27.1. BASIC THEORY 979
This establishes the first part of the proposition.Next consider the claim that LA (Ω) is always a vector space. First note KA (Ω) is
always convex due to convexity of A. Let λu,αv ∈ LA (Ω) where u,v ∈ KA (Ω) and let a,bbe scalars in F. Then
aλu+bαv = |aλ |ωu+ |bα|θv
where |ω|= |θ |= 1. Then
= (|aλ |+ |bα|)(|aλ |ωu+ |bα|θv|aλ |+ |bα|
)which exhibits aλu + bαv as a multiple of a convex combination of two elements ofKA (Ω) ,ωu and θv. Thus LA (Ω) is closed with respect to linear combinations. This showsit is a vector space. This proves the proposition.
The following norm for LA (Ω) is due to Luxemburg [94]. You might compare this tothe definition of a Minkowski functional. The definition of LA (Ω) above was cooked up sothat the following norm does make sense.
Definition 27.1.6 Define
||u||A = ||u||A,Ω ≡ inf{
t > 0 :∫
Ω
A(|u(x)|
t
)dµ ≤ 1
}.
If two functions of LA (Ω) are equal a.e. they are considered to be the same in the usualway.
Proposition 27.1.7 The number defined in Definition 27.1.6 is a norm on LA (Ω) . Also, ifΩ1 ⊆Ω, then
||u||A,Ω1≤ ||u||A,Ω .
Proof: Clearly ||u||A≥ 0. Is ||u||A finite for u∈ LA (Ω)? Let u∈ LA (Ω) so u= λv wherev ∈ KA (Ω) . Then for s > 0∫
Ω
A(|u|
s |λ |
)dµ =
∫Ω
A(|v|s
)dµ < ∞
whenever s > 1. Therefore, from the dominated convergence theorem, if s is large enough,∫Ω
A(|u|
s |λ |
)dµ ≤ 1
and this shows there are values of t > 0 such that∫Ω
A(|u(x)|
t
)dµ ≤ 1.
Thus ||u||A is finite as hoped.