unit square, we define


 f(s,t) = t + s(1 t).


Each one-dimensional path, for fixed t, is a straight line connecting

the points (0,t,t) and (1,t,1). Thus, for t < t', the slope of the

section f(-,t) is greater than that of f(-,t'), and so the second

difference is larger than the first, so A ,f < 0 for any z < z' in
 zz

the unit square. The following computations bear this out:

 1) For (0,0) 5 (s,t) S (s',t') D (1,1), f(s',t')-f(s,t) > 0. In

 fact,


 f(s',t')-f(s,t) = t'+s'(1-t')-(t+s(1-t))

 = t's '-s't'-t-s+st

 = (t'-t)+(s'-s)+st-s't'

 = (t'-t)+(s'-s)+st-s't+s't-s't'

 = (t'-t)+(s'-s)-t(s'-s)-s'(t'-t)

 = (1-s')(t'-t)+(1-t)(s'-s)

 2 0,


 since each of the four factors is nonnegative.

 2) Denoting z (s,t), z' = (s',t'),(O0,0) (s,t) 5 (s',t')

 S(1,1),


 S,.f = f(s',t')-f(s',t)-f(s,t')+f(s,t)
 zz
 t'+s'(1-t')-(t+s'(1-t))-(t'+s(1-t'))+t+s(1-t)

 = t'+s'(1-t')-t-s'(1-t)-t'-s(1-t')+t+s(1-t)