 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 $1_{[-\pi /2,\pi /2)}$. On the support of $f$, its product with $f$ is $1_{[-\pi /2,\pi /2)}(t)=f(2t)$; outside that support both sides vanish. Thus the scaling refinement equation holds in frequency. The Fourier coefficients give its precise filter normalization:These coefficients are square summable, so the interpretation in part (b) applies.

The Fourier inversion theorem gives the Shannon scaling functionThis 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.