20.6. THE DUAL SPACE OF C0 (X) 629

This proves the lemma.Let L ∈ C0(X)′. Also denote by C+

0 (X) the set of nonnegative continuous functionsdefined on X . Define for f ∈C+

0 (X)

λ ( f ) = sup{|Lg| : |g| ≤ f}.

Note that λ ( f ) < ∞ because |Lg| ≤ ||L||||g|| ≤ ||L|||| f || for |g| ≤ f . Then the followinglemma is important.

Lemma 20.6.3 If c≥ 0, λ (c f ) = cλ ( f ), f1 ≤ f2 implies λ f1 ≤ λ f2, and

λ ( f1 + f2) = λ ( f1)+λ ( f2).

Also0≤ λ ( f )≤ ||L|| || f ||

∞

Proof: The first two assertions are easy to see so consider the third.For f j ∈C+

0 (X) , there exists gi ∈C0 (X) such that |gi| ≤ fi and

λ ( f1)+λ ( f2) ≤ |L(g1)|+ |L(g2)|+2ε

= L(ω1g1)+L(ω2g2)+2ε

= L(ω1g1 +ω2g2)+2ε

= |L(ω1g1 +ω2g2)|+2ε

where |gi| ≤ fi and |ω i|= 1 and ω iL(gi) = |L(gi)|. Now

|ω1g1 +ω2g2| ≤ |g1|+ |g2| ≤ f1 + f2

and so the above shows

λ ( f1)+λ ( f2)≤ λ ( f1 + f2)+2ε.

Since ε is arbitrary, λ ( f1)+λ ( f2)≤ λ ( f1 + f2) . It remains to verify the other inequality.Now let |g| ≤ f1 + f2, |Lg| ≥ λ ( f1 + f2)− ε . Let

hi (x) =

{fi(x)g(x)

f1(x)+ f2(x)if f1 (x)+ f2 (x)> 0,

0 if f1 (x)+ f2 (x) = 0.

Then hi is continuous and h1(x)+ h2(x) = g(x), |hi| ≤ fi. The reason it is continuous ata point where f1 (x)+ f2 (x) = 0 is that at every point y where f1 (y)+ f2 (y) > 0, the topdescription of the function gives ∣∣∣∣ fi (y)g(y)

f1 (y)+ f2 (y)

∣∣∣∣≤ |g(y)|Therefore,

−ε +λ ( f1 + f2) ≤ |Lg| ≤ |Lh1 +Lh2| ≤ |Lh1|+ |Lh2|≤ λ ( f1)+λ ( f2).