The plane wave as a generalized momentum eigenfunctional

[Well-definedness and continuity] For $\phi \in \mathcal{S}$, [L3] gives $\mathrm{\mid }{e}^{ik\cdot x}\mathrm{\mid }=1$, and [L5] plus ${\prod }_j(1+x_j^2)={\sum }_{\alpha \in \{0,2{\}}^d}{x}^{\alpha }$ gives ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }\phi (x)\mathrm{\mid }dx\le C_d{\sum }_{\alpha \in \{0,2{\}}^d}{p}_{\alpha ,0}(\phi )<\mathrm{\infty }\mathrm{.}$ Thus the defining integral is absolutely convergent and has the bound in (a). Since this bound is by a finite sum of defining Schwartz seminorms, $u_k$ is continuous in the Schwartz topology. Conjugate-linearity is immediate from the definition, and $u_k(\phi )=\int \overline{\phi (x)}{e}^{-2\pi ix\cdot (-k\mathrm{/}2\pi )}dx={\mathcal{F}}_2(\bar{\phi })(-k\mathrm{/}(2\pi ))$ by [L1]. This proves (a).

[Eigenrelation] Let $\phi \in \mathcal{S}$ and put $\psi :=\bar{\phi }$, a Schwartz class. Since $\overline{P_j\phi }=\overline{-i\mathrm{\hslash }{\mathrm{\partial }}_j\phi }=i\mathrm{\hslash }{\mathrm{\partial }}_j\bar{\phi }=i\mathrm{\hslash }{\mathrm{\partial }}_j\psi$, one has $u_k(P_j\phi )=\int e^{ik\cdot x}\overline{P_j\phi }dx=i\mathrm{\hslash }\int e^{ik\cdot x}{\mathrm{\partial }}_j\psi (x)dx=i\mathrm{\hslash }{\mathcal{F}}_2({\mathrm{\partial }}_j\psi )(-\frac{k}{2\pi }),$ where the last step evaluates the $L^1$ transform at $-k\mathrm{/}(2\pi )$. By [L2], ${\mathcal{F}}_2({\mathrm{\partial }}_j\psi )(\xi )=2\pi i{\xi }_j{\mathcal{F}}_2\psi (\xi )$ for every $\xi$, so at $\xi =-k\mathrm{/}(2\pi )$, $u_k(P_j\phi )=i\mathrm{\hslash }\cdot 2\pi i(-\frac{k_j}{2\pi }){\mathcal{F}}_2\psi (-\frac{k}{2\pi })=\mathrm{\hslash }{k}_j{u}_k(\phi ),$ using step 1.1. Applying the same identity to $P_j\phi$ (again a Schwartz class) gives $u_k(P_j^2\phi )=\mathrm{\hslash }{k}_j{u}_k(P_j\phi )=(\mathrm{\hslash }{k}_j{)}^2{u}_k(\phi )$. This proves (b).

[No $L^2$ representative] Suppose $F\in L^2$ satisfies $\int F\bar{\phi }=u_k(\phi )$ for all $\phi \in \mathcal{S}$, and put $G:=e^{-ik\cdot x}F\in L^2$, a class of the same norm because the multiplier has modulus one [L3]. Multiplication by $e^{ik\cdot x}$ preserves the Schwartz class, since its derivatives are constant multiples of itself and Leibniz's rule expresses every derivative of the product as a finite sum of a polynomial times a derivative of $\phi$. Replacing $\phi$ by $e^{ik\cdot x}\phi$ in the assumed identity gives $\int G\bar{\phi }=\int F\overline{e^{ik\cdot x}\phi }=u_k\left(e^{ik\cdot x}\phi \right)=\int \bar{\phi },$ so $\int (G-1)\bar{\phi }=0$ for every $\phi \in \mathcal{S}$. For each $n\ge 1$ let ${\mathrm{\Omega }}_n:=(-n,n{)}^d$; the restriction $(G-1){\mathrm{\mid }}_{{\mathrm{\Omega }}_n}$ belongs to $L^2({\mathrm{\Omega }}_n)$ because $G\in L^2({\mathbb{R}}^d)$ and $1\in L^2({\mathrm{\Omega }}_n)$. Its pairing with every $v\in C_c^{\mathrm{\infty }}({\mathrm{\Omega }}_n)$ is zero, so [L4] gives $G=1$ almost everywhere on ${\mathrm{\Omega }}_n$. Since ${\bigcup }_{n\ge 1}{\mathrm{\Omega }}_n={\mathbb{R}}^d$, it follows that $G=1$ almost everywhere on ${\mathbb{R}}^d$. But then ${\int }_{{\mathbb{R}}^d}\mathrm{\mid }G{\mathrm{\mid }}^2={\int }_{{\mathbb{R}}^d}1=+\mathrm{\infty }$ by monotone convergence on expanding boxes [L4], contradicting $G\in L^2$. Hence no such $F$ exists. For $z\mathrm{\ne }0$, $\mathrm{\mid }z{e}^{ik\cdot x}\mathrm{\mid }=\mathrm{\mid }z\mathrm{\mid }$ on a set of infinite Lebesgue measure, so no nonzero scalar multiple of the plane wave belongs to $L^2$ either. This proves (c).

[Conclusion] Statement (d) records that only Schwartz classes and the functional $u_k$ were used; no normalizable class and no probability law is attached to $u_k$ by this example.