1$\ we have $F(c_1 t)\ge c_2 F(t)$\ for all $0\le t<\infty$. We
will say that $F$\ satisfies the {\dt \Deltacond\/} if $F^{-1}$\ is dilatory.
\moreproclaim
If $F$\ is a \phifunction, we will define the function $\tilde F(t)$\ to
be $1/F(1/t)$\ if $t>0$, and $0$\ if $t=0$.
The definition of a \phifunction\ is slightly more restrictive than that of an
Orlicz function in that we insist that $F$\ be strictly increasing.
The notion of dilatory replaces
the notion of convexity. The Orlicz spaces generated by dilatory
functions are only quasi-Banach spaces, in contrast to those generated by Orlicz
functions, which are Banach spaces.
The \Deltacond\ appears widely in literature
about Orlicz spaces.
\Definition: If $(\Omega,\F,\mu)$\ is a measure space, and $F$\ is a
\phifunction, then we define the {\dt Luxemburg functional\/} by
$$ \normo{f}_F = \inf\left\{\, c :
\int_\Omega F\bigl(\modo{f(\omega)}/c\bigr) \,d\mu(\omega) \le 1 \,\right\}
,$$
for every measurable function $f$. We define the {\dt Orlicz space},
$L_F(\Omega,\F,\mu)$\ (or $L_F(\mu)$, $L_F(\Omega)$ or $L_F$\ for short), to be
the vector space of measurable functions $f$\ for which $\normo f_F < \infty$,
modulo functions that are zero almost everywhere.
Now we define the Orlicz--Lorentz spaces.
\Definition: If $(\Omega,\F,\mu)$\ is a measure space, and
$F$\ and $G$\ are \phifunction s, then we define the {\dt
Orlicz--Lorentz functional\/} of a measurable function $f$\ by
$$ \normo{f}_{F,G} = \normo{f^*\circ\tilde F\circ\tilde G^{-1}}_G .$$
We define the {\dt Orlicz--Lorentz space},
$L_{F,G}(\Omega,\F,\mu)$\ (or $L_{F,G}(\mu)$, $L_{F,G}(\Omega)$ or $L_{F,G}$\ for
short), to be the vector space of measurable functions $f$\ for which $\normo
f_{F,G} <\infty$, modulo functions that are zero almost everywhere.
\Definition: If $(\Omega,\F,\mu)$\ is a measure space, and
$F$\ is a \phifunction, then we define the ({\dt weak-}){\dt
Orlicz--Lorentz functional\/} by
$$ \normo{f}_{F,\infty} = \sup_{x\ge0} \tilde F^{-1}(x) f^*(x) .$$
We define the {\dt Orlicz--Lorentz space},
$L_{F,\infty}(\Omega,\F,\mu)$\ (or $L_{F,\infty}(\mu)$, $L_{F,\infty}(\Omega)$ or
$L_{F,\infty}$\ for short), to be the vector space of measurable functions $f$\
for which $\normo f_{F,\infty} <\infty$, modulo functions that are zero almost
everywhere.
We see that $L_{F,F} = L_F$\ with equality of norms, and that if $F(t) =
t^p$\ and $G(t) = t^q$, then $L_{F,G} = L_{p,q}$, and $L_{F,\infty} =
L_{p,\infty}$, also with equality of norms. Thus, if $F(t) = t^p$, we shall
write $L_{p,G}$\ for $L_{F,G}$, and $L_{G,p}$\ for $L_{G,F}$.
Also, if $A$\ is any measurable set, then $\normo{\chi_A}_{F,G} =
\normo{\chi_A}_{F,\infty} = \normo{\chi_A}_F = \tilde F^{-1}\bigl(\mu(A)\bigr)$.
The Orlicz--Lorentz spaces defined here are equivalent to the
definition given in the introduction, as we now describe.
\Definition: A {\dt weight function\/} is a function $w:(0,\infty) \to
(0,\infty)$\ such that
$$ W(t) = \int_0^t w(s)\,ds $$
is a \phifunction.
Then if $w$\ is a weight function, and $G$\ is a \phifunction, then
$\Lambda_{w,G} = L_{\tilde W^{-1} \circ G,G}$, where
$$ W(t) = \int_0^t w(s) \, ds .$$
Now let us provide some examples. We define the {\dt modified
logarithm\/} and the {\dt modified exponential\/} functions by
$$ \eqalignno{
\lm(t)
&=\cases{ 1+\log t & if $t\ge1$\cr
1/\bigl(1+\log(1/t)\bigr) & if $0*2$. Then for any function of the
above form, we have that
$$ 1 <
\int_0^\infty G\bigl(2^{-1} f^*\circ\tilde G^{-1}(x)\bigr) \,dx
\le \sum_{i=1}^N \tilde G(d^{k_i}) G(2^{-1} d^{-k_i}). $$
Therefore, the cardinality of the set of $n\in\N$\ such that
$$ \tilde G(d^n) G(2^{-1} d^{-n}) \le N^{-1} $$
is less than $N$. By Lemma~5.2.1, this shows that $G$\ satisfies the
\Deltacond.
\endproof
\proclaim Lemma 5.2.5. Suppose that $F$, $G_1$\ and $G_2$\ are \phifunction s
such that one of $G_1$\ and $G_2$\ is dilatory. If
$G_1$\ satisfies the \Deltacond, and for some $c<\infty$\ we have that
$\normo f_{F,G_2} \ge \invc \normo f_{F,G_1}$\ for all $f\in\T_F$, then
$G_2$\ satisfies the \Deltacond.
\Proof: By Lemma~5.2.3, we have that $G_1$\ is dilatory. Now the
result follows immediately from Lemmas~5.1.2 and~5.2.4.
\endproof
\beginsection 5.3.\enspace Comparison Conditions for $L_{1,G}$
In this subsection, we give the key lemma that demonstrates the relationship
between the almost convexity of $G$, and the comparison between $L_{1,G}$\ and
$L_1$. As a corollary, we will also obtain results that show that in the
definition of the `almost properties' that we can take the value of $a$\ to be
arbitrarily large.
\proclaim Lemma 5.3.1. Suppose that $G$\ is a \phifunction. Then
the following are equivalent.
\itemi For some $c<\infty$, we have that $\normo f_{1,G} \le c\, \normo f_1 $\
for all measurable $f$.
\itemii For some $c<\infty$, we have that $\normo f_{1,G} \le c\, \normo f_1 $\
for all $f\in\T_1$.
\itemiii For all sufficiently large $a$, there are numbers $b>1$\ and
$N\in\N$\ such that for all $m\in\N$, the cardinality of the set of $n\in\Z$\
such that we do {\bf not} have
$$ G(a^{n+m}) \ge a^{m-N} G(a^n) $$
is less than $b^m$.
\itemiv $G$\ is almost convex.
The proof will require the next lemma.
\proclaim Lemma 5.3.2. Let $G$\ be a \phifunction. If $G$\ is almost convex, then
given $a'>1$,
there are numbers $a>1$, $b>1$, $c<\infty$\ and $N\in\N$\ such that
$a>\max\{a',b\}$\ and such that for all $m\in\N$, the cardinality of the set
of $n\in\Z$\ such that we do {\bf not} have
$$ G(a^{n+m}) \ge a^{m-N} G(a^n) $$
is less than $cb^m$.
\moreproclaim
There are similar results if $G$\ is almost concave or almost constant.
\Proof: There are numbers $a>1$, $b>1$\ and $N\in\N$\ such that for all $m\in\N$,
the cardinality of the set
$$ A_m = \{\, n\in\Z : G(a^{n+m}) < a^{m-N} G(a^n) \} $$
is less than $b^m$. Pick $c\in\N$\ such that $a^c > b$\ and $a^c>a'$, and let
$$ A'_m = \{\, n\in\Z : G(a^{c(n+m)}) < a^{c(m-N)} G(a^{cn}) \} .$$
Then, if $n\in A'_m$, then at least one of $cn$, $cn+m,\ldots,$\ or $cn+(c-1)m
\in A_m$, and hence $\modo{A'_m} < cb^m$.
\endproof
\Proofof Lemma 5.3.1: Clearly, (i)\implies(ii) and (iii)\implies(iv). We will
show (ii)\implies(iii). By Lemma~5.2.3, we know that $G$\ is dilatory. Thus we
suppose that $a>2$\ and $G(at) \ge 2G(t)$\ for all $t\ge0$. Choose $N$\
so that $a^{N-1}>c$. We will prove the result by showing that there cannot be
numbers $m\in\N$\ and $n_1 1 ,\cr}$$
where the penultimate inequality follows because $\tilde G(a^{-n_i}) > 2
\tilde G(a^{-n_{i+1}})$.
Thus $\normo f_{1,G} \ge a^m \ge a^{N-1} \normo f_1$,
which is a contradiction.
Now we show that (iv)\implies(i).
By Lemma~5.3.2, there are numbers $a>b>1$, $c<\infty$\ and $N\in\N$\ such that
for all $m\in\N$, the cardinality of the set
$$ A_m = \{\, n\in\Z : G(a^n) < a^{m-N} G(a^{n-m}) \} $$
is less than $cb^m$.
Let $\{\lists k\}$\ be the (possible finite or
empty) set of integers not in $\bigcup_{m=1}^\infty A_m$. Define
the sequence of sets $B_m$\ by setting $B_1 = A_1$, and
$$ B_m = \{k_{m-1}\} \cup A_m \setminus \bigcup_{m'*