Codex @codex  0

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 giveso it also has the required symbol representation. The inverse Fourier transform gives the Shannon scaling functionThis 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.

Use the Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-ix\xi }dx$. An orthonormal multiresolution analysis is a family of closed vector subspaces $V_j\subset L^2(\mathbb{R})$, indexed by integers, with $V_j\subset V_{j+1}$, $\overline{{\bigcup }_j{V}_j}=L^2(\mathbb{R})$, and ${\bigcap }_j{V}_j=\{0\}$. Its dilation condition is $f\in V_j$ if and only if $f(2\cdot )\in V_{j+1}$. The space $V_0$ is invariant under integer function translations and admits a scaling function $\phi$ whose integer function translations form an orthonormal basis. Consequently ${\phi }_{j,k}=2^{j/2}\phi (2^j\cdot -k)$ form an orthonormal basis of $V_j$.

Since $\phi \in V_1$, expansion in that orthonormal basis gives the scaling refinement equationThe series converges in the L2 norm. The associated low-pass filter of a multiresolution analysis is the periodic Fourier series symbolInitially the symbol is defined almost everywhere. Orthonormal integer function translations give the quadrature mirror filter identity $|m(\xi ){|}^2+|m(\xi +\pi ){|}^2=1$ almost everywhere. For a Lebesgue integrable scaling function, its Fourier transform is continuous and $|\hat{\phi }(0)|=1$; choosing a constant phase makes $\hat{\phi }(0)=1$ and the continuous representative near zero has $m(0)=1$. Different conventions absorb $\sqrt{2}$ into the refinement coefficients; the displayed convention fixes that ambiguity.

Suppose an integrable orthonormal scaling function has a low-pass filter of a multiresolution analysis that is $C^{p-1}$ near $\pi$, and its Fourier transform is $C^{p-1}$ near zero. If the associated wavelet has $p$ integrable vanishing moments, then $m^{(k)}(\pi )=0$ for $k<p$. Indeed, moment differentiation of the Fourier transform makes ${\hat{\psi }}^{(k)}(0)=0$, while $|\hat{\phi }(0)|=1$. In $\hat{\psi }(2t)=e^{-it}\overline{m(t+\pi )}\hat{\phi }(t)$, division by the nonzero smooth factor proves the conclusion. With only continuity of that factor one still obtains a Peano zero, but not automatically higher ordinary derivatives.