 Solution

The density axiom and continuity of the Fourier transform at zero imply $|\hat{\phi }(0)|=1$. 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, isSince $\phi \in L^1$, its Fourier transform is continuous and bounded, so this tends to $|\hat{\phi }(0){|}^2\Vert f{\Vert }_2^2$. Density and nesting give $P_{j}f\to f$ in $L^2$. Thus $|\hat{\phi }(0)|=1$, in particular it is nonzero. Evaluating the scaling refinement equation in frequency at zero now giveswhich 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 $m$.