The complementarity product needs extra regularity

The function $u$ is continuous on $B\setminus \{0\}$: there $\mathrm{\mid }x\mathrm{\mid }$ is continuous and positive, so $x\mapsto \log \log (1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$ is continuous on the punctured inner ball, and $u$ equals the constant $\log \log 4$ on the outer annulus, with matching limiting value at $\mathrm{\mid }x\mathrm{\mid }=1\mathrm{/}4$. Moreover $\log \log (1\mathrm{/}r)\to +\mathrm{\infty }$ as $r\downarrow 0$, because $\log (1\mathrm{/}r)\to +\mathrm{\infty }$ and $\log s\to +\mathrm{\infty }$ as $s\to +\mathrm{\infty }$; hence for every $M$ there is $\rho \in (0,1\mathrm{/}4)$ with $u(x)>M$ whenever $0<\mathrm{\mid }x\mathrm{\mid }<\rho$.

For $x\mathrm{\ne }0$, rationalising the norm difference gives $(\mathrm{\mid }x+t{e}_i\mathrm{\mid }-\mathrm{\mid }x\mathrm{\mid })\mathrm{/}t=(2x_i+t)\mathrm{/}(\mathrm{\mid }x+t{e}_i\mathrm{\mid }+\mathrm{\mid }x\mathrm{\mid })\to x_i\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ as $t\to 0$, hence ${\mathrm{\partial }}_i\mathrm{\mid }x\mathrm{\mid }=x_i\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$. On the punctured inner ball $0<\mathrm{\mid }x\mathrm{\mid }<1\mathrm{/}4$ the chain rule [F3] gives ${\mathrm{\partial }}_{i}u(x)=\frac{d}{dr}\log \log (1\mathrm{/}r){\mid }_{r=\mathrm{\mid }x\mathrm{\mid }}\cdot \frac{x_i}{\mathrm{\mid }x\mathrm{\mid }}=-\frac{x_i}{\mathrm{\mid }x{\mathrm{\mid }}^2\log (1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })}$, so $\mathrm{\mid }Du\mathrm{\mid }(x)=\frac{1}{\mathrm{\mid }x\mathrm{\mid }\log (1\mathrm{/}\mathrm{\mid }x\mathrm{\mid })}$; on $1\mathrm{/}4<\mathrm{\mid }x\mathrm{\mid }<1\mathrm{/}2$ the gradient vanishes. The joining circle is Lebesgue-null by [F1] applied to its indicator, since the radial integral is supported at $r=1\mathrm{/}4$; the origin is null by [F6]. Define the gradient to be zero on these exceptional sets. By polar coordinates [F1], the substitution $r=e^{-t}$ [F2] and the convergence facts there, ${\int }_B{u}^{2}dx=\sigma (S^1){\int }_0^{1\mathrm{/}4}(\log \log (1\mathrm{/}r){)}^{2}rdr+O(1)=\sigma (S^1){\int }_{\log 4}^{\mathrm{\infty }}(\log t{)}^2{e}^{-2t}dt+O(1)<+\mathrm{\infty },$ ${\int }_B\mathrm{\mid }Du{\mathrm{\mid }}^{2}dx=\sigma (S^1){\int }_0^{1\mathrm{/}4}\frac{dr}{r{\log }^2(1\mathrm{/}r)}=\sigma (S^1){\int }_{\log 4}^{\mathrm{\infty }}{t}^{-2}dt=\frac{\sigma (S^1)}{\log 4}<+\mathrm{\infty },$ the first integral converging because $(\log t{)}^2{e}^{-2t}\le t^{-2}$ for all large $t$ and the remaining compact piece is finite. Hence $u\in L^2(B)$ and its classical gradient lies in $L^2(B)$.

The function $u$ is Borel measurable, since it is continuous on the open set $B\setminus \{0\}$; along almost every coordinate line — all lines except the single line through $0$ in each of the two coordinate directions — the section is continuous and piecewise $C^1$ with bounded derivatives on each compact subinterval away from $0$ (the joining circle meets a coordinate line in at most two points), hence Lipschitz and absolutely continuous there, and the exceptional lines form a null set. By step 1.2 the classical coordinate derivatives exist almost everywhere, are measurable and lie in $L^2(B)$, so the ACL characterisation [F4] gives $u\in H^1(B)$ and identifies $Du$ with the classical gradient almost everywhere.

The class of $u$ has no continuous representative. Suppose $w:B\to \mathbb{R}$ were continuous with $w=u$ almost everywhere. On the compact ball ${\bar{B}}_{1\mathrm{/}8}(0)$ the function $w$ is bounded, say $\mathrm{\mid }w\mathrm{\mid }\le M$ [F5]. By step 1.1 choose $\rho \in (0,1\mathrm{/}8)$ with $u(x)>M+1$ for all $0<\mathrm{\mid }x\mathrm{\mid }<\rho$; the punctured ball $B_{\rho }(0)\setminus \{0\}$ has positive Lebesgue measure [F6], so it contains a point $x$ with $w(x)=u(x)>M+1$, contradicting $\mathrm{\mid }w\mathrm{\mid }\le M$.

The functions $w_1:=u$ and $w_2:=u+{\mathbf{1}}_{\{0\}}$ (that is, $w_2(0)=1$ and $w_2(x)=u(x)$ for $x\mathrm{\ne }0$) are both representatives of the same $L^2$ class, because they differ only on the Lebesgue-null singleton $\{0\}$ [F6, F4].

By [F7], $\mu ={\delta }_0$ is a nonnegative Radon measure on the locally compact space $B$ and $\int fd{\delta }_0=f(0)$; let $\phi \in C_c^{\mathrm{\infty }}(B)$ with $\phi (0)=1$ be the cutoff of [F8]. Evaluating the formula $\phi \mapsto {\int }_{B}u\phi d\mu$ with the representative $w_1$ gives $w_1(0)\phi (0)=0$, while evaluating it with $w_2$ gives $w_2(0)\phi (0)=1$; the two candidates differ by the nonzero distribution $\phi \mapsto \phi (0)$. Hence the formula is not independent of the representative of the $H^1$ class, and no distribution $u\cdot \mu$ is defined by it: the product is not a well-defined distribution of the class $u$ alone.

Consequently an $H^1$ class alone does not define its product with an arbitrary Radon measure. Sufficient hypotheses supplied by the obstacle corollaries are continuity of the representatives, as in The obstacle reaction is supported on the contact set under measure regularity, or the $L^2$ representation of the reaction, as in clause 2 of Obstacle complementarity in distribution form; neither follows from $u\in H^1$ (and the class here has no continuous representative by step 2.2). This contrasts with Multiplication of a distribution by a smooth function, where the multiplier is a genuine function and the product is representative-independent.