= Solution
The half-open intervals $[-\pi+2\pi k,\pi+2\pi k)$ tile the line, so exactly one term contributes to $\sum_k|f(t+2\pi k)|^2$. It equals one almost everywhere.
Choose the <Shannon scaling mask>, a $2\pi$-periodic <low-pass filter of a multiresolution analysis> which equals one on $[-\pi/2,\pi/2)$ and zero on the rest of $[-\pi,\pi)$. On the support of $f$ its product with $f(t)$ equals $f(2t)$; outside that support both sides vanish. Its <Fourier coefficients> give
$$
a_0=1,\qquad a_n=\frac{2\sin(n\pi/2)}{\pi n}\quad(n\ne0),
$$
so it also has the required symbol representation. The inverse <Fourier transform> gives \b[the <Shannon scaling function>]
$$
\boxed{\phi(x)=\frac1{2\pi}\int_{-\pi}^{\pi}e^{ixt}\,dt
=\frac{\sin(\pi x)}{\pi x},\qquad\phi(0)=1.}
$$
This <sinc function> has $L^2$ norm one. It illustrates why the <Fourier transform> convention in part B must allow $L^2$ transforms: the <Shannon scaling function> is not absolutely integrable on the line.
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