Delta-square is outside the product contract

Choose a compactly supported smooth cutoff $\chi$ equal to one near $0$. By [F1], $\chi {\delta }_0={\delta }_0$ and $\widehat{\chi {\delta }_0}(\xi )=⟨{\delta }_0,\chi (x)e^{-2\pi ix\cdot \xi }⟩=1$ for every $\xi$. This constant function is not rapidly decreasing in any open cone about a nonzero covector, so [F2] gives $(0,\xi )\in \mathrm{WF}({\delta }_0)$ for every $\xi \mathrm{\ne }0$.

Since $\rho \ge 0$ and $\int \rho =1$ by [F4], it is positive at some point. Continuity gives a ball on which ${\rho }^2$ is bounded below by a positive constant, so $\int {\rho }^2>0$. Compact support and smoothness make $\rho$ bounded, and $0\le {\rho }^2\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_{\mathrm{\infty }}\rho$ with $\int \rho =1$ gives ${\rho }^2\in L^1$.

For every $\xi \mathrm{\ne }0$, both $(0,\xi )$ and $(0,-\xi )$ belong to $\mathrm{WF}({\delta }_0)$ by step 1.1. The hypothesis of the product criterion [F3] therefore fails for $(u,v)=({\delta }_0,{\delta }_0)$; the theorem does not define ${\delta }_0^2$.

Let $f\in \mathcal{D}({\mathbb{R}}^n)$ equal one on a neighborhood of $0$. Since $\mathrm{supp}\rho \subset B_1(0)$, for all sufficiently small $\epsilon$, $f(\epsilon y)=1$ on $\mathrm{supp}\rho$. Substitution $x=\epsilon y$ in the Lebesgue integral of the smooth compactly supported square gives $⟨{\rho }_{\epsilon }^2,f⟩=\int {\epsilon }^{-2n}\rho (x\mathrm{/}\epsilon {)}^{2}f(x)dx={\epsilon }^{-n}\int \rho (y{)}^{2}dy$. By step 1.2 this tends to $+\mathrm{\infty }$, whereas convergence in ${\mathcal{D}}^{\mathrm{\prime }}$ would make the pairing with the fixed test $f$ converge to a finite scalar. Thus these unrenormalized squares have no distributional limit.