= Solution
Use localization and cancellation of the assumed <interval-adapted wavelet basis>: at scale $j$, every detail <wavelet> has support diameter at most $B2^{-j}$, <L1 norm> at most $B'2^{-j/2}$, 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 $0<\beta\le1$ 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> gives
$$
|f(x)-T_{x_0}(x)|\le C_\alpha\|f\|_{C^\alpha}|x-x_0|^\alpha.
$$
For $r=0$ this is the <Hölder seminorm> bound. For $r\ge1$, 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 functions>
$$
|\langle f,\psi_{j,n}\rangle|\le\sup_{\operatorname{supp}\psi_{j,n}}|f-T_{x_0}|\,\|\psi_{j,n}\|_1\le C_\alpha B^\alpha B'\|f\|_{C^\alpha}2^{-j\alpha}2^{-j/2}.
$$
Thus
$$
\boxed{|\langle f,\psi_{j,n}\rangle|\le C2^{-j(\alpha+1/2)}\|f\|_{C^\alpha}.}
$$
Here $C$ 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.
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