= Solution
Use the printed <Fourier transform> convention $\widehat\phi(t)=\int\phi(x)e^{-ixt}\,dx$, interpreted by the <Plancherel theorem> when $\phi$ is only square integrable. By the scaling and <function translation> rules,
$$
\widehat{\phi(2\cdot-n)}(t)=\frac12e^{-int/2}f(t/2).
$$
Taking the <Fourier transform> of the <scaling refinement equation> gives $f(t)=m(t/2)f(t/2)$, or equivalently
$$
\boxed{f(2t)=m(t)f(t),\qquad m(t)=\frac12\sum_na_ne^{-int}.}
$$
Conversely 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, $\{\sqrt2\phi(2\cdot-n)\}$ 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 $L^2(\mathbb R)$:
$$
\int_{\mathbb R}|m(t)f(t)|^2\,dt=\int_0^{2\pi}|m(t)|^2\sum_k|f(t+2\pi k)|^2\,dt=\int_0^{2\pi}|m(t)|^2\,dt.
$$
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 $\phi$.
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