· 6,019 chars · 15 deps · depth 30 Reason: Proof of the Fourier coefficient properties lemma (Stage 4 foundations).
Linearity is the bilinearity of the inner product; injectivity is the completeness of the trigonometric system; Parseval, the expansion and the Riesz-Fischer statement are the orthonormal expansion theorem applied to the trigonometric orthonormal basis along the enumeration; and the realisation clause follows from injectivity together with the invertibility of the weights.
Claim 1. Let U,U′∈L2(Tn), t∈R and k∈Zn. By condition (b) of Real Inner Product Space §inner-product, U+U′(k)=⟨U+U′,Ek⟩L2=⟨U,Ek⟩L2+⟨U′,Ek⟩L2=U^(k)+U^′(k)=(U^+U^′)(k), and by condition (c), tU(k)=t⟨U,Ek⟩L2=(tU^)(k). As k was arbitrary, U+U′=U^+U^′ and tU=tU^, which are the two conditions of Linear Map.
Claim 5. Let κ be an enumeration and c∈Map(Zn,R) with ∑j=1∞c(κ(j))2 convergent. Applying Orthonormal Expansions in a Real Hilbert Space §riesz-fischer to the sequence cj=c(κ(j)) of real numbers, the series ∑j=1∞c(κ(j))Eκ(j) converges in L2(Tn); writing U for its sum, (∥U∥L2)2=∑j=1∞c(κ(j))2 and ⟨U,Eκ(j)⟩L2=c(κ(j)) for every j∈N. Given k∈Zn, by Bijection of Sets there is j∈N with κ(j)=k, so U^(k)=c(k); hence U^=c. If U′∈L2(Tn) also satisfies U^′=c, then U′=U by claim 2. This proves claim 5.
Uniqueness. If W,W′∈L2(Tn) both satisfy W^(k)=ρkmc(k)=W^′(k) for every k, then W^=W^′ and W=W′ by claim 2.
Subspace. By Elementary Identities in a Real Inner Product Space §zero, 0L2(k)=⟨0L2,Ek⟩L2=0=ρkm⋅0 for every k, so the zero family lies in Hm, with Λm of it equal to 0L2. Let c,d∈Hm, t∈R, W=Λmc and W′=Λmd. By claim 1 and distributivity in R, for every k,
so c+d and tc lie in Hm, with Λm(c+d)=W+W′=Λmc+Λmd and Λm(tc)=tW=tΛmc by uniqueness. Thus Hm satisfies the three conditions of Linear Subspace, so it is a linear subspace of Map(Zn,R) and, by The Real Vector Space of Real-Valued Functions on a Set §subspace, a real vector space under the pointwise operations; and Λm satisfies the two conditions of Linear Map for that vector space structure.
Bijection. Let W∈L2(Tn). Define c∈Map(Zn,R) by c(k)=(ρkm)−1W^(k); then ρkmc(k)=W^(k) for every k, so c∈Hm and Λmc=W. If d∈Hm also satisfies Λmd=W, then ρkmd(k)=W^(k)=ρkmc(k) for every k, and multiplying by (ρkm)−1 gives d(k)=c(k), so d=c. Hence every W has exactly one preimage under Λm, which is the condition of Bijection of Sets.