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The multiplication of a distribution by a smooth function is $⟨\psi T,\phi ⟩=⟨T,\psi \phi ⟩$. If $\psi$ is a Schwartz function, the Leibniz rule shows that $\phi \mapsto \psi \phi$ is a continuous map $\mathcal{S}\to \mathcal{S}$, so $\psi T$ is a tempered distribution. When both $\psi$ and $T$ are radial, $\psi \circ R^t=\psi$, andThus Multiplication by a radial Schwartz function preserves radial tempered distributions.

For the convolution of a tempered distribution with a Schwartz function, setTranslations of $\psi$ depend smoothly on $x$ in the Schwartz space, so this is a smooth function with ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T_y,{\partial }^{\alpha }\psi (x-y)⟩$. The bound for $T$ by finitely many seminorms, together with $1+|y|\le (1+|x|)(1+|x-y|)$, provesThus the function also defines a tempered distribution. For a radial function $\psi$, put $g_x(y)=\psi (x-y)$. Then $\psi (Rx-y)=\psi (x-R^{t}y)=g_x(R^{t}y)$, soHence Convolution with a radial Schwartz function preserves radial tempered distributions.

Smoothing alone need not give a Schwartz function: for the constant tempered distribution $T=1$ and a Schwartz function with integral one, $T\ast \psi =1$. The radial Schwartz approximation of tempered distributions therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial mollifier $\rho \in C_c^{\infty }$, supported in the unit ball with integral one, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one on the unit ball. PutEach $f_j$ is a smooth function of compact support, hence a Schwartz function, and the two invariance calculations above make it radial.

It remains to prove convergence, including the simultaneous changes of both scales. With $\check{\rho }(x)=\rho (-x)$, the distributional convolution pairing isFor every fixed $m$, the Leibniz rule, rapid decay outside the radius-$j$ ball, and the chain rule for $\chi (x/j)$ giveConvolution by ${\check{\rho }}_{\epsilon }$ is uniformly bounded in $q_m$ for $0<\epsilon \le 1$, because its shifts have size at most one. The mean value theorem applied to $\phi (x-y)-\phi (x)$ similarly givesSplitting the error into the convolved cutoff error and the approximate identity error provesIf $|⟨T,\theta ⟩|\le C{q}_m(\theta )$, this yieldsConsequentlyweakly, and even in the strong dual topology, since $q_{m+1}$ is uniformly bounded on every subset of the Schwartz space that is a bounded set in a topological vector space. No assertion that $T\ast {\rho }_{1/j}$ itself is rapidly decreasing is needed.

Under the smoothness hypotheses in the smooth-mask vanishing-moment criterion, write any nonzero integer $j$ as $j=2^r\ell$, where $r\ge 0$ and $\ell$ is odd, possibly negative. Iterating the scaling refinement equation exactly $r+1$ times givesThe last MRA low-pass filter factor is $m(\pi \ell +t/2^{r+1})$. Its derivatives of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the MRA low-pass filter is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the product rule forces every derivative of order less than $p$ of the product to vanish. In particular compact support and a finite MRA low-pass filter supply all these hypotheses. Therefore the integer-frequency zeros of a scaling function are

Under only the printed assumptions, the exact same finite refinement argument gives the Peano zero $\hat{\phi }(2\pi j+t)=o(|t{|}^{p-1})$. Indeed, the last MRA low-pass filter factor has that estimate, the other MRA low-pass filter factors are bounded by one almost everywhere by the quadrature mirror filter identity, and the last Fourier transform factor is bounded because $\phi \in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by continuity of $\hat{\phi }$. It does not imply arbitrary ordinary higher derivatives. In the lacunary scaling-phase regularity counterexample, $\hat{\phi }=a{\hat{\phi }}_0$ is nondifferentiable at dense dyadic points next to the isolated integer-frequency zeros of the smooth base transform. Thus the same regularity qualification is necessary here.

Let $M$ bound the piecewise Hölder norms, and let $B=\Vert f{\Vert }_{\infty }$. At scale $j$, classify a wavelet as bad when its localized support of a function meets a discontinuity, and as good otherwise. Compact support and bounded overlap imply at most $C_{0}K$ bad wavelets per scale and at most $C_1{2}^j$ total wavelets. A good support lies in a single smooth piece, so the preceding wavelet coefficient decay for Hölder functions gives $|c_{j,n}|\le CM{2}^{-j(\alpha +1/2)}$. The permitted bounded-function estimate gives $|c_{j,n}|\le CB{2}^{-j/2}$ for bad coefficients.

Consider a specific N-term approximation: retain all coarse scaling functions and all wavelets through scale $J$, and also retain the bad coefficients at scales $J<j\le L$, where $L=\lceil 2\alpha J\rceil$. Since $\alpha >1/2$, $L\ge J$ for large $J$. Its term count is bounded byBy the Parseval identity, its squared L2 norm error is the sum of omitted squared coefficients. The good tail satisfiesand the remaining bad tail satisfiesChoose $J=\lfloor {\log }_2(N/(2C_2^{\prime }))\rfloor$ with a fixed sufficiently large $C_2^{\prime }$; then the $O(KJ)$ extra terms fit the remaining budget for all sufficiently large $N$, while $2^J$ is comparable to $N$. Padding to $N$ terms does not increase the error. The best N-term approximation is at least as accurate as this constructed selection, soThe constant may depend on the finite number of jumps, the piecewise Hölder norms, $B$ and the fixed wavelet family. Values exactly at the jumps are immaterial in $L^2$. The improvement comes from retaining only a bounded number of jump-crossing coefficients at each finer scale, as in best N-term wavelet approximation of piecewise Hölder functions.

For real $u\in L^1(\Omega )$, its total variation seminorm on a domain isThe vector fields have compact support inside $\Omega$, so no boundary term is charged. Taking both $\xi$ and $-\xi$ makes this equivalent to a supremum of absolute pairings. When finite, the distributional derivative $Du$ is a finite vector-valued Radon measure, and the displayed supremum is its total variation norm of a measure. For a smooth function it equals ${\int }_{\Omega }|\nabla u{|}_{2}dx$.

The function of bounded variation on a domain belongs to the Banach spaceThe L1 norm is needed because variation alone vanishes on constant functions and is only a seminorm. No smoothness of $u$ is part of this definition.

For nonnegative insect density assume $N_0>0$. Substitution of $N=e^{\alpha t}R$ gives $R_t=D{e}^{2\alpha t}(R^2{R}_x{)}_x$, so chooseThen ${\tau }^{\prime }=D{e}^{2\alpha t}>0$ and the porous medium equation is $R_{\tau }=(R^2{R}_x{)}_x$. Its total mass is conserved when the flux vanishes at infinity. For the mass-preserving similarity solution, direct differentiation givesOn the positive support, integrating once with zero boundary flux gives $F^2{F}^{\prime }=-\xi F/4$ and hence $F^2=({\xi }_0^2-{\xi }^2)/4$. The semicircle mass integral is $\pi {\xi }_0^2/4=N_0$, confirming the profile and its symmetric compact support.

The front is at $|x|={\xi }_0{\tau }^{1/4}$. When $\alpha <0$, $\tau (t)$ increases towards $D/(2|\alpha |)$, so the supremum of the distance reached isIt is approached as $t\to \infty$, rather than attained at a finite time. If $N_0=0$, the nonnegative solution is identically zero and the corresponding distance is zero.

Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$ and $D_j=-i{\partial }_{x_j}$. Thus $D_j$ has Fourier symbol ${\xi }_j$. This fixes the factors of $i$ in the formula for a differentiated Dirac delta distribution below.

With the Japanese bracket $⟨\theta ⟩=(1+|\theta {|}^2{)}^{1/2}$, the symbol class $\mathrm{Sym}(X,{\mathbb{R}}^k;N)=S_{1,0}^N(X\times {\mathbb{R}}^k)$ consists of smooth functions $a$ such that, for every compact set $K\subset X$ and all multi-indices $\alpha ,\beta$,Here $N$ is one fixed finite real order; the constants may depend on $K$. Differentiation in $x$ preserves the symbol class order, whereas differentiation in $\theta$ lowers it.

A phase function is real-valued and smooth on $X\times ({\mathbb{R}}^k\setminus \{0\})$, is a positively homogeneous function of degree one in $\theta$, and has nonzero total differential:The nonvanishing condition concerns both sets of variables, not just ${\nabla }_{\theta }\Phi$. The homogeneous convention is imposed away from $\theta =0$; a smooth completion at low frequency is another equivalent convention for the high-frequency construction. Low-frequency changes contribute a smooth function of $x$.

Choose a cutoff function $\rho (\theta )$ equal to one near zero. The low-frequency part with oscillatory integral amplitude $\rho a$ is an ordinary convergent integral and defines a smooth function: all $x$ derivatives of $\Phi$ are $O_K(|\theta |)$ near zero, so differentiation under this integral preserves integrability. For the remaining part define, when $\theta \ne 0$,Then $L{e}^{i\Phi }=e^{i\Phi }$. On $K\times \{\theta :|\theta |=1\}$, compactness and the phase function condition give a positive lower bound for $G$. Homogeneity consequently gives $G\ge c_K|\theta {|}^2$. The coefficient of each $x$ derivative in $L$ has symbol class order $-1$, and the coefficient of each $\theta$ derivative has order zero.

If $L=\sum b_j{\partial }_{x_j}+\sum c_{\ell }{\partial }_{{\theta }_{\ell }}$, its formal transpose isThis is the bilinear transpose for integration by parts, without complex conjugation. It lowers the symbol class order by one. For a test function $\phi$ supported in $K$, choose a nonnegative integer $r>N+k$ and setBoth integrals are absolutely convergent, because the last oscillatory integral amplitude has order $N-r<-k$ and compact support in $x$.

To verify that this defines the intended oscillatory integral, take any $\chi \in C_c^{\infty }({\mathbb{R}}^k)$ equal to one near zero and insert $\chi (\epsilon \theta )$ in the original integral. Repeated integration by parts gives the preceding expression with $(L^t{)}^r$ applied also to this cutoff function. Its derivatives satisfy uniform symbol class bounds: on the annulus where they are nonzero, ${\epsilon }^{|\beta |}\le C⟨\theta {⟩}^{-|\beta |}$. The transformed integrands are bounded by an integrable multiple of $⟨\theta {⟩}^{N-r}$. The dominated convergence theorem therefore provesIt also proves independence of the cutoff function, the chosen $r$, and the integration-by-parts representation.

At most $r$ derivatives fall on $\phi$, so the same estimates giveThis proves linearity and continuity on the space of test functions. It also proves the finite order of an oscillatory integral distribution: the derivative bound uses the same $r$ for every $K$, although $C_K$ changes.

For $x_0\in X$, take $k=n$, $\Phi =(x-x_0)\cdot \theta$, and $a=(2\pi {)}^{-n}{\theta }^{\alpha }$. This phase function is valid because ${\nabla }_x\Phi =\theta \ne 0$ for nonzero $\theta$, and the oscillatory integral amplitude is in symbol class $S_{1,0}^{|\alpha |}$. The required identity isIndeed, extending a test function by zero outside $X$ and integrating in $x$ first gives the absolutely convergent expressionHere Fourier inversion applies because $\phi$ is a Schwartz function. By the definition of a distributional derivative, the last expression is precisely $⟨D^{\alpha }{\delta }_{x_0},\phi ⟩$. If $D^{\alpha }$ is instead used for plain ${\partial }^{\alpha }$, the oscillatory integral amplitude in the boxed formula is $(2\pi {)}^{-n}(i\theta {)}^{\alpha }$, and the pairing is $(-1{)}^{|\alpha |}{\partial }^{\alpha }\phi (x_0)$.

Not every distribution is one oscillatory integral of the stated class. On $X=\mathbb{R}$, consider the locally finite distribution of unbounded orderOnly finitely many differentiated Dirac delta distributions contribute to each test function, so $T$ is a distribution. Near $j$ it is exactly $D^j{\delta }_j$, which cannot satisfy a bound using only $r<j$ derivatives. Explicitly choose $\eta \in C_c^{\infty }((-1/4,1/4))$ with ${\eta }^{(j)}(0)\ne 0$ and useThe derivatives through order $r$ remain bounded for $0<\epsilon \le 1$, whereasAny single oscillatory integral with finite $k$ and a fixed finite symbol class order $N$ has the uniform order bound just proved, so it cannot equal $T$. The same construction works on any nonempty open set in positive dimension by choosing points that leave every compact set and taking locally finite differentiated Dirac delta distributions there. This obstruction concerns a single fixed-order oscillatory integral, rather than local representations or an infinite sum of them.

Let $T_q\phi (x)={\sum }_{k=0}^q{\phi }^{(k)}(0)x^k/k!$ denote the Taylor polynomial at zero. On $0<|x|<1$, the integrand defining the Hadamard finite-part reciprocal-power distribution isThe first term is odd, so it cancels under symmetric truncation. The Taylor remainder in the second term has absolute value at most $|x{|}^m\Vert {\phi }^{(m)}{\Vert }_{\infty }/m!$. The integral near zero therefore has a finite limit. For $|x|\ge 1$, the original terms are integrable: $x^{-m}\phi$ has compact support, and each subtracted power has exponent $k-m\le -2$.

This proves existence and linearity. It also gives the global estimateThusIn particular the order of a distribution is bounded by one integer on all compact sets, not merely separately on each of them.

For the recurrence, write the truncated Hadamard finite-part integral aswhere parity evaluates all subtraction integrals explicitly:Set $C_1=0$ and ${\Lambda }_1=\mathrm{pv}(1/x)$. On the two truncated intervals, integration by parts givesThere are no boundary contributions at infinity. The coefficients of the subtraction terms satisfyThis is exactly the negative of the boundary term's Taylor polynomial through degree $m$; the degree-$m$ term itself vanishes by parity. HenceTaking the limit proves the full distributional identityThe same calculation includes $m=1$. Starting with the distributional derivative of the logarithmic modulus from part (i), induction yieldsThis identity holds on all of $\mathbb{R}$ as a distributional identity. It is stronger than matching the ordinary derivatives away from zero, which would leave possible differentiated Dirac delta distributions at zero undetermined.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when all functions eventually have compact support in one compact set $K$ and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set $K$, there are $C_K$ and a finite integer $m_K$ such thatThe integer $m_K$ is permitted to depend on $K$. The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.

For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is $O(x)$ near zero by the mean value theorem, and the integrand vanishes for large $x$ because $\phi$ has compact support. For $\mathrm{supp}\phi \subset [-R,R]$,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.

The function $\log |x|$ has local integrability, since ${\int }_0^1|\log x|dx<\infty$, and therefore defines a distribution. For its distributional derivative, remove $(-\epsilon ,\epsilon )$ and use integration by parts on both remaining intervals:The boundary term is $O(\epsilon |\log \epsilon |)$ and tends to zero. The omitted integral of $\log |x|{\phi }^{\prime }$ tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.

Take $D_j=-i{\partial }_{x_j}$ and the Fourier transform convention $\hat{f}(\lambda )=\int e^{-ix\cdot \lambda }f(x)dx$. Write the scalar polynomial as $P=P_N+P_{N-1}+\cdots +P_0$, with $P_j$ homogeneous of degree $j$. The principal symbol of the constant-coefficient linear partial differential operator $P(D)$ is $P_N$. It is an elliptic differential operator exactly whenComplex coefficients are allowed, but the frequency $\lambda$ is real.

By compactness of the unit sphere, $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Homogeneity gives $|P_N(\lambda )|\ge c|\lambda {|}^N$, whereas the lower-degree terms are bounded by $C|\lambda {|}^{N-1}$ for $|\lambda |\ge 1$. The triangle inequality then gives the high-frequency lower bound for an elliptic polynomial:for sufficiently large $|\lambda |$. ThusThe Japanese bracket is $⟨\lambda ⟩=(1+|\lambda {|}^2{)}^{1/2}$. If $N=0$, a nonzero constant $P$ satisfies the same assertion directly.

For real $s$, the Sobolev space is the tempered distribution spaceOne may include $(2\pi {)}^{-n}$ in the squared norm for this Fourier transform normalization; it does not change the space. The Local Sobolev space iswhere multiplication of a distribution by a smooth function is followed by extension by zero.

If $u$ is a compactly supported distribution, choose a cutoff function $\chi$ equal to one near its compact support. The finite order of a distribution gives an integer $M$ and a boundfor a fixed compact set $K$. This also makes $u$ a tempered distribution. Its Fourier transform of a compactly supported distribution is the smooth functionThe Leibniz rule gives the last estimate, and parameter differentiation under the finite-order pairing proves smoothness of this Fourier transform. Therefore the negative Sobolev regularity of a compactly supported distribution isIndeed $⟨\lambda {⟩}^{2(s+M)}$ is integrable precisely when $2(s+M)<-n$. This proves the claimed existence of a sufficiently negative Sobolev space index.

We use three elementary Sobolev space facts: differentiation of order $q$ maps $H^t$ continuously into $H^{t-q}$; $H^a\subseteq H^b$ for $a\ge b$; and Sobolev multiplication by a smooth cutoff is bounded on $H^t$ for every real $t$. The first two follow directly from the frequency weights. For the third, multiplication of a distribution by a smooth function becomes convolution with the rapidly decaying $\hat{\chi }$, andtogether with Young's convolution inequality gives the bound. These facts apply after localization as well.

The high-frequency lower bound for an elliptic polynomial gives a useful global implication. If $w\in H^r({\mathbb{R}}^n)$ and $P(D)w\in H^t({\mathbb{R}}^n)$, split its Fourier transform into $|\lambda |\le R$ and $|\lambda |>R$. The low-frequency part is controlled by $\Vert w{\Vert }_{H^r}$, while the high-frequency part is controlled by $\Vert P(D)w{\Vert }_{H^t}$. Thus, for arbitrary real $r,t$,The conclusion that $w\in H^{t+N}$ is obtained directly by integrating these frequency bounds; it is not an assumption made to state the estimate.

For the cutoff bootstrap for local elliptic regularity, fix $U⋐X$. A cutoff function equal to one near $\overline{U}$ makes $u$ a compactly supported distribution after multiplication, so the preceding negative-index argument gives $u\in H_{\mathrm{loc}}^r(U)$ for some finite $r$. Suppose inductively that $u\in H_{\mathrm{loc}}^q(U)$. For any test function $\chi \in C_c^{\infty }(U)$,The Leibniz rule shows that this commutator has order at most $N-1$, with smooth functions as coefficients, all with compact support:A second cutoff function equal to one near $\mathrm{supp}\chi$ lets us apply the stated Sobolev space bounds to every term. Hence $[P(D),\chi ]u\in H^{q-N+1}({\mathbb{R}}^n)$, while $\chi P(D)u\in H^s({\mathbb{R}}^n)$. For $N\ge 1$, the global estimate yieldsThis holds for every such $\chi$, so it improves the Local Sobolev space index by one until $s+N$ is reached. A finite number of iterations starting at $r$ provesSince $U⋐X$ was arbitrary, the conclusion holds throughout $X$. For $N=0$, division by the nonzero constant $P$ proves it immediately. This is the asserted elliptic regularity, proved without assuming an initial nonnegative Sobolev space index.

A first-order example on ${\mathbb{R}}^2$ is the first-order Cauchy-Riemann operatorIt is an elliptic differential operator with complex coefficients. In one real variable, $D=-id/dx$ is already a first-order elliptic differential operator.

There is no scalar elliptic differential operator of odd order in three variables. If its order $N$ were odd, its principal symbol would satisfy $P_N(-\omega )=-P_N(\omega )$ on $S^2$. Regard the continuous mapThe Borsuk-Ulam theorem gives $f(\omega )=f(-\omega )$ for some $\omega$. Oddness then gives $f(\omega )=0$, contradicting ellipticity. This is the odd-order obstruction for scalar elliptic operators in at least three dimensions; restricting to a three-dimensional subspace proves the higher-dimensional case too. For real coefficients alone, the same obstruction follows from the intermediate value theorem along a path between antipodal points.

The scalar qualification matters. An elliptic system of differential equations can be first order in three variables: with the Pauli matrices, the matrix symbol $A(\lambda )={\sum }_{j=1}^3{\sigma }_j{\lambda }_j$ satisfies $A(\lambda {)}^2=|\lambda {|}^{2}1$ and is invertible for $\lambda \ne 0$. This does not contradict the scalar polynomial obstruction.

Let $P_j$ be the orthogonal projection onto $V_j$. The dilation property and the orthonormal basis property give the orthonormal basis ${\phi }_{j,k}(x)=2^{j/2}\phi (2^{j}x-k)$ of $V_j$. The periodized nonnegative functionhas period one and ${\int }_0^{1}F=\Vert \phi {\Vert }_2^2=1$, by Tonelli theorem. For a function $g\in L^2(\mathbb{R})$ with compact support contained in a bounded interval $I$, Cauchy-Schwarz inequality and the Parseval identity giveas $j\to -\infty$. The last step uses absolute continuity of the integral of a locally $L^1$ function, since the interval's length tends to zero. If $f\in {\bigcap }_j{V}_j$, thenFunctions with compact support are dense in $L^2$, so $f=0$. ConsequentlyThis proof needs neither the density axiom nor any compact support assumption on the scaling function.

With $N$ vanishing moments, a wavelet annihilates every polynomial of degree below $N$. A Taylor polynomial then explains small wavelet coefficients on smooth parts of a signal, and efficient best N-term approximation. Compact support localizes coefficients near a feature, limits the number of boundary interactions, and permits a finite filter implementation. Increasing the number of vanishing moments while keeping an orthonormal basis generally requires a larger support of a function: a finite orthonormal filter with $N$ vanishing moments needs at least $2N$ taps, and the minimal-support Daubechies wavelet has support length $2N-1$ in the standard normalization.

The Haar wavelet, $\psi ={\chi }_{[0,1/2)}-{\chi }_{[1/2,1)}$, has one vanishing moment, unit support length and discontinuities. It is inexpensive and particularly suitable for piecewise constant data with sharp jumps. A Daubechies wavelet of a larger order has more vanishing moments and a longer finite filter; sufficiently large orders also provide greater regularity. It is useful when smooth trends should yield small coefficients, although the wider support of a function can spread a jump across more coefficients. Choose Haar for compact jump localization; choose a higher-order Daubechies wavelet for smooth polynomial structure. More vanishing moments alone does not make every low-order wavelet highly differentiable.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int \phi (x)e^{-ix\xi }dx$. Nesting and the orthonormal basis of $V_1$ give the scaling refinement equationHere $h_k=⟨\phi ,\sqrt{2}\phi (2\cdot -k)⟩$. Compact support makes these inner products zero for all but finitely many $k$, so $m$ is a trigonometric polynomial. Expanding the orthogonality of the integer translates of $\phi$ in the orthonormal basis of $V_1$ yieldsConsequentlyThis holds everywhere because $m$ is continuous. The invoked properties are nesting, dyadic dilation, orthonormal integer translates, and compact support; mere finite-energy refinement would not imply the quadrature mirror filter identity.

The factor $1+e^{-i\xi }$ has a simple zero at $\pi$, and $L(\pi )\ne 0$ makes the zero of $m$ there exactly order $N$. The quadrature mirror filter construction gives a high-pass symbol $m_1(\xi )=e^{-i\xi }\overline{m(\xi +\pi )}$, up to a constant phase, andThe normalized scaling function has $\hat{\phi }(0)=1$, so $\hat{\psi }$ has a zero of exactly order $N$ at zero. Compact support of the Daubechies wavelet permits differentiation under its Fourier transform:HenceThere are exactly $N$ vanishing moments, rather than just at least $N$.

Use the usual localized, compact support construction of an interval-adapted wavelet basis, including boundary wavelets with the stated vanishing moments, and order the linear N-term approximation by increasing resolution. Also interpret a piecewise polynomial function as having finitely many pieces. These conventions matter: regularity and vanishing moments alone, or an arbitrary enumeration, do not establish the asserted rates.