= Solution
The half-open translates $[-\pi,\pi)+2\pi k$ partition the real line. Exactly one summand is therefore nonzero in the periodization of $|f|^2$, giving $\sum_k|f(t+2\pi k)|^2=1$.
Choose the <Shannon scaling mask> to be the $2\pi$-periodic extension of $\mathbf1_{[-\pi/2,\pi/2)}$. On the <support> of $f$, its product with $f$ is $\mathbf1_{[-\pi/2,\pi/2)}(t)=f(2t)$; outside that <support> both sides vanish. Thus the <scaling refinement equation> holds in frequency. The <Fourier coefficients> give its precise filter normalization:
$$
a_n=\frac1\pi\int_{-\pi/2}^{\pi/2}e^{int}\,dt
=\begin{cases}1,&n=0,\\\dfrac{2\sin(n\pi/2)}{\pi n},&n\ne0.\end{cases}
$$
These <coefficients> are square summable, so the interpretation in part (b) applies.
The <Fourier inversion theorem> gives the <Shannon scaling function>
$$
\phi(x)=\frac1{2\pi}\int_{-\pi}^{\pi}e^{ixt}\,dt,
\qquad\boxed{\phi(x)=\frac{\sin(\pi x)}{\pi x}\ (x\ne0),\qquad\phi(0)=1.}
$$
This is the normalized <sinc function>. Its <Fourier transform> is taken in the <Plancherel theorem> sense: the <Shannon scaling function> belongs to $L^2$ but not $L^1$, so its Fourier integral is not absolutely convergent. The inverse integral above is an ordinary integral because the rectangular transform belongs to $L^1$. The resulting refinement spaces are exactly the $L^2$ <functions> with <Fourier transform> supported in $[-2^j\pi,2^j\pi]$; their union is <dense> and their intersection is zero. Thus this example also gives a full <multiresolution analysis>.
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