980 CHAPTER 27. ORLITZ SPACES
Now suppose ||u||A = 0 and let
En ≡{
x : |u(x)| ≥ 1n
}.
Then for arbitrarily small values of t,∫En
A((1/n)
t
)dµ ≤
∫Ω
A(|u(x)|
t
)dµ ≤ 1
and so for arbitrarily small values of t,
A((1/n)
t
)µ (En)≤ 1.
Letting t→ 0+ yields a contradiction unless µ (En) = 0. Now
µ ([|u(x)|> 0])≤∞
∑n=1
µ (En) = 0.
Thus u = 0 as claimed.Consider the other axioms of a norm. Let u,v ∈ LA (Ω) and let α,β be scalars. Then
||αu+βv||A ≡ inf{
t > 0 :∫
Ω
A(|u(x)+ v(x)|
t
)dµ ≤ 1
}Without loss of generality ||u||A , ||v||A < ∞ since otherwise there is nothing to prove.
||u+ v||A ≡ inf{
t > 0 :∫
Ω
A(|αu(x)+βv(x)|
t
)dµ ≤ 1
}.
≤ inf{
t > 0 :∫
Ω
A(|α| |u|+ |β | |v|
t
)dµ ≤ 1
}
= inf
{t > 0 :
∫Ω
A
(|α| (|α|+|β |)|u|t + |β | (|α|+|β |)|v|t
(|α|+ |β |)
)dµ ≤ 1
}
≤ inf{
t > 0 :|α|
(|α|+ |β |)
∫Ω
A(
|u|t/(|α|+ |β |)
)dµ ≤ 1
}+ inf
{t > 0 :
|β |(|α|+ |β |)
∫Ω
A(
|v|t/(|α|+ |β |)
)dµ ≤ 1
}
= |α| inf{
t/(|α|+ |β |)> 0 :∫
Ω
A(
|u|t/(|α|+ |β |)
)dµ ≤ 1
}+ |β | inf
{t/(|α|+ |β |)> 0 :
∫Ω
A(
|v|t/(|α|+ |β |)
)dµ ≤ 1
}