= Solution
Use the unnormalized <Fourier transform> in the question and the <Plancherel theorem>, with inverse factor $1/(2\pi)$. For a general $L^2$ <scaling function>, the transform is interpreted in the $L^2$ sense; the printed integral need not be absolutely convergent. All frequency identities are almost-everywhere statements.
Changing variables in the <Fourier transform> gives
$$
\widehat{\phi(2\cdot-n)}(\xi)=\frac12e^{-in\xi/2}f(\xi/2).
$$
Thus taking the transform of the <scaling refinement equation> gives $f(\xi)=m(\xi/2)f(\xi/2)$, or
$$
\boxed{f(2t)=m(t)f(t),\qquad m(t)=\frac12\sum_na_ne^{-int}.}
$$
Conversely the same calculation and injectivity of the <Fourier transform> recover refinement, with convergence understood in $L^2$.
To justify both the <orthogonality> assertion and this convergence precisely, put $P(t)=\sum_{k\in\mathbb Z}|f(t+2\pi k)|^2$. It belongs to $L^1[-\pi,\pi]$ by monotone integration. The <Plancherel theorem> gives
$$
\langle\phi,\phi(\cdot-j)\rangle
=\frac1{2\pi}\int_{-\pi}^{\pi}P(t)e^{-ijt}\,dt.
$$
Hence orthonormality of all integer translates is equivalent, by <uniqueness of Fourier coefficients in L1>, to \b[the periodized-energy condition]
$$
\boxed{P(t)=1\quad\text{almost everywhere}.}
$$
When this holds, the squared norm of $\sum_na_n\phi(2\cdot-n)$ is $\tfrac12\sum_n|a_n|^2$, so square-summable coefficient series converge in $L^2$. In the converse direction the refinement identity and $P=1$ give, by splitting the periodization of $|f(2t)|^2$ into even and odd translates,
$$
1=|m(t)|^2+|m(t+\pi)|^2.
$$
Thus $m$ is bounded and its <Fourier coefficients> $a_n/2$ are square summable. If $m_N$ are its <Fourier partial sum>, then
$$
\int_{\mathbb R}|m_N(\xi/2)-m(\xi/2)|^2|f(\xi/2)|^2\,d\xi
=2\int_{-\pi}^{\pi}|m_N(t)-m(t)|^2P(t)\,dt\longrightarrow0.
$$
This verifies the transformed refinement series converges to $f$, completing the equivalence of the two pairs of conditions without imposing an unnecessary $L^1$ assumption on $\phi$.
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