 Solution

Use the printed Fourier transform convention $\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$, interpreted by the Plancherel theorem when $\varphi$ is only square integrable. By the scaling and function translation rules,Taking the Fourier transform of the scaling refinement equation gives $f(t)=m(t/2)f(t/2)$, or equivalentlyConversely this identity, the same scaling rule and injectivity of the Fourier transform recover the scaling refinement equation.

For infinite filters, these statements are equalities in $L^2$, with Fourier series understood in $L^2[0,2\pi ]$. Under the accompanying orthonormality condition, $\{\sqrt{2}\varphi (2\cdot -n)\}$ is an orthonormal sequence, so the refinement series converges in $L^2$ precisely for $a\in {\ell }^2$. Moreover the condition in part (b)(2) makes multiplication by $f$ an isometry from periodic $L^2$ symbols into $L^2(\mathbb{R})$:Applying this to finite filter truncations justifies passage to the limit in both directions. Thus the two paired assertions have a rigorous equivalent interpretation without requiring absolute integrability of $\varphi$.