984 CHAPTER 27. ORLITZ SPACES

Now suppose µ (Ω)< ∞. In this case, only assume A(rt)≤ KrA(t) for t large enough,say for t ≥Mr. However, this is enough to conclude A(r | f |)∈ L1 (Ω) for any r > 0 becauseµ (Ω)< ∞ and f ∈ KA (Ω) . Let sn→ f pointwise with |sn| ≤ | f | , and s simple. Then

A(| f − sn|

ε

)≤ A

(2ε| f |)∈ L1 (Ω)

and so the dominated convergence theorem implies

limn→∞

∫Ω

A(| f − sn|

ε

)dµ = 0.

Hence ∫Ω

A(| f − sn|

ε

)dµ < 1

for all n large enough and so for such n,

|| f − sn||A ≤ ε

which proves the proposition.It turns out EA (Ω) is the largest linear subspace of KA (Ω) .

Proposition 27.1.11 EA (Ω) is the maximal linear subspace of KA (Ω) .

Proof: Let M be a subspace of KA (Ω) . Is M ⊆ EA (Ω)? For f ∈M, f/ε ∈ KA (Ω) forall ε > 0 because of the fact that M is a subspace and f ∈M. Thus A(| f |/ε) is in L1 (Ω).Let ε > 0 be given, choose δ > 0 and let

Fδ ≡ {x : | f (x)| ≤ δ}

By the dominated convergence theorem there exists δ small enough that∫Fδ

A(

2 | f |ε

)dµ <

12.

Let |sn| ≤ | f |XFCδ

and sn→ f XFCδ

pointwise for sn a simple function. Thus sn = 0 on Fδ

and so sn ∈ S because µ(FC

δ

)< ∞. Now∫

Ω

A(| f − sn|

ε

)dµ =

∫Fδ

A(| f |ε

)dµ +

∫FC

δ

A(| f − sn|

ε

)dµ

<12+∫

FCδ

A(| f − sn|

ε

)dµ.

The integrand in the last integral is no larger than 2| f |ε

and so by the dominated convergencetheorem, this integral converges to 0 as n→ ∞. In particular, it is eventually less than 1

2 .Therefore, for such n,

|| f − sn||A ≤ r.