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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.

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 localized tensor-product wavelets and finitely many bounded polynomial pieces, with the boundary wavelets adapted as in part (b). A rectifiable smooth curve of length $\ell$ meets $O(1+\ell 2^j)$ dyadic squares of side $2^{-j}$: subdividing an arclength parametrization into pieces of length at most $2^{-j}$ covers it by that many balls, each meeting only a bounded number of squares. Enlarging squares by the fixed support diameter preserves the count.

Each normalized two-dimensional wavelet has $L^1$ norm $O(2^{-j})$. A coefficient meeting the curve is therefore $O(2^{-j})$, and the total squared energy of these coefficients at level $j$ is $O(2^{-j})$. If $p<q$, all other coefficients vanish by the vanishing moments. Keeping the curve coefficients through level $J$ costs $O(2^J)$ and leaves squared error $O(2^{-J})$.

The printed part (e) does not repeat $p<q$. The bound still holds for any fixed polynomial degree when $q\ge 1$: on a smooth piece, a Taylor polynomial in the variable carrying a wavelet gives coefficient size $O(2^{-(q+1)j})$. There are $O(4^j)$ such coefficients, so their squared energy is $O(2^{-2qj})$. Retain all coefficients through $K=\lfloor J/2\rfloor$, and curve coefficients through $J$. The cost is $O(4^K+2^J)=O(2^J)$ and the omitted squared energy isThe best N-term approximation is at least as good as this selection, provingThis argument covers the unqualified finite-degree clause without adding an unnecessary restriction $p<q$.

Let $x^{\ast }(\lambda )=\sqrt{1-1/\lambda }$ and center the map at this parameter-dependent fixed point:A Taylor polynomial in $X$, with the coefficients expanded at $(x,\lambda )=(1/\sqrt{2},2)$, givesunder the period-doubling scaling $X=O(\sqrt{\mu })$. The coefficients areHere we used $F_x(x^{\ast }(\lambda ),\lambda )=3-2\lambda$, $F_{xx}=-6\lambda x$, and $F_{xxx}=-6\lambda$.

Composing the local map with itself and retaining the terms of order $\mu X$ and $X^3$ givesBesides the fixed solution $X=0$, the two leading solutions therefore satisfyThus the period-two points born at $\lambda =2$ areto leading order. With the coefficients above, the radicand is positive for $\mu >0$.

The boundary Schauder estimate iswhere $C$ depends only on the dimension, $\alpha$, the ellipticity constants, and the coefficient Hölder norms.

Subtract a $C^{2,\alpha }$ extension of $\phi$ to reduce to a function $v$ with zero data on the flat boundary; this changes the forcing by a controlled $C^{0,\alpha }$ term. The key local estimate is that for every $\delta >0$,To prove it, argue by contradiction. A failing normalized sequence has points $x_k,y_k$ at which the second-derivative Hölder quotient stays nonzero. Set $r_k=|x_k-y_k|$, subtract the appropriate second-order Taylor polynomial, and rescale space by $r_k$ and the functions by $r_k^{2+\alpha }[D^2{v}_k{]}_{\alpha }$. Necessarily $r_k\to 0$.

If the rescaled distance to the flat boundary tends to infinity, the domains converge to all of ${\mathbb{R}}^n$; otherwise they converge to a half-space. Coefficient compactness freezes the principal matrix, while the normalized right-hand sides converge locally uniformly to zero. A linear change of variables turns every limiting equation into the Laplace equation. In the whole-space case the polynomial-growth Liouville theorem for harmonic functions makes the limit a polynomial of degree at most two. In the half-space case the zero boundary data permit odd reflection across the flat boundary, after which the same theorem applies. The subtracted Taylor normalization forces that polynomial to vanish to second order, contradicting the nonzero limiting Hölder oscillation.

Apply the local estimate on all interior balls and boundary half-balls. The interior and boundary forms of the Simon absorption lemma absorb the term multiplied by $\delta$, while the Hölder interpolation inequality absorbs the remaining lower $C^2$ norm into the top seminorm and the $C^0$ norm. Restoring $\phi$ yields the displayed estimate.

A Peano zero through order $r$ means that the function admits the zero Taylor polynomial with remainder $o(|t{|}^r)$ at $x_0$. This does not assert the existence of ordinary iterated derivatives on a neighborhood. For example, multiplying $t^{r+1}$ by a bounded continuous function with dense points of nondifferentiability preserves this estimate while destroying neighborhood differentiability.

Subtracting the quadratic Taylor polynomial at the rescaling centre makes the function, gradient and Hessian matrix vanish there. Dividing by $A{r}^{2+\mu }$ preserves the normalized Hölder oscillation of the Hessian matrix while turning a small forcing-to-oscillation ratio into a harmonic limiting equation. The three vanished Taylor coefficients provide local bounds needed by the Arzela-Ascoli theorem.

The Taylor remainder is the difference between a function and a finite Taylor polynomial. Bounds on higher derivatives or their Hölder continuity control this error.

The Taylor theorem expresses a sufficiently differentiable function as a finite Taylor polynomial plus a remainder.

A wavelet has $q$ vanishing moments when these integrals vanish for $0\le r<q$. For compact support, this is equivalent to a zero of order at least $q$ at the origin of its Fourier transform. Such a wavelet annihilates polynomials of degree less than $q$, and a Taylor polynomial bounds coefficients on smooth regions.

Localized wavelets of support diameter $O(2^{-j})$ and $L^1$ norm $O(2^{-j/2})$, with enough vanishing moments, satisfySubtract a Taylor polynomial annihilated by the wavelet, apply the Hölder-Taylor remainder bound on its support of a function, and multiply by its $L^1$ norm. The same argument applies to localized boundary wavelets with the same cancellation; finitely many coarse scaling functions are handled separately.