 Solution

ID: past-exam-of-the-real-and-harmonic-analysis-course-of-the-university-of-oxford/2008/3/d/solution

μ (n) = \int_{T} e^{- in x} d μ (x), (μ * ν) (E) = \int_{T} \int_{T} 1_{E} (x + y) d μ (x) d ν (y) .

2 π

. This is the pushforward of the product complex measure under addition; equivalently its integral against a continuous

f

\iint f (x + y) d μ (x) d ν (y)

. The integrands are bounded and the variation measures finite, so Fubini applies. In particular,

μ * ν (n) = \iint e^{- in (x + y)} d μ (x) d ν (y) = (\int e^{- in x} d μ (x)) (\int e^{- in y} d ν (y)) .

Thus $μ * ν (n) = μ (n) ν (n)$ . These measure coefficients use no additional Haar normalization factor: a probability has zeroth coefficient one.

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