= Solution
Normalize $\widehat\varphi(0)=1$, as in the displayed product convention. Iterating the <scaling refinement equation> gives $\widehat\varphi(\xi)=\prod_{j\ge1}m(\xi/2^j)$. The finite-product identity
$$
\prod_{j=1}^J\frac{1+e^{-i\xi/2^j}}2
=\frac{1-e^{-i\xi}}{2^J(1-e^{-i\xi/2^J})}
\longrightarrow\frac{1-e^{-i\xi}}{i\xi}
$$
has the value one at zero by continuity. Therefore
$$
\boxed{\widehat\varphi(\xi)=\left(\frac{1-e^{-i\xi}}{i\xi}\right)^N\prod_{j\ge1}L(\xi/2^j).}
$$
Write $M=\sup|L|$. Since $L(0)=1$, we have $M\ge1$. For $|\xi|>1$, take $J=\lceil\log_2|\xi|\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi|^{\log_2M}$. For the remaining factors, the <trigonometric polynomial> satisfies $|L(t)|\le1+C|t|$ on $|t|\le1$, hence
$$
\prod_{j>J}|L(\xi/2^j)|\le\exp\left(C\sum_{j>J}|\xi|2^{-j}\right)\le e^C.
$$
Using $|(1-e^{-i\xi})/(i\xi)|\le\min(1,2/|\xi|)$ gives
$$
|\widehat\varphi(\xi)|\le C'(1+|\xi|)^{-N+\log_2M}.
$$
The strict hypothesis permits $\varepsilon>0$ with $N-\log_2M\ge\alpha+1+\varepsilon$. The given Fourier-decay criterion proves \b[uniform Hölder regularity of exponent $\alpha$]. For the ordinary increment definition of <Hölder continuity>, take $0<\alpha\le1$: Fourier inversion and $|e^{ih\xi}-1|\le C_\alpha|h\xi|^\alpha$ directly give the required bound. If $\alpha>1$, “Lipschitz-$\alpha$” must mean higher-order <Hölder space> regularity; an ordinary increment bound of exponent greater than one forces a function to be constant. An unnormalized <scaling function> contributes its constant phase to the product formula.
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