A normalized square-integrable vector can have infinite kinetic form energy

In spherical coordinates [F4], $J={\int }_{{\mathbb{R}}^3}(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^{-2}d\xi =4\pi {\int }_0^{\mathrm{\infty }}\frac{r^2}{(1+r^2{)}^2}dr\mathrm{.}$ The integrand is bounded by $r^2$ near zero and by $r^{-2}$ for $r\ge 1$; thus $0<J<\mathrm{\infty }$ and $c=J^{-1\mathrm{/}2}$ is defined. By [F2], $\widehat{f}$ is the Fourier transform of an $L^2$ vector and its norm is $c^{2}J=1$.

Its formal kinetic form integral equals $4{\pi }^2{c}^2{\int }_{{\mathbb{R}}^3}\frac{\mathrm{\mid }\xi {\mathrm{\mid }}^2}{(1+\mathrm{\mid }\xi {\mathrm{\mid }}^2{)}^2}d\xi =16{\pi }^3{c}^2{\int }_0^{\mathrm{\infty }}\frac{r^4}{(1+r^2{)}^2}dr=+\mathrm{\infty },$ because the radial integrand tends to $1$ as $r\to \mathrm{\infty }$. By [F1], finite form energy is equivalent to this weighted integral being finite, so $f\mathrm{\notin }{H}^1({\mathbb{R}}^3;\mathbb{C})=Q(H_0)$.

The vector is normalized in $L^2$ by step 1.1 but outside the exact form domain by step 2.1. Thus it cannot be inserted into the min-max variational family described in [F3], which requires its subspaces to lie in $Q(H_0)$; the universal trial-vector claim is false. This counterexample concerns domain admissibility and assigns no finite value to the divergent integral.