 Solution

First use the usual real-valued, measurable interpretation of the transition data. The complementary squares then imply $|f|\le 1$, and the compact support gives $f\in L_2$. Consider the periodization on $[-\pi ,\pi ]$. For $|t|<2\pi /3$, only $f(t)=1$ contributes. For $2\pi /3\le t\le \pi$, only $f(t)$ and $f(t-2\pi )$ can contribute, and their squares sum to one. For $-\pi \le t\le -2\pi /3$, apply the complementary identity at $t+2\pi$ to obtain $f(t{)}^2+f(t+2\pi {)}^2=1$. Therefore the orthonormal translates and Fourier periodization identity holds almost everywhere.

Define a periodic MRA low-pass filter by prescribing it on $[-\pi ,\pi ]$:In the central band, $f(t)=1$, so $f(2t)=m(t)f(t)$. In the rest of this fundamental interval, both $m(t)$ and $f(2t)$ vanish. If $\pi <|t|<4\pi /3$, reducing $t$ modulo $2\pi$ puts it in an outer band where $m=0$, and again $f(2t)=0$. For $|t|\ge 4\pi /3$, both $f(t)$ and $f(2t)$ vanish. Endpoint choices affect only a null set. Thus the complementary transition bands for a scaling function give both Fourier conditions, with this explicitly constructed periodic mask.

A concrete real-valued choice, illustrating that the transition constraints are consistent, isOn the positive transition band put $u=3t/(2\pi )-1\in [0,1]$. The other translate has parameter $1-u$, so the sum is ${\cos }^2(\pi u/2)+{\sin }^2(\pi u/2)=1$.

There is a genuine qualification in the printed formulation: it uses ordinary squares without explicitly declaring $f$ real-valued. If arbitrary complex values are allowed, the claimed verification is false. Take $f=1$ for $|t|\le 2\pi /3$, $f=0$ for $|t|\ge 4\pi /3$, and setThe printed complementary-square rule holds, including its endpoints: in the interior $i^2+(\sqrt{2}{)}^2=1$, while the endpoints pair one with zero. Nevertheless, for $2\pi /3<t<\pi$,This compactly supported $L_2$ function is the Fourier transform of an $L_2$ function, so it is a counterexample within the natural complex Hilbert space. The verification is valid for real-valued $f$, or with absolute squares in the transition rule; it is not valid for unrestricted complex $f$ with ordinary squares.