1. As f is continuous and O is compact, sup{e ?2f (y) | y ? O} ? and H(x, p) ? p 2 ? sup{e ?2f (y) | y ? O} for every x ? O. Then, H(x, p) ? ? uniformly in x ? O as p ? ?. Given this coercivity property of H, Proposition 4.1 in Chapter II of B-C [2] implies that u ? L. Using the facts that O is compact and H is convex in p, a comparison principle (B-C [2], Theorem 5.9 in Chapter II) implies that u is the only viscosity solution of (8). (A) Suppose u is a generalized solution of (8);of Theorem 6.12. Define H : O � n ? by H(x, p) = p 2 ? e ?2f (x) . The PDE in (8) is equivalent to H(., Du(.)) = 0. Suppose u is a viscosity solution of
2. Given u ? U, denote the mapping � ? O �(dz) u(z) by U . As u is continuous and ?(O) is given the weak * topology, U is continuous;B) Hold;By Theorem 5.5, ? ?1 (A) ? U and A = ? ? ? ?1 (A), and by Corollary 3.7, ? ?1 (F ) ? U a and F = ? ? ? ?1 (F )
3. Closed Media and ∪-Closed Families
4. We now show that A is lower hemicontinuous at x ? O. As O and ?(O) are metrizable, it is sufficient to show that, for every sequence (x n ) ? O converging to x and � ? A(x), there exists a sequence (� n ) ? ?(O) converging to � such that � n ? A(x n ) for every n ? N . So, consider a sequence (x n ) ? O converging to x ? O and let � ? A(x);Therefore, A(x) is closed in ?(O) for every x ? O. As ?(O) is compact metric, A is upper hemicontinuous and has compact values
5. If n ? N is such that u(x n ) ? U (�), then set t n = 1 and � n = t n �. Clearly, � n ? A(x n ). Now consider n ? N such that u(x n ) > U (�). Then, U (?) ? U (? xn ) = u(x n ) > U (�);Suppose U (�) = U (?). Then, u(x n ) = U (? xn ) ? U (�) for every n ? N . Set � n = � for every n ? N . Then, � n ? A(x n ) for every n ? N and (� n ) converges to �