bulk Euler-Lagrange equation and boundary conditions therefore carry different parts of the same variation.

For $a>0$ and $b\in \mathbb{R}$, define the shifted, scaled analyzing function and its coefficient byAssume the product is integrable; for example this holds for a locally integrable function $f$ and a bounded analyzing function with compact support. The bar is complex conjugation. This normalization preserves the L2 norm of the analyzing function. If $p$ is a polynomial of degree of a polynomial less than $q$, then $p(b+au)$ is a sum of monomials $u^r$ with $r<q$. Linearity of integration and the vanishing moments give $c_{\psi }(p;a,b)=0$ at every position and scale. Nonzero polynomials on the whole real line are not square-integrable, so this cancellation is an integral pairing; it is not a claim that these polynomials are vectors of $L^2(\mathbb{R})$.

Here is a quantitative reason that smooth regions yield small details. Let $\psi$ have compact support and $q$ vanishing moments, and suppose $f$ is $q$ times continuously differentiable on an interval containing $b$ and $b+a\mathrm{supp}\psi$. Subtract its degree-$(q-1)$ Taylor polynomial at $b$. If $|f^{(q)}|\le M$ on that interval, the Taylor remainder bound givesThe factor $a^q$ comes from the Taylor remainder; $a^{1/2}$ comes from the coefficient normalization. Thus the analyzing function measures the departure from the local polynomial trend, rather than the height of $f$ itself. The wavelet coefficient decay for Hölder functions uses the same argument with a Hölder-Taylor remainder bound. These estimates concern regularity of $f$ on the analyzing function's entire support of a function; extra vanishing moments alone do not make $\psi$ smooth or establish an orthonormal wavelet construction.^[ref]

A wave packet is a localized superposition of Fourier modes. In a narrow packet, nearby wavenumbers interfere to produce a slowly varying wave envelope around a rapid carrier wave. Its leading translation is governed by group velocity; the carrier's wave phase follows phase velocity. Localized does not necessarily mean compact support: a Gaussian function decays rapidly but extends over the whole line. The Gaussian envelope in a Klein-Gordon medium gives an original example with both translation and spreading.