27.2. DUAL SPACES IN ORLITZ SPACE 987

Therefore, LB (Ω) is a linear space contained in KA (Ω) . It follows from Proposition 27.1.11that LB (Ω)⊆ EA (Ω) . This proves the theorem.

The norm of EA (Ω) is the same as the norm on LA (Ω) so this shows LB (Ω) ↪→ LA (Ω) .Note that for 1 < p < q and A(t) = t p/p,B(t) = tq/q,

A(rt) = rpt p

and

limt→∞

A(rt)B(t)

= 0

showing this case is covered by the above theorem.If A is increasing essentially more slowly than B and µ (Ω) < ∞, this has shown the

following inclusions

EB (Ω)⊆ KB (Ω)⊆ LB (Ω) ↪→ EA (Ω)⊆ KA (Ω)⊆ LA (Ω) .

In the case of A(t) = t p/p,B(t) = tq/q both (A,Ω) and (B,Ω) are ∆ regular and so in thiscase or any other case where the N functions are ∆ regular, the above sequence of inclusionsreduces to

EB (Ω) = KB (Ω) = LB (Ω) ↪→ EA (Ω) = KA (Ω) = LA (Ω) .

27.2 Dual Spaces In Orlitz SpaceRecall that for s, t ≥ 0,

st ≤ A(t)+ Ã(s) .

Let v ∈ LÃ (Ω) and u ∈ LA (Ω) . Then there is a version of Holder’s inequality as follows.For ε > 0,

|v|||v||Ã + ε

∈ KÃ (Ω) ,|u|

||u||A + ε∈ KA (Ω) .

Therefore, ∫Ω

(|u|

||u||A + ε

)(|v|

||v||Ã + ε

)dµ ≤

∫Ω

A(

|u|||u||A + ε

)dµ

+∫

Ω

Ã(

|v|||v||Ã + ε

)dµ ≤ 2

and so uv ∈ L1 (Ω) and∣∣∣∣∫Ω

uvdµ

∣∣∣∣≤ ∫Ω

|u| |v|dµ ≤ 2(||v||Ã + ε

)(||u||A + ε) .

Since ε is arbitrary this shows∣∣∣∣∫Ω

uvdµ

∣∣∣∣≤ ∫Ω

|u| |v|dµ ≤ 2 ||v||Ã ||u||A . (27.2.14)