Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 318 / 5 / b

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- Solution b

Solution

Fourier transforming the refinement equation and changing variables gives

f (2 t) = m (t) f (t), m (t) = \frac{1}{2} n \sum a_{n} e^{- in t} .

By Parseval identity, orthonormality of the translates is equivalent to

\frac{1}{2 π} \int_{R} ∣ f (t) ∣^{2} e^{in t} d t = δ_{n 0} .

These are precisely the Fourier coefficients of the periodization

P (t) = \sum_{k} ∣ f (t + 2 πk) ∣^{2}

. Therefore

{ϕ (\cdot - n)} is orthonormal ⟺ k \sum ∣ f (t + 2 πk) ∣^{2} = 1 a.e.

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