F4 Bare Nodal Evaluation Supercharge Not Closable

For $f\in L^2(B;\mathbb{C})$, let $\tilde{f}$ be its zero extension to ${\mathbb{R}}^2$. Then $\tilde{f}\in L^2({\mathbb{R}}^2;\mathbb{C})$ with the same norm. For each $\epsilon >0$, [F2] with $n=2$, $p=2$ gives $g\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2;\mathbb{C})$ such that $\mathrm{\parallel }g-\tilde{f}{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}<\epsilon$. Its restriction to $\bar{B}$ lies in $\mathcal{D}$, and $\mathrm{\parallel }g{\mathrm{\mid }}_B-f{\mathrm{\parallel }}_{L^2(B)}\le \mathrm{\parallel }g-\tilde{f}{\mathrm{\parallel }}_{L^2({\mathbb{R}}^2)}<\epsilon$. Thus $\mathcal{D}$ is dense in $H$.

Define $\rho (t)=0$ for $t\le 0$ and $\rho (t)=e^{-1\mathrm{/}t}$ for $t>0$, and put $\chi (t)=\rho (t)\mathrm{/}(\rho (t)+\rho (1-t))$. The denominator is positive, $\chi$ is smooth, $\chi =0$ on $(-\mathrm{\infty },0]$, $\chi =1$ on $[1,\mathrm{\infty })$, and ${\chi }^{\mathrm{\prime }}$ is supported in $[0,1]$. For $0<\epsilon <1$, set $L_{\epsilon }=\log (1\mathrm{/}\epsilon )$ and ${\eta }_{\epsilon }(r)=\chi \text{\hspace{0.17em}⁣}\left(\frac{\log (r\mathrm{/}{\epsilon }^2)}{L_{\epsilon }}\right)(r>0),{\eta }_{\epsilon }(0)=0.$ It is smooth and radial because it is constant for $r\le {\epsilon }^2$ and for $r\ge \epsilon$. It equals zero near $0$ and one for $r\ge \epsilon$. The function ${\psi }_{\epsilon }=1-{\eta }_{\epsilon }$ has support in ${\bar{B}}_{\epsilon }$, so $\mathrm{\parallel }{\psi }_{\epsilon }{\mathrm{\parallel }}_2^2\le \pi {\epsilon }^2$. In polar coordinates, with $t=\log (r\mathrm{/}{\epsilon }^2)\mathrm{/}{L}_{\epsilon }$, ${\int }_B\mathrm{\mid }\mathrm{\nabla }{\eta }_{\epsilon }{\mathrm{\mid }}^2=\frac{2\pi }{L_{\epsilon }}{\int }_0^1\mathrm{\mid }{\chi }^{\mathrm{\prime }}(t){\mathrm{\mid }}^{2}dt⟶0.$ Therefore ${\psi }_{\epsilon }\to 0$ in $H^1(B)$, while ${\psi }_{\epsilon }(0)=1$.

The graph closure criterion in the Statement is equivalent to the usual closed-extension definition: a graph closure that is a graph is a closed extension, and any closed extension contains the graph closure. Its being a graph is equivalent to having no nonzero vector $(0,v)$; such a vector belongs to the graph closure exactly when a sequence of domain elements tends to zero while their outputs tend to $v$. Put ${\psi }_{\epsilon }$ on the plus branch and zero on the minus branch. The inputs tend to zero in $L^2\oplus L^2$, and graph-norm equivalence gives $D{\psi }_{\epsilon }\to 0$ in $K$. Their point-difference output is always $1$. Hence the graph of $A$ contains $(({\psi }_{\epsilon },0),(D{\psi }_{\epsilon },0,1))\to (0,(0,0,1))$. It also contains the zero graph pair $(0,(0,0,0))$. The closure of the graph therefore contains two distinct outputs over the same zero input, so it is not the graph of an operator. Thus $A$ is not closable.

For smooth $u$ on $\bar{B}$, $\mathrm{\parallel }(1-{\eta }_{\epsilon })u{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}^2\pi {\epsilon }^2,$ and the product rule gives $\mathrm{\parallel }\mathrm{\nabla }((1-{\eta }_{\epsilon })u){\mathrm{\parallel }}_2^2\le 2\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{\mathrm{\infty }}^2\pi {\epsilon }^2+\frac{4\pi \mathrm{\parallel }u{\mathrm{\parallel }}_{\mathrm{\infty }}^2}{L_{\epsilon }}{\int }_0^1\mathrm{\mid }{\chi }^{\mathrm{\prime }}(t){\mathrm{\mid }}^{2}dt\to 0.$ Thus ${\eta }_{\epsilon }u\to u$ in $H^1(B)$. For a smooth pair $(u_{+},u_{-})$, $({\eta }_{\epsilon }{u}_{+},{\eta }_{\epsilon }{u}_{-})$ is zero at both origins, so it satisfies the matching condition and converges to the pair in $H^1\oplus H^1$. By [F1], smooth pairs are dense in $H^1(B)\oplus H^1(B)$. The closure of the matched smooth pairs is consequently the whole direct-sum space. This proves the claimed loss of point matching under completion.