Example

Let $n\in \mathbf{N}$. We endow $\mathbf{R}^n$ with Lebesgue measure, and we identify $\mathbf{R}^n$ and its dual group as in corollary 3 of II, p. 236. Let N denote the euclidean norm on $\mathbf{R}^n$ and $\mathscr{F}$ the Fourier transform on $\mathscr{S}(\mathbf{R}^n)$ and on $\mathscr{S}'(\mathbf{R}^n)$ (n$^o12$ of IV, p. 217).

Let $\Delta_{\mathscr{S}}$ be the partial operator on $L^2(\mathbf{R}^n)$ whose domain is the space $\mathscr{S}(\mathbf{R}^n)$ and such that

for every $\varphi \in \mathscr{S}(\mathbf{R}^n)$. One then has $\mathscr{F}(\Delta_{\mathscr{S}}(\varphi )) = 4\pi^2N^2\mathscr{F}(\varphi )$ for every function $\varphi \in \mathscr{S}(\mathbf{R}^n)$ (remark 12 of IV, p. 220). Since the Fourier transform is an automorphism of $\mathscr{S}(\mathbf{R}^n)$ (prop. 18 of IV, p. 219), this means that the partial operator $u=\mathscr{F}\circ \Delta_{\mathscr{S}}\circ \mathscr{F}^{-1}$ is the restriction of the multiplication operator $m_{4\pi^2N^2}$ to the space $\mathscr{S}(\mathbf{R}^n)$. The partial operator $u$ is closable and symmetric; its closure is the positive self-adjoint partial operator $m_{4\pi^2N^2}$ (prop. 6 of IV, p. 232, applied to the space $\mathscr{D}(\mathbf{R}^n)\subset \mathscr{S}(\mathbf{R}^n)$ with the aid of prop. 4 of IV, p. 202). Consequently, the partial operator $\Delta_{\mathscr{S}}$ is essentially self-adjoint. Its closure is denoted by Δ; it is a positive self-adjoint operator, and it is the unique Laplacian on $\mathbf{R}^n($cf. example of No.$^o6$ of IV, p. 243 and cor. to prop. 26 of IV, p. 257).

According to prop. 21 of IV, p. 223, the Sobolev space $H^2(\mathbf{R}^n)$ is the set of $f\in L^2(\mathbf{R}^n)$ such that $(1 + N^2)\mathscr{F}(f)$ belongs to $L^2(\mathbf{R}^n)$, that is to say such that $\mathscr{F}(f)$ belongs to the domain of $m_{4\pi^2N^2}$. Since the Fourier transform is an isometric isomorphism of the space $L^2(\mathbf{R}^n)$ onto itself (th. 1 of II, p. 215), the domain of the Laplacian Δ is $H^2(\mathbf{R}^n)$. One has $\Delta =\mathscr{F}^{-1}\circ m_{4\pi^2N^2}\circ \mathscr{F}$. In particular, the spectrum of Δ is equal to $\mathbf{R}_+$.