RN such that for all x,y E RN.

16a,b) because a lower bound for maxaG(p) up emerges less easily than the upper bound that we shall find. 18) we reduce SE, if necessary, in order that 6, < dist(c, 892). 15), we observe that, for every 3s > 0, If lu(g) - up()1 = < kP(b - if p<8E J = u(y)} dy E. 18)]. 16a). 16b). 16b) are zero. 19) shows that vp(x) < E for every E > 0 and for all x E aG(p), if 3p < 5g; again 50 2 Some Maximum Principles for Elliptic Equations Fig. 3. 16b) are zero. 16b) once more. 12 (the boundary-point lemma for balls).

2, (c), now shows that (Lov)(xo) = Y. aij(xo) (ajatv)(xo) + 0 trace (a(xo) H(xo)) < 0, which contradicts the result of step (i). (iii) Accordingly, for every e > 0 and all x E S2, u(x) < v(x) < maxan v < maxan u + eK1, where Kt := maxxEa2 e Kxj It follows that u(x) < maxan u for all x E S2. 2 The weak maximum principle 43 maxan u < 0. 4) is impossible. 6 (the weak maximum principle for L). Suppose that (a) f2 is bounded, u e C(S2); (b) u is a C2-subsolution relative to L and SZ. Then maxi u < maxaf, u+.