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For unit-spaced order-three B-splines sampled at support midpoints $i+3/2$, the B-spline collocation matrix has diagonal $3/4$ and neighboring entries $1/8$. Conjugating by alternating diagonal signs reduces its inverse absolute values to the nonnegative inverse of $\mathrm{tridiag}(-1,6,-1)/8$. With $\tau =3-2\sqrt{2}$, the absolute row sums are $r_i=2[1-({\tau }^i+{\tau }^{n+1-i})/(1+{\tau }^{n+1})]$. This follows by solving $6r_i-r_{i-1}-r_{i+1}=8$ with $r_0=r_{n+1}=0$. The maximum occurs at the middle row or rows and tends to two as $n$ grows. Nonnegativity and subpartition of unity of the B-splines transfer this bound to the spline interpolation operator.

Let the real spline $s={\sum }_{i=p}^q{c}_i{N}_i$ have distinct knots, order $k\ge 2$, nonzero endpoint coefficients, and no identically zero knot cell in its support. It and its first $k-2$ derivatives vanish at both support endpoints. If $Z$ is its number of interior distinct zeros, there are initially $Z+2$ zero components. Each differentiation through order $k-2$ increases this count by at least one: a nonzero gap between zero components has an interior extremum, and the two endpoint zeros remain separate. The final derivative is continuous piecewise linear on $q-p+k$ cells, and each cell can meet at most one zero component unless it is identically zero, in which case all its zeros belong to the same component. It therefore has at most $q-p+k$ zero components. Thus $Z+k\le q-p+k$, proving the bound.

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.

This linear functional extracts one coefficient of a B-spline expansion using local polynomial data. With ${\psi }_i(x)={\prod }_{\ell =1}^{k-1}(x-t_{i+\ell })/(k-1)!$ and a point $\xi$ inside a knot cell in the support, it is ${\sum }_{r=0}^{k-1}(-1{)}^r{s}^{(r)}(\xi ){\psi }_i^{(k-1-r)}(\xi )$. The local polynomial identity for the Marsden dual functional gives the desired coefficient. Its bounded restriction to the local spline space can be extended to continuous data on $[t_i,t_{i+k}]$ by the Hahn-Banach theorem, with the same bound. Such extensions supply spline quasi-interpolation coefficients without requiring derivatives of the data.

Let $A$ be a doubly stochastic matrix, let $q$ be a probability vector, and put $w=Aq$. In row $i$ with $w_i>0$, define $r_i(j)=a_{ij}{q}_j/w_i$ on the support where $a_{ij}>0$. The positive row entries ${\alpha }_i=(a_{ij})$ form a probability vector. Expanding relative entropy and using the column sums givesZero-output rows contribute zero because all their $a_{ij}{q}_j$ vanish. The row sums supply the normalized comparison laws, while the column sums preserve the input entropy contribution. Equality holds exactly when $q_j=w_i$ for every positive matrix entry $a_{ij}$, including the automatic zero values in zero-output rows.

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.

The front velocity is the propagation speed of the earliest boundary of a newly introduced disturbance's support. It asks when an initially unaffected observer can first respond, whereas group velocity asks how a narrow wave envelope translates. A detector threshold defines a practical arrival time, but can miss arbitrarily small earlier responses; a packet peak is another distinct timing convention. In a causal dielectric model, a high-angular frequency contour argument gives exactly zero field for $t<x/c$, even when phase velocity exceeds $c$. This is a statement about that model's relativistic causality, not a universal identification of group velocity with front velocity.^[ref]

For the stipulated Klein-Gordon equation with compactly supported real Cauchy data, front velocity is bounded by $c$, and $c$ is the model's maximal speed of influence. A short energy method outline usesIntegration outside an expanding interval shows that zero initial energy there remains zero. Thus initial support in $[-a,a]$ cannot influence $x>a$ before $t=(x-a)/c$. The positive energy density includes the restoring term ${\omega }_0^2{u}^2/2$: this adds oscillation behind the front without extending its domain of dependence. A particular disturbance may fail to excite the leading boundary, so the maximal speed is not a promise of a detectable response for every source.

A Gaussian function is nonzero arbitrarily far away, so the Gaussian envelope in a Klein-Gordon medium has no sharp initial support boundary. Its small distant tails are already in the initial data. Likewise, keeping only a narrow interval of the frequency spectrum cannot reproduce an exact abrupt onset. Cutting off a Gaussian tail changes its Fourier transform and the accompanying initial velocity; it is a different initial-value problem. The exponential initial-velocity tail for the Klein-Gordon equation is another concrete example of a pre-existing tail outside the main disturbance.

The water-tank measurement in phase velocity records free surface elevation with calibrated resistance probes and uses a ruler and stopwatch for an independent check. The inspected protocol calls the front or back of a group a timing feature. That visible edge establishes an envelope transit estimate, not the first nonzero disturbance in the sense of front velocity.^[ref] A proposed additional control would repeat the transit fit at several detection thresholds and train durations, using matched probe calibration and the same sampling rate. Dependence on the threshold diagnoses wave envelope reshaping or limited sensitivity; records should be compared before reflections return. Such a test could expose limitations of a group velocity estimate, but would not establish an exact causal front or validate the stipulated Klein-Gordon equation for water waves.

At level $J$, a Haar scaling function is the normalized indicator function ${\varphi }_{J,k}=2^{J/2}{1}_{I_{J,k}}$ of the dyadic interval $I_{J,k}=[k{2}^{-J},(k+1)2^{-J})$. Its L2 norm is one because its squared height is $2^J$ and its support length is $2^{-J}$.

For a real square-integrable function $f$, its coefficient isMultiplying the coefficient by the Haar scaling function restores the cell average on that interval. The disjoint supports make the functions at one level an orthonormal basis of the cellwise-constant subspace. This normalization is also what turns coefficient variances in the Gaussian white noise model into $1/n$ and cellwise variances into $2^J/n$ in the Haar projection estimator in Gaussian white noise.

Let $\psi$ be an integrable function with at least one vanishing moment. For a unit jump discontinuity $f(x)=H(x-x_0)$, where $H$ is the Heaviside step function, substitution into the coefficient givesThe signed integral of the whole analyzing function is zero, but the jump discontinuity cuts through its support of a function and usually leaves an unbalanced partial integral. At fixed relative position $\theta$ with nonzero partial integral, the coefficient decays as $\sqrt{a}$, slower than the $a^{q+1/2}$ Taylor remainder bound on a smooth region. “Edges survive” means this relative contrast: the normalized coefficients still tend to zero as $a\to 0$.

For an original exact comparison, use the smooth polynomial $f(x)=x+x^2$ at $b=0$. The Haar wavelet $h$ and the second-difference analyzing function $\eta$ giveThe linear term cancels for $\eta$; it remains for $h$. For the fixed unit jump discontinuity $H(x)$, choose $b=-a/4$ so that $\theta =1/4$ at every scale. Then

Position matters. If an analyzing function's support of a function lies entirely on one constant side of the jump discontinuity, its coefficient is zero. Even a crossing can cancel: $\eta$ has zero partial integral at $\theta =3/2$. A jump discontinuity on a dyadic interval boundary can likewise escape finer Haar wavelets whose supports do not cross it. These examples describe particular functions and normalizations, not a universal ranking of wavelets for signal processing.

A distribution which vanishes on a neighborhood of each point vanishes globally. To prove this, choose a smooth partition of unity subordinate to the neighborhoods. Every test function is a finite sum of its products with the partition members, since its support is compact. The distribution vanishes on each summand. Applying the result to a difference proves that two distributions agreeing locally agree globally. This justifies defining the support of a distribution by removing the union of all open regions on which it is zero.

Target: Periodic primitive with zero mean. Parent: Periodic extension. Intended file: function.bigb.

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.

Write $w_i=(Aq{)}_i$. For each row put $S_i=\{j:a_{ij}>0\}$; the positive row entries ${\alpha }_i=(a_{ij}{)}_{j\in S_i}$ form a probability distribution because the row sum is one. When $w_i>0$, define the probability distributionPart (a), applied on this row support, gives $d(r_i,{\alpha }_i)\ge 0$. Its weighted sum expands asRows with $w_i=0$ contribute zero, since all $a_{ij}{q}_j$ in such a row vanish. Using the convention that zero-weight logarithmic terms are zero, the result isBoth stochasticity conditions have a role: row sums normalize the comparison distributions, while column sums preserve the total contribution from $q$.

This is entropy increase under doubly stochastic mixing.

For $x\in X$, let $e_x$ be the indicator function of $\{x\}$. In the ring of functions $R={\mathbb{R}}^X$, multiplication is pointwise, soSuppose first that $X$ is finite, and let $J$ be an ideal of $R$. Define $S=\{x\in X:f(x)\ne 0\text{ for some }f\in J\}$. If $x\in S$, choose such an $f$; multiplying it by $e_x/f(x)$ shows that $e_x\in J$. Every $f\in J$ satisfieswhere the real scalars are constant functions. Thus $J=(e_x:x\in S)$ has a finite generating set of an ideal, so $R$ is a Noetherian ring. If $X$ is empty, $R$ is the zero ring, and the conclusion still holds.

If $X$ is infinite, consider the idealIt is an ideal because a sum has support contained in the union of the two supports, and multiplication by any function cannot enlarge support. If $f_1,\dots ,f_r$ generated $F$, their supports would have a finite union $T$. Every element of $(f_1,\dots ,f_r)$ would vanish outside $T$, whereas $e_x\in F$ for any $x\in X\setminus T$. This contradiction proves

Convergence ${\varphi }_j\to \varphi$ in the space of test functions means that all supports lie in one compact set $K\subset \mathbb{R}$ and $\Vert {\varphi }_j^{(r)}-{\varphi }^{(r)}{\Vert }_{\infty }\to 0$ for every integer $r\ge 0$. An eventual common compact support is equivalent, since finitely many initial supports can be added to $K$.

Write the standard mollifier as ${\rho }_{\epsilon }(x)={\epsilon }^{-1}\rho (x/\epsilon )$, where $\rho \in \mathcal{D}(\mathbb{R})$ is nonnegative, supported in $[-1,1]$, and has integral one. For $0<\epsilon \le 1$, the convolution ${\rho }_{\epsilon }\ast \varphi$ has support in the fixed compact set $\mathrm{supp}\varphi +[-1,1]$. The given convolution derivative identity yieldsEach ${\varphi }^{(r)}$ is uniformly continuous; more explicitly, the mean value theorem bounds the absolute value by $\epsilon \Vert {\varphi }^{(r+1)}{\Vert }_{\infty }\int |y|\rho (y)dy$, uniformly in $x$. Thus

If ${\varphi }_j\to 0$ in the space of test functions and $K$ is a common compact support, then $\Vert {\varphi }_j{\Vert }_{L^1}\le |K|\Vert {\varphi }_j{\Vert }_{\infty }\to 0$, where $|K|$ is its Lebesgue measure.

The mollifiers themselves have no limit in the L1 space. If ${\rho }_{\epsilon }\to g$ in that norm, then for each $\delta >0$ the integral of $|g|$ on $\{|x|>\delta \}$ is zero, because ${\rho }_{\epsilon }$ vanishes there for $\epsilon <\delta$. Hence $g=0$ almost everywhere. This contradicts $\Vert {\rho }_{\epsilon }{\Vert }_{L^1}=1$. Therefore $\boxed{{\rho }_{\epsilon }\text{ does not converge in }{L}^1(\mathbb{R})}$.

Let $\Omega \subset {\mathbb{R}}^n$ be an open set and $(U_{\alpha }{)}_{\alpha \in A}$ an open cover of $\Omega$. There exist nonnegative smooth functions $({\chi }_j{)}_{j\in J}$ and choices $\alpha (j)$ such thatand the family of supports is locally finite: every point has a neighborhood meeting only finitely many of them. This is a smooth partition of unity subordinate to the cover, allowing a refinement of its indexing. In Euclidean open sets, one can choose $J$ countable and the individual supports compact in their assigned cover members. Local finiteness ensures that the sum is smooth and that only finitely many terms meet any given compact set.

The $2\pi$ periodization of a Schwartz function isFor every nonnegative integer $j$, the Schwartz seminorms give $|{\varphi }^{(j)}(x+2\pi k)|\le C_j(1+|k|{)}^{-2}$ uniformly for $x\in [0,2\pi ]$. The Weierstrass M-test gives absolute and uniform convergence there of the defining series and all its differentiated series. Termwise differentiation proves that $P\varphi$ is a smooth function, with $(P\varphi {)}^{(j)}=P({\varphi }^{(j)})$. Reindexing the absolutely convergent series proves $P\varphi (x+2\pi )=P\varphi (x)$, so it is a periodic function.

For an explicit smooth periodic partition of unity, choose the bump functionIt is a test function, positive on $[-\pi ,\pi ]$. The smooth function $S(x)=P\eta (x)$ is $2\pi$ periodic and strictly positive, since every $x$ has a translate in $[-\pi ,\pi ]$. SetDivision preserves smoothness because $S>0$, and the support remains contained in $[-2\pi ,2\pi ]$.

On the unbounded domain ${\mathbb{R}}^3$, translations already prevent compactness in every indicated Hölder space. Choose a bump function $\eta \in C_c^{\infty }(B_1)$ with $\eta (0)=1$, put $e_1=(1,0,0)$, and defineA change of variables in the Lebesgue integrals defining the Sobolev norm gives $\Vert u_m{\Vert }_{W^{1,6}({\mathbb{R}}^3)}=\Vert \eta {\Vert }_{W^{1,6}({\mathbb{R}}^3)}$, so the sequence is bounded. Each $u_m$ is a smooth function of compact support and belongs to every Hölder space $C^{0,\gamma }({\mathbb{R}}^3)$, $0<\gamma \le 1$: its bounded gradient gives the Lipschitz estimate for distances at most one, and its bounded values control larger distances.

For $m\ne \ell$, the supports are disjoint, and at $x=3m{e}_1$ we have $u_m(x)=1$ and $u_{\ell }(x)=0$. ThereforeNo subsequence can be a Cauchy sequence in any of these Hölder norms, hence none converges. This uses the translation mechanism for noncompactness of a Sobolev embedding by translation; concentration at the critical Hölder exponent alone would not rule out convergence at every smaller Hölder exponent.

First repeat the Caccioppoli inequality calculation on two concentric balls. If $h$ is a harmonic function on $B_R$, choose a cutoff function equal to one on $B_r$, supported in $B_R$, and with gradient bounded by $C/(R-r)$. The argument of (ii), with $A=I$ and $k=0$, givesHere $C_{\ast }$ depends at most on the dimension. When $h$ is only known to be smooth inside $B_R$ and in $L^2(B_R)$, the weak formulation and all integrals involving the cutoff function are still legitimate, since its support is strictly inside $B_R$.

Every classical partial derivative of the given smooth harmonic function $u$ is again a harmonic function, since constant-coefficient partial derivatives commute with the Laplacian. Choose indices $i_1,\dots ,i_m$ representing the multi-index $\alpha$, and defineFor each $j\ge 1$, the preceding Caccioppoli inequality yieldsStarting with $u\in L^2(B_1)$, these estimates also establish the integrability needed at each subsequent step. Iterating $m$ times givesThis is the desired interior derivative estimate for a harmonic function, with a constant depending only on $n$ and $m$, uniformly over every multi-index of size $m$.

There is a small incompatibility in the printed cutoff function assumptions. No $\phi \in C_c^{\infty }(B_1)$ can equal one on $B_{1/2}$ and satisfy $|\nabla \phi |\le 2$: for a unit vector $e$, choose $t<1$ beyond its support. Continuity gives $\phi (e/2)=1$, while the fundamental theorem of calculus would give $1\le 2(t-1/2)<1$. Replace the number $2$ by a fixed sufficiently large absolute constant $C_0$; such a cutoff function exists, and the requested Caccioppoli inequality is unchanged apart from its constant. We keep $M=\Vert \nabla \phi {\Vert }_{\infty }$ in the calculation, which also yields the formal bound under the printed assumptions.

The weak formulation is ${\int }_{B_1}A\nabla u\cdot \nabla \zeta =0$ for test functions $\zeta$. Since $A$ is bounded, this extends by density of smooth functions in a Sobolev space to every $\zeta \in H_0^1(B_1)$. Part (a)(ii), followed by mollification inside the compact support, makes ${\phi }^2(u-k)$ an admissible choice. The product rule givesPut $X=\int {\phi }^2|\nabla u{|}^2$ and $Y=\int (u-k{)}^2|\nabla \phi {|}^2$. The lower and upper uniform ellipticity bounds, together with Cauchy-Schwarz inequality, implyIf $X=0$ the conclusion is immediate. Otherwise divide by $X^{1/2}$ and square to obtain $X\le 4{\Lambda }^2{\lambda }^{-2}Y$. Since $\phi =1$ on $B_{1/2}$,For a fixed valid cutoff function, $M\le C_0$, this has the required constant independent of $k$. Formally inserting the printed value $M=2$ gives $16{\Lambda }^2/{\lambda }^2$. No derivative of $A$ was taken, so differentiability of the coefficients is unnecessary.