976 CHAPTER 27. ORLITZ SPACES

Next consider 27.1.4. It must be shown that

A(t0) = max{

t0s− Ã(s) : s≥ 0}.

To do so, first note

Ã(s) = max{st−A(t) : t ≥ 0} ≥ st0−A(t0) .

Hencemax

{t0s− Ã(s) : s≥ 0

}≤max{t0s− [st0−A(t0)]}= A(t0) .

Now let

s0 ≡ inf{

A(t)−A(t0)t− t0

: t > t0

}.

By convexity, the above difference quotients are nondecreasing in t and so

s0 (t− t0)≤ A(t)−A(t0)

for all t ̸= t0. Hence for all t,

s0t−A(t)≤ s0t0−A(t0)

and soÃ(s0) = s0t0−A(t0)

implyingA(t0) = s0t0− Ã(s0)≤max

{st0− Ã(s) : s≥ 0

}≤ A(t0) .

Therefore, 27.1.4 holds.Consider 27.1.6 next. To do so, let a = A′ so that

A(t) =∫ t

0a(r)dr, a increasing.

This is possible by Rademacher’s theorem, Corollary 26.4.3 and the fact that since A isconvex, it is locally Lipshitz found in Lemma 27.1.2 above. That a is increasing followsfrom convexity of A. Here is why. For a.e. s, t ≥ 0, and letting λ ∈ [0,1] ,

A(s+λ (t− s))−A(s)λ

≤ (1−λ )A(s)+λA(t)−A(s)λ

= A(t)−A(s)

Then passing to a limit as λ → 0+,

a(s)(t− s)≤ A(t)−A(s) .

Similarlya(t)(s− t)≤ A(s)−A(t)