57.5. BLASCHKE PRODUCTS 1807

Theorem 57.5.5 Let λ 1 < λ 2 < λ 3 < · · · be an increasing list of positive real numbersand let a > 0. If

∞

∑n=1

1λ n

= ∞, (57.5.23)

then linear combinations of 1, tλ 1 , tλ 2 , · · · are dense in C ([0,b]).

Proof: Let X denote the closure of linear combinations of{

1, tλ 1 , tλ 2 , · · ·}

in C ([0,b]) .If X ̸= C ([0,b]) , then letting f ∈ C ([0,b]) \X , define Λ ∈ C ([0,b])′ as follows. First letΛ0 : X +C f be given by Λ0 (g+α f ) = α || f ||

∞. Then

sup||g+α f ||≤1

|Λ0 (g+α f )| = sup||g+α f ||≤1

|α| || f ||∞

= sup||g/α+ f ||≤ 1

|α|

|α| || f ||∞

= sup||g+ f ||≤ 1

|α|

|α| || f ||∞

Now dist( f ,X) > 0 because X is closed. Therefore, there exists a lower bound, η > 0 to||g+ f || for g ∈ X . Therefore, the above is no larger than

sup|α|≤ 1

η

|α| || f ||∞=

(1η

)|| f ||

∞

which shows that ||Λ0|| ≤(

1η

)|| f ||

∞. By the Hahn Banach theorem Λ0 can be extended to

Λ ∈C ([0,b])′ which has the property that Λ(X) = 0 but Λ( f ) = || f || ̸= 0. By the Weier-strass approximation theorem, Theorem 9.1.7 or one of its cases, there exists a polynomial,p such that Λ(p) ̸= 0. Therefore, if it can be shown that whenever Λ(X) = 0, it is the casethat Λ(p) = 0 for all polynomials, it must be the case that X is dense in C ([0,b]).

By the Riesz representation theorem the elements of C ([0,b])′ are complex measures.Suppose then that for µ a complex measure it follows that for all tλ k ,∫

[0,b]tλ k dµ = 0.

I want to show that then ∫[0,b]

tkdµ = 0

for all positive integers. It suffices to modify µ is necessary to have µ ({0}) = 0 since thiswill not change any of the above integrals. Let µ1 (E) = µ (E ∩ (0,b]) and use µ1. I willcontinue using the symbol, µ .

For Re(z)> 0, define

F (z)≡∫[0,b]

tzdµ =∫(0,b]

tzdµ