There is a classical example that the set of exponential functions of the form $\{ e^{2\pi i \lambda t } : \lambda \in \mathbb{Z}\}$ for $t\in \mathbb{R}$ is an orthonormal basis (ONB) for $L^2[0,1]$. And due to the $1$-periodicity of the expoenential it is also an ONB for $L^2(\mathbb{R})$.
Is it correct to assume that $\{ e^{2\pi i \langle \lambda, t \rangle } : \lambda \in \mathbb{Z}^d \}$ is an ONB for edit ${L^2[0,1]^d}$ (not ${L^2(\mathbb{R}^d)}$) where $\langle \lambda, t \rangle$ is the euclidian inner product between $\lambda = (\lambda_1,\ldots,\lambda_d) \in \mathbb{Z}$ and $t = (t_1,\ldots,t_d) \in \mathbb{R}^d$ using vector notation.
I'm thinking that one can restrict the domain from $\mathbb{R}^d$ to $[0,1]^d$ due to the $1$-periodicity of the exponential, and I have shown orthogonality and basis property up to dimension 2, but the terms double in size when i increase the dimension.
Is there a nice general way this can be proven for $d$ dimensions for both orthogonality and the basis property, or is my assumption simply wrong to begin with?
Note that I'm not allowing for any flexibility in terms of shifting $\mathbb{Z}$ with a real number $\alpha$, i.e $\lambda \in \mathbb{Z}+\alpha$, as that increases the complexity dramatically in the higher dimensions. I only consider $\lambda \in \mathbb{Z}$
Let $\lambda, \mu\in \mathbb{Z}^d.$ Then $$\int\limits_{[0,1]^d} e^{i\pi\langle \lambda , x\rangle }e^{-i\pi\langle, \mu x\rangle }\,dx=\prod_{j=1}^d\int\limits_0^1 e^{i\pi x_j\lambda _j}e^{-i\pi x_j\mu _j}\,dx_j$$ If $\lambda=\mu$ we get $1.$ If $\lambda\neq \mu$ then $\lambda_j\neq \mu_j$ for some $j.$ Then the integral with index $j$ vanishes. Thus we got orthonormality.
Next let $\mathcal{A}$ denote the linear span of the functions $[0,1]^d\ni x\mapsto e^{i\pi \langle x,\lambda\rangle }.$ Then $\mathcal{A}$ forms a subalgebra in the space of $C_{per}([0,1]^d).$ The subalgebra is closed under complex conjugation as $\overline{e^{i\pi\langle x,\lambda\rangle }}=e^{i\pi\langle x, -\lambda\rangle}.$ It contains constants, as $e^{i\pi\langle x,0\rangle}=1,$ and separates points, as if $u\neq v\in [0,1)^d$ then $u_j\neq v_j$ for some $j.$ Then the function $x\mapsto e^{i\pi x_j}$ gives different values at $u$ and $v.$ Therefore by the Stone-Weierstrass theorem the algebra $\mathcal{A}$ is uniformly dense in $C_{per}([0,1]^d).$ Furthermore $C_{per}([0,1]^d)$ is dense in $L^2([0,1]^d)$ with respect to $L^2$ norm. Thus $\mathcal{A}$ is dense in $L^2([0,1]^d)$ with respect to $L^2$ norm. Hence $L^2([0,1]^d)$ does not contain a nonzero function orthogonal to all the functions $[0,1]^d\ni x\mapsto e^{i\pi \langle x,\lambda\rangle }.$
Remark Applying the Stone-Weierstrass theorem is a slight overkill. The proof can be performed by applying the Weierstrass theorem, or the Cesaro means, but it requires more explanations.