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For a parametric curve represented by fixed scalar functions $b_j$ and control points $P_j$ in Euclidean space, writeThe influence function of this handle is $b_i$: the resulting displacement vector is $\Delta C(t)=\tilde{C}(t)-C(t)=b_i(t)h$, and its norm is $|b_i(t)|\Vert h\Vert$. These are exact finite changes, not only a first-order estimate. Where $b_i$ is negative the displacement points opposite to $h$. Where $b_i(t)=0$, the curve is unchanged at that parameter.

For the cubic Bézier curve above, the Bernstein basis influence function is $b_{1,3}(t)=3t(1-t{)}^2$. Its derivative isIt increases up to $t=1/3$ and decreases thereafter on $[0,1]$. Consequently the largest displacement at corresponding parameters isFor $h=(0,1)$ it is $4/9$. This is a comparison at the same parameter. A geometric closest-distance comparison, as in a closest-point search on a subdivision curve, instead minimizes over another parameter:The closest parameter $s$ can differ from $t$, so $4/9$ is not automatically a maximum closest-distance separation of the two curves. Neither parameter is assumed to be arc length.

The support of a function $b_i$ is the closure of the parameter values where $b_i\ne 0$. It bounds where a nonzero handle move can act, but may include boundary points where the actual displacement is zero. Here the support of a function $b_{1,3}$ is all of $[0,1]$ relative to the parameter domain, while the endpoints are unaffected. In contrast, a B-spline of order $k$ has support within $[t_i,t_{i+k}]$ in its spline knot sequence, so its handle affects only that parameter range; the support and Gram bandwidth of B-splines describes the same locality in another application. Changing the spline knots or the basis along with the handle requires a different comparison.

When the basis functions form a partition of unity, ${\sum }_j{b}_j(t)=1$, each evaluation is an affine combination of the control points. Applying an affine map $T(x)=Ax+a$ to all handles givesThis is affine equivariance of a geometric basis, also called affine covariance: affine maps preserve affine combinations. In particular, translating every handle by $h$ translates the whole curve by $h$, whereas moving just one handle produces $b_i(t)h$. The Bernstein basis has this partition of unity property.^[ref] Nonnegative weights further make evaluations convex combinations, giving the convex hull bound illustrated by the control polygon; nonnegativity is not needed for affine covariance.

For a proof outline of the influence formula, subtract the two basis representations: every unchanged coefficient cancels, leaving $b_i(t)(P_i+h)-b_i(t)P_i=b_i(t)h$. The same subtraction with several moved handles gives $\Delta C(t)={\sum }_i{b}_i(t)h_i$. No claim about interpolation, closest-point search on a subdivision curve or endpoint motion follows until the actual basis values are examined.

Particles keep their velocity and move along straight trajectories. The method of characteristics gives $f(t,x,v)=f_0(x-tv,v)$. The shear preserves Lebesgue measure and every Lp norm. Compact initial support remains compact at each finite time, but need not remain in one compact set for all times.

At level $J$, a Haar scaling function is the normalized indicator function $2^{J/2}{1}_{I_{J,k}}$ of a dyadic interval. The functions at that level form an orthonormal basis of the cellwise-constant subspace. Their normalization is what turns white-noise coefficient variances into $1/n$ and cellwise variances into $2^J/n$.

An orthonormal wavelet $\psi$ generates an orthonormal basis of $L^2(\mathbb{R})$ throughThe factor $2^{j/2}$ compensates for the shrinking support of a function, keeping the L2 norm equal to one. Orthogonality makes different coefficients separate squared-norm contributions, and completeness allows the wavelet series to recover any square-integrable function in the L2 norm.

For the Haar wavelet, wavelets at one scale have disjoint supports. At different scales their dyadic intervals are either disjoint or nested. On the support of the finer wavelet, the coarser one is constant, while the finer one has zero integral, so their L2 inner product is zero. Together with the unit L2 norm, this explains orthonormality. On the bounded interval $[0,1]$, add the constant Haar scaling function and use $j\ge 0$, $0\le k<2^j$; the constant function is not an $L^2(\mathbb{R})$ basis element on the whole line.

In three dimensions set $w=ru$ for a radial function satisfying $\square u=F$. If the source has support in $s\le r\le s+1$ and $|F(s,r)|\le (1+s{)}^{-2}$, the Duhamel principle bounds its contribution to $w(t,t+a)$ by $\frac{1}{2}\log (1+t)$ for $1/2\le a\le 1$. Its backward characteristic interval meets the source in a set of length at most one, and $r|F|\le (1+s{)}^{-1}$. Division by $r\simeq 1+t$ gives $|u|\lesssim \log (2+t)/(1+t)$ for compactly supported initial data.