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Finite Parseval and Plancherel identity for the unitary DFT

Statement

Let N≥1 and f,g∈CZ/N. Then

⟨FNf,FNg⟩=⟨f,g⟩,

where both pairings are the counting inner products of The counting inner product on CZ/NZ. In particular

∑k=0N−1∣(FNf)(k)∣2=∑x=0N−1∣f(x)∣2,

so FN preserves the counting inner product; invertibility of FN is not asserted here.

Facts & Assumptions

Given: A natural number N≥1, functions f,g∈CZ/N, and classes x,y∈Z/NZ.

[F1]

FNh(k)=N−1/2∑x=0N−1h([x]N)e−2πikx/N (The unitary discrete Fourier transform on Z/NZ); ⟨u,v⟩=∑z∈Z/Nu(z)v(z)‾ is the counting inner product (The counting inner product on CZ/NZ).

[F2]

∑k=0N−1e2πi(a−b)k/N=N when a≡b(modN) and 0 otherwise (Orthogonality of the characters x↦e2πikx/N on Z/NZ).

[F3]

Finite sums over Z/NZ are computed from any enumeration, are invariant under reindexing along a bijection, split over disjoint unions, satisfy the finite Fubini rule, and carry scalar factors (A finite sum in a commutative monoid indexed by an arbitrary finite set, Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule); the classes [0]N,…,[N−1]N enumerate the group (For n≥1, every class in Z/n has one representative r with 0≤r<n, so ∣Z/n∣=n; while Z/0 is in bijection with Z).

[L2]

Conjugation and modulus: z+w‾=z‾+w‾, zw‾=z‾ w‾, zz‾=∣z∣2, ∣z∣=0  ⟺  z=0 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive); ∣exp⁡(x+iy)∣=ex (exp⁡(x+iy)=ex(cos⁡y+isin⁡y), ∣exp⁡(x+iy)∣=ex, and eiπ+1=0).

Proof

technique · direct
1.1F1F2F3L1L2L3

Conjugation of the second factor: for each class k, (FNg)(k)‾=N−1/2∑y=0N−1g([y]N)‾ e2πiky/N. Indeed, conjugation is additive and multiplicative and fixes the real scalar N−1/2 (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive, Laws of rational exponents); and e−2πiky/N‾=e2πiky/N, because for ζ=eiθ with θ=−2πky/N one has ∣ζ∣=1 by [L2] and hence ζ‾=∣ζ∣2/ζ=ζ−1=e−iθ.

1.2F2

The orthogonality sum: ∑k=0N−1e−2πik(x−y)/N=N when [x]=[y] and 0 otherwise, by [F2] with the integers y and x in the roles of the two congruence parameters.

1.3F3

Collapsing a sum supported at one class: for any c:Z/NZ→C and class a, ∑y∈Z/Nc(y)⋅(N if y=a, 0 otherwise)=N c(a); split the index set into {a} and its complement by [F3], the complement contributing 0 because every term there has the factor 0C.

2.1F1F3L1L3step 1.1

Expanding the pairing: substituting [F1] for both transforms, step 1.1 for the conjugate factor, and interchanging the finite sums by [F3] gives ⟨FNf,FNg⟩=N−1∑x=0N−1∑y=0N−1f([x]N)g([y]N)‾∑k=0N−1e−2πik(x−y)/N, where the exponentials combine by the addition law [L1] and the scalar is N−1/2N−1/2=N−1 by [L3].

3.1F1F3L2L3step 1.2step 1.3step 2.1∎

Evaluating the inner sum by step 1.2 and then collapsing the outer sum by step 1.3 gives ⟨FNf,FNg⟩=N−1∑x=0N−1N f([x]N)g([x]N)‾=∑x=0N−1f([x]N)g([x]N)‾=⟨f,g⟩ by [L3] and the standard-representative form of the counting inner product [F1]. Taking g=f and using zz‾=∣z∣2 from [L2], the same computation gives ∑k∣(FNf)(k)∣2=∑x∣f([x]N)∣2; both assertions are proved.

Remarks

  • Isometry versus unitary isomorphism. The identity proved here says that FN preserves the counting inner product; it does not by itself assert that FN is bijective, and none of its steps uses inversion. Invertibility is the separate content of the inversion theorem on this page, and [L2] alone does not supply it.

  • Both sides use the same weight. There is no factor 1/N on either side of the displayed identity, and the cancellation of the two N−1/2 factors against the orthogonality value N is the only place where the normalisation is used. In Taylor's convention the same computation reads as the unitarity of Φn between the (1/n)-weighted space on Γn and the counting-measure space on Zn (11.5).

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Sources