Theorem 8.6.668BE4F589DCE

Proof. Choose smooth $h_ k\downarrow u$ by 8.6.4 and put $\varepsilon _ j=8^{-j}$. We construct smooth strictly admissible $u_ j$ with $u_ j\geq u+\varepsilon _ j$. For $j=1$, take $k_1=1$. At a subsequent step choose $k_ j\geq \max \{ j,k_{j-1}+1\} $ so large that

Such an index exists: the open sets $\{ h_ k<u_{j-1}-\varepsilon _{j-1}/2\} $ increase with $k$ and cover $X$, since $h_ k\downarrow u\leq u_{j-1}-\varepsilon _{j-1}$. Compactness gives a single index for all of $X$.

For this fixed obstacle $h_{k_ j}$ choose $b_ j$ as in 8.6.5, and then choose $N_ j$ so large that its error $E_{N_ j}+2M_{N_ j}/N_ j$ is at most $\varepsilon _ j$. The resulting smooth solution $w_ j$ satisfies

Put $u_ j=w_ j+2\varepsilon _ j$. Adding a constant preserves strict admissibility, and $u_ j\geq u+\varepsilon _ j$. For $j\geq 2$,

Finally, $u\leq u_ j\leq h_{k_ j}+2\varepsilon _ j$ and $k_ j\geq j$ give pointwise convergence to $u$. The bounds $\inf _ X u\leq u_ j\leq \sup _ X u_1$ give a common uniform bound. $\square$