Combining(34)and(35)we find that )-o(x0)≤th, 1<0 Send t」0 to obtain h l Vo(xo)h-1≤ Since h E Rn is arbitrary we obtain (32)as required. This proves that V is a viscosity subsolution a e. supersolution proerty, Let E C(Q)and suppose that V-p attains a local minimum at o E Q; so there exists r>0 such that the ball B(o, r)CQ and V(x)-o(x)≥V(xo)-o(x0)x∈B(xo,r) 39 We want to show that Vo(xo)|-1≥0. Suppose that(40) is false, so that Vo(xo)|-1≤-a<0 (41) for some 1>a>0. By making r>0 smaller if necessary, we may assume V(x)-1≤-a/2<0yx∈B(xo,r) By the fundamental theorem of calculus. we have o(x)=0(x0)+/Va(x+(1-7)xo)(x-xo)dhy Now from the dynamic programming relation(22), pick z*E B(ao, r),2*+ ro, such that V(xo0)=|xo-2|+V(2) Using this and(39)we have ((2*)-0(xo)≥|z*-xo However, from( 42) and(43)we must have (0(2)-0(x0)≤(1-a/2)z*-ol Inequalities(45)and(46)are in contradiction, so in fact(40)holds. This proves that V is a supersolution It can be seen here that the dynamic programming principle provided the key inequal ities to derive the sub- and supersolution relationsCombining (34) and (35) we find that −(φ(x0 + th) − φ(x0)) ≤ t|h|, (36) and so −( φ(x0 + th) − φ(x0) t|h| ) − 1 ≤ 0 (37) Send t ↓ 0 to obtain − 1 |h| ∇φ(x0)h − 1 ≤ 0. (38) Since h ∈ Rn is arbitrary we obtain (32) as required. This proves that V is a viscosity subsolution. Supersolution property. Let φ ∈ C 1 (Ω) and suppose that V −φ attains a local minimum at x0 ∈ Ω; so there exists r > 0 such that the ball B(x0, r) ⊂ Ω and V (x) − φ(x) ≥ V (x0) − φ(x0) ∀ x ∈ B(x0, r). (39) We want to show that |∇φ(x0)| − 1 ≥ 0. (40) Suppose that (40) is false, so that |∇φ(x0)| − 1 ≤ −α < 0 (41) for some 1 > α > 0. By making r > 0 smaller if necessary, we may assume |∇φ(x)| − 1 ≤ −α/2 < 0 ∀ x ∈ B(x0, r). (42) By the fundamental theorem of calculus, we have φ(x) = φ(x0) + Z 1 0 ∇φ(γx + (1 − γ)x0)(x − x0)dγ (43) Now from the dynamic programming relation (22), pick z ∗ ∈ B(x0, r), z ∗ 6= x0, such that V (x0) = |x0 − z ∗ | + V (z ∗ ). (44) Using this and (39) we have −(φ(z ∗ ) − φ(x0)) ≥ |z ∗ − x0|. (45) However, from (42) and (43) we must have −(φ(z ∗ ) − φ(x0)) ≤ (1 − α/2)|z ∗ − x0|. (46) Inequalities (45) and (46) are in contradiction, so in fact (40) holds. This proves that V is a supersolution. It can be seen here that the dynamic programming principle provided the key inequal￾ities to derive the sub- and supersolution relations. 13