Kenneth Kuttler

27.2. DUAL SPACES IN ORLITZ SPACE 989

a contradiction since r < 1. Therefore, from 27.2.15,

||Lv|| ≥ |Lv (un)| ≡∫

v(x)u(x)dµ = ||Lv||∫

Ã(

r |v(x)|||Lv||

)dµ

and so

1≥∫

Ã(

r |v(x)|||Lv||

)dµ

By the monotone convergence theorem, letting n→ ∞,

1≥∫

Ω

Ã(

r |v(x)|||Lv||

)dµ

showing that

||v||Ã ≤||Lv||

Since this holds for all r ∈ (0,1), it follows ||Lv|| ≥ ||v||Ã as claimed. This proves theproposition.

Now what follows is the Riesz representation theorem for the dual space of EA (Ω) .

Theorem 27.2.2 Suppose µ (Ω)< ∞ and suppose L ∈ EA (Ω)′ . Then the map v→ Lv fromLÃ (Ω) to EA (Ω)′ is one to one continuous, linear, and onto. If (Ω,A) is ∆ regular thenv→ Lv is one to one, linear, onto and continuous as a map from LÃ (Ω) to LA (Ω)′ .

Proof: It is obvious this map is linear. From Proposition 27.2.1 it is continuous andone to one. It remains only to verify that it is onto. Let L ∈ EA (Ω)′ and define a complexvalued function, λ , mapping the measurable sets to C as follows.

λ (F)≡ L(XF) .

In case µ (F) ̸= 0,∫Ω

XF (x)A−1(

1µ (F)

))dµ =

∫F

A−1(

1µ (F)

))dµ

=∫

1µ (F)

dµ = 1

and so

||XF ||A ≤1

A−1(

1µ(F)

) . (27.2.18)

In fact, λ is actually a complex measure. To see this, suppose Fi ↑ F. Then from the formulajust derived,

||XFi −XF ||A =∣∣∣∣XF\Fi

∣∣∣∣A ≤

A−1(

1µ(F\Fi)

)

27.2. DUAL SPACES IN ORLITZ SPACE 989aa contradiction since r < 1. Therefore, from 27.2.15,eel > Maa = ff wCsdu(o)aqe= [ill fa (Se|, > | a(“re) djn I|Lv||By the monotone convergence theorem, letting n + ©,a mt)1> | a( dua \ ||Lrl||Z~.and soshowing thatlIvlla<Since this holds for all r € (0,1), it follows ||L,|| > ||v||z as claimed. This proves theproposition.Now what follows is the Riesz representation theorem for the dual space of E, (Q).Theorem 27.2.2 Suppose U (Q) < and suppose L € E,(Q)'. Then the map v —> L, fromLz (Q) to Ea (Q)' is one to one continuous, linear, and onto. If (Q,A) is A regular thenv — L, is one to one, linear, onto and continuous as a map from Lx (Q) to La (Q)'.Proof: It is obvious this map is linear. From Proposition 27.2.1 it is continuous andone to one. It remains only to verify that it is onto. Let L € E, (Q)’ and define a complexvalued function, 2, mapping the measurable sets to C as follows.A (F)=L(¥7).In case U(F) £0,Beane a)and so1(ay)In fact, A is actually a complex measure. To see this, suppose F; + F. Then from the formulajust derived,| 2e|l4 < (27.2.18)1|| 2% — Pla = || Ze\e| |, $< —Z—>Av! (an)