TheoremBase

Indexing the trigonometric system along a single coordinate axis produces an orthonormal sequence, to which the criterion for a Hilbert space to fail to be finite-dimensional applies.

Proof

Each result cited is universally quantified over the data in its own statement, and is applied to the data named here.

By The Flat Torus: Standing Notation (iv) the space L2(Tn)L^{2}(\mathbb{T}^{n}) is a real Hilbert space for the inner product ⟨ ⋅ ,⋅ ⟩L2(Tn)\langle\,\cdot\,,\cdot\,\rangle_{L^{2}(\mathbb{T}^{n})} of L²(μ) Is a Real Hilbert Space (i); write ∣ ⋅ ∣|\,\cdot\,| for its norm, so that 0≤∣X∣0\le|X| and ∣X∣2=⟨X,X⟩L2(Tn)|X|^{2}=\langle X,X\rangle_{L^{2}(\mathbb{T}^{n})} for every X∈L2(Tn)X\in L^{2}(\mathbb{T}^{n}). Let N\mathbb{N} be the set of natural numbers, let Z\mathbb{Z} be the set of integers, let ι:N→R\iota:\mathbb{N}\to\mathbb{R} be the canonical map of R\mathbb{R}, written without a subscript as in The Integers as a Subset of the Real Numbers, and let EkE_{k} for k∈Znk\in\mathbb{Z}^{n} be the classes of The Trigonometric System on the Torus is Orthonormal (i).

For j∈Nj\in\mathbb{N} let k(j)∈Rnk(j)\in\mathbb{R}^{n} be the point whose first coordinate is ι(j)\iota(j) and whose iith coordinate is 00 for every i∈[n]i\in[n] with i≠1i\ne1. By The Integers as a Subset of the Real Numbers every member of the image of ι\iota lies in Z\mathbb{Z}, and 0∈Z0\in\mathbb{Z} by claim 2 of Arithmetic, Order, Discreteness and Intervals of the Integers; so every coordinate of k(j)k(j) lies in Z\mathbb{Z} and therefore k(j)∈Znk(j)\in\mathbb{Z}^{n} by Lattice-Periodic Functions and the Periodic Function Classes (i). Put fj=Ek(j)f_{j}=E_{k(j)}, which lies in L2(Tn)L^{2}(\mathbb{T}^{n}) by The Trigonometric System on the Torus is Orthonormal (i); this defines a sequence (fj)j∈N(f_{j})_{j\in\mathbb{N}} in L2(Tn)L^{2}(\mathbb{T}^{n}).

The map j↦k(j)j\mapsto k(j) is injective. Indeed, points of Rn\mathbb{R}^{n} are nn-tuples of real numbers, so k(j)=k(l)k(j)=k(l) forces ι(j)=ι(l)\iota(j)=\iota(l), and ι\iota is injective by claim 7 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, whence j=lj=l.

Let i,j∈Ni,j\in\mathbb{N} with i≠ji\ne j. Then k(i)≠k(j)k(i)\ne k(j) by the injectivity just proved, so ⟨fi,fj⟩L2(Tn)=0\langle f_{i},f_{j}\rangle_{L^{2}(\mathbb{T}^{n})}=0 by The Trigonometric System on the Torus is Orthonormal (iii). Let i∈Ni\in\mathbb{N}. The same clause, taken with k=m=k(i)k=m=k(i), gives ∣fi∣2=⟨fi,fi⟩L2(Tn)=1|f_{i}|^{2}=\langle f_{i},f_{i}\rangle_{L^{2}(\mathbb{T}^{n})}=1. Since 0≤∣fi∣0\le|f_{i}|, since 0≤10\le1 by claim 1 of Elementary Arithmetic in an Ordered Field, and since 1⋅1=11\cdot1=1, the uniqueness in Existence and Uniqueness of the Nonnegative Square Root of a Nonnegative Real Number, applied to the nonnegative real number 11, gives ∣fi∣=1|f_{i}|=1. Hence (fj)j∈N(f_{j})_{j\in\mathbb{N}} is an orthonormal sequence in L2(Tn)L^{2}(\mathbb{T}^{n}).

Therefore A Real Hilbert Space Containing an Orthonormal Sequence Is Not Finite-Dimensional (i), applied to the real Hilbert space L2(Tn)L^{2}(\mathbb{T}^{n}) and to this sequence, gives that L2(Tn)L^{2}(\mathbb{T}^{n}) is not finite-dimensional.

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