 Solution

Use the Fourier transform convention $f(t)=\hat{\varphi }(t)=\int \varphi (x)e^{-ixt}dx$ and the Plancherel theorem normalization $\Vert \varphi {\Vert }_2^2=(2\pi {)}^{-1}\Vert f{\Vert }_2^2$. 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 $\widehat{\varphi (2\cdot -n)}(t)=\frac{1}{2}{e}^{-int/2}f(t/2)$. Thus the scaling refinement equation transforms to $f(t)=m(t/2)f(t/2)$, or $f(2t)=m(t)f(t)$, with the $2\pi$-periodic MRA low-pass filter $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. 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 givesThese 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 equivalenceFor 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 givesHence $m$ is bounded and belongs to $L_2$ of a period. Write $m=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$ in $L_2$ and let $m_N$ denote the symmetric partial sums. Since $P=1$,After substituting $t/2$, this proves convergence of the transformed refinement sums to $f$ in $L_2$. Inverting the Fourier transform proves $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$ in $L_2$. Thus the two spatial conditions and the two Fourier conditions are equivalent, with the periodic mask and convergence conventions made explicit.