The specified Gaussian family has no finite vacuum-vector norm limit

Let $({\epsilon }_n)$ be an arbitrary sequence of positive widths in metres with ${\epsilon }_n\to 0$. For any Schwartz test $g$, the change of variables $y=(x-x_0)\mathrm{/}{\epsilon }_n$ gives ${\int }_{{\mathbb{R}}^4}{\rho }_{{\epsilon }_n,x_0}(x)g(x)d^{4}x={\int }_{{\mathbb{R}}^4}\rho (y)g(x_0+{\epsilon }_{n}y)d^{4}y\mathrm{.}$ For each $y$, the integrand tends to $\rho (y)g(x_0)$ and is bounded in absolute value by the integrable function $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\rho (y)$. The sequence-indexed dominated-convergence theorem [F2] gives convergence of these integrals to $g(x_0)$, using $\int \rho =1$. Since the sequence was arbitrary, the sequential criterion for a limit gives the full limit as $\epsilon \downarrow 0$; hence these are point-mass approximants.

For $n=0,1,2,\dots$, take the explicit sequence ${ϵ}_n=(1\mathrm{m})\mathrm{/}(n+1)$. Every width is positive and the sequence decreases to zero. By [F1], $\mathrm{\parallel }{\mathrm{\Phi }}_{\mu }({\rho }_{{ϵ}_n,x_0})\mathrm{\Omega }\mathrm{\parallel }\to \mathrm{\infty }$. A norm-convergent sequence in a Hilbert space has bounded norms, so this sequence cannot converge to a finite vacuum vector. Hence the claimed norm-limit point field does not exist in this representation. This refutes only the stated norm limit and does not exclude distributional or renormalized constructions.