(a)

Suppose {gm,n}(m,n)∈Z2 constitutes a windowed Fourier frame of L2(R) with ∆x∆ξ = 1 (which includes the case of an orthonormal basis). Then, either σx(g) = ∞ or σξ(g) =∞. Proof. We only prove here the orthonormal basis case due to Battle [1]. For the general non-orthogonal case, which includes the Gabor frame, see [2]. Our strategy here is the following: Assume σx(g) < ∞ and σξ(g) <∞, then lead to contradiction. Let us consider the inner product, 〈xg, g′〉, which also appeared in the proof of the inequality of the Heisenberg uncertainty principle. Note that xg is in L2(R) so as g′, because ‖xg‖2 = x2|g(x)|2 dx = σ2x(g) <∞, since the mean of g is 0 and ‖g‖2 = 1. Recognizing that Fg ′ = 2piiξ ĝ(ξ) and σ2ξ (g) < ∞, we can show g ′ ∈ L2(R). Now, we have the following:〈 xg, g′ m∈Z n∈Z 〈xg, gm,n〉 gm,n,