27.1. BASIC THEORY 975

so ˜̃A = A. Alsost ≤ A(t)+ Ã(s) for all s, t ≥ 0 (27.1.5)

and for all s > 0,

A

(Ã(s)

s

)≤ Ã(s) . (27.1.6)

Proof: First consider the claim à is convex. Let λ ∈ [0,1] .

Ã(λ s1 +(1−λ )s2)≡max{[s1λ +(1−λ )s2] t−A(t) : t ≥ 0}

≤ λ max{s1t−A(t) : t ≥ 0}+(1−λ )max{s2t−A(t) : t ≥ 0}

= λ Ã(s1)+(1−λ ) Ã(s2) .

It is obvious à is stictly increasing because st is strictly increasing in s. Next consider27.1.2.

For s > 0 let ts denote the number where the maximum is achieved. That is,

Ã(s)≡ sts−A(ts) .

ThusÃ(s)

s= ts−

A(ts)s≥ 0. (27.1.7)

It follows from this thatlim

s→0+ts = 0

since otherwise, a contradiction results to 27.1.7, the expression becoming negative forsmall enough s. Thus

ts ≥Ã(s)

s≥ 0

and this shows

lims→0+

Ã(s)s

= 0.

which shows 27.1.2.To verify the second part of 27.1.2, let ts be as just described. Then for any t > 0

Ã(s)s

= ts−A(ts)

s≥ t− A(t)

s

It follows

lim infs→∞

Ã(s)s≥ t.

Since t is arbitrary, this proves the second part of 27.1.2.The inequality 27.1.5 follows from the definition of Ã(s) .