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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 27K / a / Solution

Use the convention

f (u) = \int_{R^{d}} e^{i u \cdot x} f (x) d x .

The Fourier inversion theorem says that, when

f, f \in L^{1} (R^{d})

,

f (x) = \frac{1}{( 2 π ) ^{d}} \int_{R^{d}} e^{- i u \cdot x} f (u) d u

for almost every

x

.

Since

\overline{f (u)} = \int_{R^{d}} e^{- i u \cdot x} \overline{f (x)} d x,

the Fubini's theorem and Fourier inversion give

\int_{R^{d}} ∣ f (u) ∣^{2} d u = \int_{R^{d}} f (u) \int_{R^{d}} e^{- i u \cdot x} \overline{f (x)} d x d u = \int_{R^{d}} \overline{f (x)} \int_{R^{d}} e^{- i u \cdot x} f (u) d u d x = (2 π)^{d} \int_{R^{d}} ∣ f (x) ∣^{2} d x .

Thus the Plancherel theorem identity in this convention is

∥ f ∥_{2}^{2} = (2 π)^{d} ∥ f ∥_{2}^{2} .

The inverse integral is a continuous function of

x

, since

f \in L^{1}

permits dominated convergence. If

f

is continuous, it and the inverse integral are continuous and agree almost everywhere. They must agree everywhere: otherwise their continuous difference would be nonzero on an open set of positive measure. This is the continuous version of Fourier inversion.

Solved by gpt-5.6-sol high.

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