m-1
Z jm (R ) E If(s+1 ,0) f(s ,0)j + Z AR (f)
l,j i=f0 IO i,j
j>k
SVar[s,s']f(0,-) + Var[(s,O),(s',t')](
m f[ (R)
(same as before). Taking supremum we get imf(R)| ml f(R). The
proof of the other inequality is the same as that for case (2).
4) Finally, assume (s,t) < (0,0). We proceed similar to the
above, but this time we add zero to both o and T, and use these
partitions in our figuring of variations (Figure 3-10). Denote
S= 0, t1 = 0. For i < k 2 or j < 1 2, we have Imf(Ri,)I = 0.
(s ',t
I
L__
(s,t)
Figure 3-10 The grid oxT
For I = k 1, j = 1 1, we have |mf(R ij)I = If(0,0) for
'3