inequality for geometric Brownian motion

(4.12) V (x; s) = sup Ex;s0S 0c1 sup Ex;s0)1S 0V3(X ; S supE x;s0V 3(X ; S )0c1 .+ supE x;s0V 3(X ; S )0c1Hence, to complete the proof, it is enough to show the following two facts:(4.13) supE x;s0V 3(X ; S )0c1 V 3(x; s)(4.14) E x;s0S3 0c3 1 = V3(x; s) .Proof of (4.13)+(4.14): By Itˆo’s formula ( see Remark 4.2 below ) we have:(4.15) V3(X t ; S t ) = V3(X 0 ; S 0 ) +Zt+ 1 2Zt00@V3@x (X r; S r ) dX r +Zt@ 2 V3@x 2 (X r; S r ) d X; Xr0@V3@s (X r; S r ) dS rwhere we set (@ 2 V3=@x 2 )(g3(s); s) = 0 . By (2.2), (2.3), and the fact that dX; Xt = 2 X 2 t dt ,this can be written as follows:(4.16) V3(X t ; S t ) = V3(X 0 ; S 0 ) +ZtL X V3(X r ; S r ) dr +Zt+Zt00@V3@x (X r; S r ) X r dB r .0@V3@s (X r; S r ) dS rNext note that L X V3(x; s) = c for g3(s) < x < s and L X V3(x; s) = 0 for 0 < x g3(s) .Moreover, due to the normal reflection of X , the set of those r > 0 for which X r = S r isof Lebesgue measure zero. Finally, since the increment dS r equals 0 outside of the diagonalx = s , and V3(x; s) at the diagonal satisfies (2.13), we see that the second integral in (4.16) isidentically zero. These facts matched together in (4.16) show that:(4.17) V3(X ; S ) V3(x; s) + c + M for any (bounded) stopping time for X , where M = (M t ) t0is the local martingale given by:(4.18) M t =Zt0@V3@x (X r; S r ) X r dB rfort 0 . Moreover, this also shows that:(4.19) V3(X ; S ) = V3(x; s) + c + M .for any stopping time for X satisfying 3 .12