Proposition 17.5.42099099BAE21

Proof. Let $B=\sup _ X|f|$. The potential $u-|t|B$ is a competitor for the obstacle $u+tf$, and every competitor lies below $u+|t|B$, so $u-|t|B\leq u_ t\leq u+|t|B$. Hence $u_ t\sim u$, and $E(u_ t)\geq E(u)-|t|Bm>-\infty $, so $u_ t\in \mathcal E^1(X,\theta ,\phi )$ by theorem 17.3.14. The obstacle $u+tf$ is upper semicontinuous, so $u_ t\leq u+tf$ everywhere, by proposition 16.11.4(1),(2). In particular $u_0\leq u$; since $u$ is itself a competitor for the obstacle $u$, also $u\leq u_0$, so $u_0=u$.

Next we show that $\mu _{u_{t+s}}\to \mu _{u_ t}$ weakly as $s\to 0$. The obstacles $u+tf$ and $u+(t+s)f$ differ by at most $|s|B$, so proposition 16.11.4(4) gives $u_ t-|s|B\leq u_{t+s}\leq u_ t+|s|B$. In particular all $u_ t$ have the pole set of $u$, and $|u_{t+s}-u_ t|\leq |s|B$ outside it. Since $|\arctan a-\arctan b|\leq |a-b|$ for real $a,b$, and adding the same smooth local potential of $\theta $ to $u_{t+s}$ and $u_ t$ does not change their difference, the distance $d_*$ of definition 13.7.2 between the local potentials of $u_{t+s}$ and $u_ t$ is at most $|s|B$ everywhere. Hence, for every $\delta >0$, the sets where it exceeds $\delta $ are empty once $|s|B<\delta $, and $u_{t+s}\to u_ t$ in extended capacity as $s\to 0$. These potentials have full mass relative to $\phi $, so corollary 13.9.4, applied along every sequence $s_ k\to 0$, gives the asserted weak convergence; in particular $\int _ Xf\, \mu _{u_{t+s}}\to \int _ Xf\, \mu _{u_ t}$.

The obstacle $u+tf$ is bounded above, and it is quasi-continuous, because $u$ is (section 16.11) and $f$ is continuous. Hence theorem 16.13.4 shows that $\mu _{u_ t}$ is carried by the contact set $\{ u_ t=u+tf\} $. On the contact set of $u_{t+s}$ we have $u_{t+s}=u+(t+s)f$ and $u_ t\leq u+tf$, so $u_{t+s}-u_ t\geq sf$ there. On the contact set of $u_ t$ we have $u_ t=u+tf$ and $u_{t+s}\leq u+(t+s)f$, so $u_{t+s}-u_ t\leq sf$ there. The differences are bounded and the pole sets carry no mass, so the tangent bounds of lemma 17.4.10 give, for every real $s$,

Dividing by $s>0$, respectively $s<0$, and letting $s\to 0$, we obtain the right and left derivatives from the weak convergence just proved. Both equal $\int _ Xf\, \mu _{u_ t}$, which depends continuously on $t$ by the same weak convergence. $\square$