= Solution
Use the <Fourier transform> convention $f(t)=\widehat\phi(t)=\int\phi(x)e^{-ixt}\,dx$ and the <Plancherel theorem> normalization $\|\phi\|_2^2=(2\pi)^{-1}\|f\|_2^2$. For a general $L_2$ <scaling function>, the transform means its $L_2$ extension; an ordinary absolutely convergent integral is not required.
A change of variables gives $\widehat{\phi(2\cdot-n)}(t)=\tfrac12e^{-int/2}f(t/2)$. Thus the <scaling refinement equation> transforms to $f(t)=m(t/2)f(t/2)$, or \b[$f(2t)=m(t)f(t)$], with the $2\pi$-periodic <MRA low-pass filter> $m(t)=\tfrac12\sum_na_ne^{-int}$. The periodicity is necessary; allowing an arbitrary quotient $f(2t)/f(t)$ would not express refinement by integer translates.
For orthonormality, let $P(t)=\sum_{k\in\mathbb Z}|f(t+2\pi k)|^2$. This is a $2\pi$-periodic $L_1$ function, since $f\in L_2$. The <Plancherel theorem> gives
$$
\langle\phi(\cdot-j),\phi(\cdot-\ell)\rangle=\frac1{2\pi}\int_{\mathbb R}|f(t)|^2e^{i(\ell-j)t}\,dt=\frac1{2\pi}\int_0^{2\pi}P(t)e^{i(\ell-j)t}\,dt.
$$
These inner products equal $\delta_{j\ell}$ precisely when the <Fourier coefficients> of $P$ agree with those of one. The <uniqueness of Fourier coefficients in L1> proves the equivalence
$$
\boxed{\{\phi(\cdot-n)\}\text{ orthonormal}\quad\Longleftrightarrow\quad\sum_k|f(t+2\pi k)|^2=1\ \text{a.e.}}
$$
For completeness, the converse refinement implication is also valid with the natural $L_2$ series interpretation. Suppose this periodization identity and $f(2t)=m(t)f(t)$ hold for a measurable $2\pi$-periodic $m$. Splitting the periodization at $2t$ into even and odd translates gives
$$
\begin{aligned}
1&=\sum_k|f(2t+2\pi k)|^2\\
&=|m(t)|^2\sum_r|f(t+2\pi r)|^2+|m(t+\pi)|^2\sum_r|f(t+\pi+2\pi r)|^2\\
&=|m(t)|^2+|m(t+\pi)|^2.
\end{aligned}
$$
Hence $m$ is bounded and belongs to $L_2$ of a period. Write $m=\tfrac12\sum_na_ne^{-int}$ in $L_2$ and let $m_N$ denote the symmetric partial sums. Since $P=1$,
$$
\int_{\mathbb R}|(m_N(t)-m(t))f(t)|^2\,dt=\int_0^{2\pi}|m_N(t)-m(t)|^2P(t)\,dt\longrightarrow0.
$$
After substituting $t/2$, this proves convergence of the transformed refinement sums to $f$ in $L_2$. Inverting the <Fourier transform> proves $\phi=\sum_na_n\phi(2\cdot-n)$ in $L_2$. Thus \b[the two spatial conditions and the two Fourier conditions are equivalent], with the periodic mask and convergence conventions made explicit.
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