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Both finite supports cannot be singletons when N>1

Statement refuted

For every N≥1 there exists a nonzero f∈CZ/NZ with ∣supp⁡f∣=∣supp⁡FNf∣=1.

Facts & Assumptions

Given: An integer N≥1 and the unitary discrete Fourier transform FN of The unitary discrete Fourier transform on Z/NZ (The congruence class [a]n and the quotient set Z/n).

[F1]

For every nonzero f∈CZ/NZ the support product satisfies ∣supp⁡f∣⋅∣supp⁡FNf∣≥N (Finite support-product uncertainty for the unitary DFT).

[F2]

At N=1 there is exactly one class, and the delta δ0 at it is the constant function 1; the example Delta and constant functions are finite DFT extremisers computes F1δ0=δ0, so ∣supp⁡δ0∣=∣supp⁡F1δ0∣=1.

Counterexample

technique · direct
1.1F1given

Impossibility for N>1. Suppose N>1 and a nonzero f satisfied ∣supp⁡f∣=∣supp⁡FNf∣=1. Then the left-hand side of the bound [F1] equals 1, so 1≥N, contradicting N>1. Hence no such f exists for N>1.

1.2F2given

The case N=1. At N=1 the group Z/1Z has the single class [0], and by [F2] the delta δ0 is the constant function 1 with F1δ0=δ0; both its support and the support of its transform equal the one-element set {[0]}.

2.1step 1.1step 1.2∎

Conclusion. The universal claim fails already at N=2, where [F1] forces a support product of at least 2; step 1.2 shows that the hypothesis N>1 is essential, since the excluded configuration does occur at N=1 and the bound of [F1] is exactly attained there.

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