25.1 Basic Theory
All the theorems about the Lp spaces have generalizations to something called an Orlitz
space. [?], [?] Instead of the convex function, A
one considers a more general
convex increasing function called an N
Definition 25.1.1 A : [0,∞) → [0,∞) is an N function if the following two conditions
For A an N function,
As an example see the following picture of a typical N function.
Note that from the assumption, 25.1.2 the maximum in the definition of
This is because for
is negative for all t large enough. On the other hand, it equals 0 when t = 0 and so it
suffices to consider only t in a compact set.
Lemma 25.1.2 Let ϕ : ℝ → ℝ be a convex function. Then ϕ is Lipschitz
Proof: Since it is convex, the difference quotients,
are increasing because by convexity, if a < t < x
and this reduces to
Also these difference quotients are bounded below by
Then A is some finite real number. Similarly there exists a real number B such that for
all t ∈
Now let a ≤ s < t ≤ b. Then
where θ is such that θa +
and so the above implies
and this proves the lemma.
The following is like the inequality, st ≤ tp∕p + sq∕q, important in the study of Lp
Proposition 25.1.3 If A is an N function, then so is
and for all s > 0,
Proof: First consider the claim
is convex. Let
It is obvious
is stictly increasing because
is strictly increasing in s.
For s > 0 let ts denote the number where the maximum is achieved. That
It follows from this that
since otherwise, a contradiction results to 25.1.7, the expression becoming negative for
small enough s. Thus
and this shows
which shows 25.1.2.
To verify the second part of 25.1.2, let ts be as just described. Then for any
t > 0
Since t is arbitrary, this proves the second part of 25.1.2.
The inequality 25.1.5 follows from the definition of
Next consider 25.1.4. It must be shown that
To do so, first note
By convexity, the above difference quotients are nondecreasing in t and so
for all t≠t0. Hence for all t,
Therefore, 25.1.4 holds.
Consider 25.1.6 next. To do so, let a = A′ so that
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
Then passing to a limit as λ →
(If you like, you can simply assume from the beginning that A
is given this way as an
integral of a positive increasing function,
and verify directly that such an A
and satisfies the properties of an N
function. There is no loss of generality in doing so.)
Thus geometrically, A
equals the area under the curve defined by
and above the x
axis from x
In the definition of
be the point where the
maximum is achieved. Then
and so at this point,
This means that
is the area to the left of
the graph of
which is to the right of the y
axis for y
between 0 and a
that in fact
The following picture illustrates the reasoning which
The inequality results from replacing a
in the last integral on the top
An example of an N function is A
0 and p >
For this example,
Definition 25.1.4 Let A be an N function and let
be a measure space.
This is called the Orlitz class. Also define
where F is the field of scalars, assumed to be either ℝ or ℂ.
Δ regular if either of the following conditions hold.
< ∞ and for all r >
0, there exists Mr and Kr >
0 such that
Note there are N functions which are not Δ regular. For example, consider
It can’t be Δ regular because
However, functions like xp∕p for p > 1 are Δ regular.
Then the following proposition is important.
Proposition 25.1.5 If
Δ regular, then KA
. In any case,
is a vector space and KA
is Δ regular. Then I claim
is a vector space. This will
Let f,g ∈ KA
so f + g ∈ KA
in this case. Now suppose
Thus f + g ∈ KA
in this case also.
Next consider scalar multiplication. First consider the case of 25.1.10. If f ∈ KA
α ∈ F
so in the case of 25.1.10 αf ∈ KA
f ∈ KA
In the case of 25.1.11
This establishes the first part of the proposition.
Next consider the claim that LA