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Use localization and cancellation of the assumed interval-adapted wavelet basis: at scale $j$, every detail wavelet has support diameter at most $B{2}^{-j}$, L1 norm at most $B^{\prime }{2}^{-j/2}$, and annihilates polynomials of degree less than $q$. These bounds also apply to the modified boundary wavelets; a restriction of an arbitrary whole-line wavelet without the cancellation-preserving boundary construction would not suffice. The finitely many coarse scaling functions can be treated separately.

Write $\alpha =r+\beta$ with $0<\beta \le 1$ as above. Because $\alpha <q$, we have $r<q$. Choose $x_0$ in the support of a function of ${\psi }_{j,n}$, and let $T_{x_0}$ be the degree-$r$ Taylor polynomial of $f$. The Hölder-Taylor remainder bound givesFor $r=0$ this is the Hölder seminorm bound. For $r\ge 1$, the integral Taylor remainder is bounded using $f^{(r)}(t)-f^{(r)}(x_0)$, whose magnitude is at most $[f^{(r)}{]}_{C^{0,\beta }}|t-x_0{|}^{\beta }$.

The vanishing moments remove $T_{x_0}$ from the inner product. Applying the remainder estimate and the L1 norm bound on the localized support yields the wavelet coefficient decay for Hölder functionsThusHere $C$ depends on the fixed wavelet family, the exponent and boundary construction, but not on $f,j,n$. The proof uses the assumed regularity and moments; it does not identify the minimal Daubechies wavelet order with its differentiability order, which need not coincide.

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.

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]