Kenneth Kuttler

27.1. BASIC THEORY 977

and so(a(t)−a(s))(t− s)≥ 0.

(If you like, you can simply assume from the beginning that A(t) is given this way as anintegral of a positive increasing function, a, and verify directly that such an A is convex andsatisfies the properties of an N function. There is no loss of generality in doing so.) Thusgeometrically, A(t) equals the area under the curve defined by a and above the x axis fromx = 0 to x = t. In the definition of Ã(s) let ts be the point where the maximum is achieved.Then

Ã(s) = sts−A(ts)

and so at this point, Ã(s)+A(ts) = sts. This means that Ã(s) is the area to the left of thegraph of a which is to the right of the y axis for y between 0 and a(ts) and that in facta(ts) = s. The following picture illustrates the reasoning which follows.

Ã(s)

A(t0)

graph of a

Therefore,

Ã(s)s

= ts−A(ts)

s= ts−

∫ ts

0a(r)dr

= ts−1

a(ts)

∫ ts

0a(r)dr =

1a(ts)

(tss−

∫ ts

0a(r)dr

)and so

(Ã(s)

∫ Ã(s)/s

0a(r)dr =

∫ 1a(ts)

∫ ts0 (s−a(r))dr

0a(τ)dτ

≤∫ ts

0s−a(r)dr = sts−A(ts) = Ã(s) .

The inequality results from replacing a(τ) with a(ts) in the last integral on the top line.An example of an N function is A(t) = t p

p for t ≥ 0 and p > 1. For this example,

Ã(s) = sp′

p′ where 1p +

1p′ = 1.

Definition 27.1.4 Let A be an N function and let (Ω,S , µ) be a measure space. Define

KA (Ω)≡{

u measurable such that∫

Ω

A(|u|)dµ < ∞

}. (27.1.8)

27.1. BASIC THEORY 977and so(a(t) —a(s))(t—s) 20.(If you like, you can simply assume from the beginning that A(t) is given this way as anintegral of a positive increasing function, a, and verify directly that such an A is convex andsatisfies the properties of an N function. There is no loss of generality in doing so.) Thusgeometrically, A (t) equals the area under the curve defined by a and above the x axis fromx=0 tox=t. Inthe definition of A (s) let t; be the point where the maximum is achieved.Then -A(s) = sts —A (ts)and so at this point, A(s) +A (t,) = sts. This means that A(s) is the area to the left of thegraph of a which is to the right of the y axis for y between 0 and a(t,) and that in facta(t,) = s. The following picture illustrates the reasoning which follows.A(s)A(to)Therefore,and soA “A(s)/s aby fg (s—a(r))dra( ") = / a(r)dr= | “0” a(t)dt00ts ~< | s—a(r)dr=st;—A(ts) =A(s).0The inequality results from replacing a(t) with a(t;) in the last integral on the top line.An example of an N function is A(t) = e for t > 0 and p > 1. For this example,A(s) — LedeA(s) = yr Where 5 +77 = 1.Definition 27.1.4 Let A be an N function and let (Q,.%, &) be a measure space. DefineK,(Q)= {" measurable such that | A(|u|)du < ae (27.1.8)Q