Point evaluation is unbounded at the two-dimensional H1 threshold

Choose $R>0$ with $\overline{B(c,R)}\subset {\widehat{S}}_2$, and choose $\chi \in C_c^{\mathrm{\infty }}(B(c,R))$ equal to one on $B(c,R\mathrm{/}2)$. For $0<ϵ<R\mathrm{/}4$, put $L_{ϵ}=\log (R\mathrm{/}ϵ)$ and $u_{ϵ}(x)=\chi (x)\frac{\log \text{\hspace{0.17em}⁣}\left(R\mathrm{/}\sqrt{\mathrm{\mid }x-c{\mathrm{\mid }}^2+{ϵ}^2}\right)}{L_{ϵ}}\mathrm{.}$ This is smooth and compactly supported in the simplex, and $u_{ϵ}(c)=1$.

In polar coordinates about $c$, $r\ge 0$ and the stated range $0<ϵ<R\mathrm{/}4$ gives $r^2+{ϵ}^2\ge {ϵ}^2>0$ and the radial derivative has no zero denominator. The undivided logarithm has uniformly bounded squared $L^2$ integral on $B(c,R)$: its absolute value is bounded by $\log (R\mathrm{/}r)+C$ for $0<r<R$ and small $ϵ$, and ${\int }_0^{R}r(\log (R\mathrm{/}r)+C{)}^{2}dr<\mathrm{\infty }$. Its radial derivative has magnitude $r\mathrm{/}(r^2+{ϵ}^2)$, so ${\int }_0^R\frac{r^3}{(r^2+{ϵ}^2{)}^2}dr\le C+\log (R\mathrm{/}ϵ)=C+L_{ϵ},$ by splitting at $r=ϵ$. After division by $L_{ϵ}$, these estimates show the $L^2$ norm and the gradient contribution from $\chi \mathrm{\nabla }\log (R\mathrm{/}\sqrt{r^2+{ϵ}^2})$ tend to zero. The term $\mathrm{\nabla }\chi$ is supported where $R\mathrm{/}2\le r<R$, on which the logarithm is uniformly bounded; its divided $L^2$ norm also tends to zero. Therefore $\mathrm{\parallel }{u}_{ϵ}{\mathrm{\parallel }}_{W^{1,2}({\widehat{S}}_2)}\to 0$.

If point evaluation at $c$ were bounded, some finite $C$ would satisfy $\mathrm{\mid }u(c)\mathrm{\mid }\le C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,2}}$ for every smooth $u$. By [F2] this is the boundedness condition for the candidate nodal functional. Applying it to $u_{ϵ}$ gives $1\le C\mathrm{\parallel }{u}_{ϵ}{\mathrm{\parallel }}_{W^{1,2}}\to 0$, a contradiction. Hence evaluation is unbounded at the critical exponent.