= Solution
The density axiom and continuity of the <Fourier transform> at zero imply $|\widehat\varphi(0)|=1$. Here is a proof that avoids assuming a normalization of the integral. Choose a nonzero $f$ whose <Fourier transform> has bounded support. The <MRA projection Fourier identity>, with no aliasing once $j$ is sufficiently large, is
$$
\|P_jf\|_2^2=\frac1{2\pi}\int_{\mathbb R}|\widehat f(\xi)|^2|\widehat\varphi(2^{-j}\xi)|^2\,d\xi.
$$
Since $\varphi\in L^1$, its <Fourier transform> is continuous and bounded, so this tends to $|\widehat\varphi(0)|^2\|f\|_2^2$. Density and nesting give $P_jf\to f$ in $L^2$. Thus $|\widehat\varphi(0)|=1$, in particular it is nonzero. Evaluating the <scaling refinement equation> in frequency at zero now gives
$$
\widehat\varphi(0)=m(0)\widehat\varphi(0),\qquad \boxed{m(0)=1},
$$
which is stronger than the requested absolute-value equality. Multiplying the <scaling function> by a constant of modulus one normalizes its integral to one without changing $m$.
Back to article page
