 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 givesThus taking the transform of the scaling refinement equation gives $f(\xi )=m(\xi /2)f(\xi /2)$, orConversely 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 givesHence orthonormality of all integer translates is equivalent, by uniqueness of Fourier coefficients in L1, to the periodized-energy conditionWhen this holds, the squared norm of ${\sum }_n{a}_n\varphi (2\cdot -n)$ is $\frac{1}{2}{\sum }_n|a_n{|}^2$, 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,Thus $m$ is bounded and its Fourier coefficients $a_n/2$ are square summable. If $m_N$ are its Fourier partial sum, thenThis 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 $\varphi$.