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)