Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 340 / 2 / d / Solution

Normalize $\hat{\phi }(0)=1$, as in the displayed product convention. Iterating the scaling refinement equation gives $\hat{\phi }(\xi )={\prod }_{j\ge 1}m(\xi /2^j)$. The finite-product identityhas the value one at zero by continuity. ThereforeWrite $M=\sup |L|$. Since $L(0)=1$, we have $M\ge 1$. For $|\xi |>1$, take $J=\lceil {\log }_2|\xi |\rceil$. The first $J$ factors are bounded by $M^J\le M|\xi {|}^{{\log }_{2}M}$. For the remaining factors, the trigonometric polynomial satisfies $|L(t)|\le 1+C|t|$ on $|t|\le 1$, henceUsing $|(1-e^{-i\xi })/(i\xi )|\le \min (1,2/|\xi |)$ givesThe strict hypothesis permits $\epsilon >0$ with $N-{\log }_{2}M\ge \alpha +1+\epsilon$. The given Fourier-decay criterion proves uniform Hölder regularity of exponent $\alpha$. For the ordinary increment definition of Hölder continuity, take $0<\alpha \le 1$: 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.