Proof of The Lebesgue Space of Square-Integrable Functions on the Torus is Not Finite-Dimensional

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 §lebesgue the space $L^{2}(\mathbb{T}^{n})$ is a real Hilbert space for the inner product $\langle\,\cdot\,,\cdot\,\rangle_{L^{2}(\mathbb{T}^{n})}$ of The Lebesgue Space of Square-Integrable Functions is a Real Hilbert Space §inner-product; write $|\,\cdot\,|$ for its norm, so that $0\le|X|$ and $|X|^{2}=\langle X,X\rangle_{L^{2}(\mathbb{T}^{n})}$ for every $X\in L^{2}(\mathbb{T}^{n})$. Let $\mathbb{N}$ be the set of natural numbers, let $\mathbb{Z}$ be the set of integers, let $\iota:\mathbb{N}\to\mathbb{R}$ be the canonical map of $\mathbb{R}$, written without a subscript as in The Integers as a Subset of the Real Numbers, and let $E_{k}$ for $k\in\mathbb{Z}^{n}$ be the classes of The Trigonometric System on the Torus is Orthonormal §classes.

For $j\in\mathbb{N}$ let $k(j)\in\mathbb{R}^{n}$ be the point whose first coordinate is $\iota(j)$ and whose $i$th coordinate is $0$ for every $i\in[n]$ with $i\ne1$. By The Integers as a Subset of the Real Numbers every member of the image of $\iota$ lies in $\mathbb{Z}$, and $0\in\mathbb{Z}$ by claim 2 of Arithmetic, Order and Discreteness of the Integers; so every coordinate of $k(j)$ lies in $\mathbb{Z}$ and therefore $k(j)\in\mathbb{Z}^{n}$ by Lattice-Periodic Functions and the Periodic Function Classes §lattice. Put $f_{j}=E_{k(j)}$, which lies in $L^{2}(\mathbb{T}^{n})$ by The Trigonometric System on the Torus is Orthonormal §classes; this defines a sequence $(f_{j})_{j\in\mathbb{N}}$ in $L^{2}(\mathbb{T}^{n})$.

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

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

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