Consider 25.1.6 next. To do so, let a = A^{′} so that
∫ t
A (t) = a (r)dr,a increasing.
0
This is possible by Rademacher’s theorem, Corollary 24.4.3 and the fact that
since A is convex, it is locally Lipshitz found in Lemma 25.1.2 above. That a is
increasing follows from convexity of A. Here is why. For a.e. s,t ≥ 0, and letting
λ ∈
[0,1]
,
A (s+ λ (t− s))− A (s) (1− λ)A (s) + λA(t)− A (s)
----------λ--------- ≤ -----------λ-------------
= A(t)− A (s)
Then passing to a limit as λ → 0+,
a(s)(t− s) ≤ A(t)− A (s).
Similarly
a(t)(s− t) ≤ A(s)− A (t)
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 an
integral of a positive increasing function, a, and verify directly that such an A is convex
and satisfies the properties of an N function. There is no loss of generality in doing so.)
Thus geometrically, A
(t)
equals the area under the curve defined by a and above the x
axis from x = 0 to x = t. In the definition of
^
A
(s)
let t_{s} be the point where the
maximum is achieved. Then
^A (s) = sts − A (ts)
and so at this point,
^A
(s)
+ A
(t)
s
= st_{s}. This means that
A^
(s)
is the area to the left of
the graph of a which is to the right of the y axis for y between 0 and a
(t)
s
and
that in fact a
(t)
s
= s. The following picture illustrates the reasoning which
follows.
PICT
Therefore,
^A (s) A(ts) 1 ∫ ts
--s-- = ts − --s--= ts − s a (r)dr
∫ ts 0 ( ∫ ts )
= ts − -1--- a(r)dr = -1--- tss− a (r)dr
a(ts) 0 a(ts) 0
and so
( ) ∫
^A(s)- ∫ ^A(s)∕s ∫ a1(ts) ts0 (s−a(r))dr
A s = 0 a(r)dr = 0 a(τ)dτ
∫ t
≤ ss − a(r)dr = sts − A (ts) = A^(s) .
0
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)
=
tpp-
for t ≥ 0 and p > 1. For this example,
^
A
(s)
=
sp′
p′
where
1
p
+
-1
p′
= 1.
Definition 25.1.4Let A be an N function and let
(Ω, S, μ)
be a measure space.Define
{ ∫ }
KA (Ω ) ≡ u measurable such that A (|u|)dμ < ∞ . (25.1.8)
Ω
(25.1.8)
This is called the Orlitz class. Also define
LA (Ω ) ≡ {λu : u ∈ KA (Ω) and λ ∈ F} (25.1.9)
(25.1.9)
where F is the field of scalars, assumed to be either ℝ or ℂ.
The pair
(A,Ω )
is called Δ regularif either of the following conditions hold.
A (rx) ≤ KrA (x) for all x ∈ [0,∞ ) (25.1.10)
(25.1.10)
or μ
(Ω)
< ∞ and for all r > 0, there exists M_{r}and K_{r}> 0 such that
A (rx) ≤ KrA (x) for all x ≥ Mr. (25.1.11)
(25.1.11)
Note there are N functions which are not Δ regular. For example, consider
A(x) ≡ ex2 − 1.
It can’t be Δ regular because
r2x2
lim e-x2-− 1-= ∞.
r→ ∞ e − 1
However, functions like x^{p}∕p for p > 1 are Δ regular.