534 CHAPTER 19. HILBERT SPACES

Then S is dense in L2 (0,1) if and only if

∞

∑i=1

1pi

= ∞.

Proof: The polynomials are dense in L2 (0,1) and so S is dense in L2 (0,1) if andonly if for every ε > 0 there exists a function f from S such that for each integer m ≥

0,(∫ 1

0 | f (x)− xm|2 dx)1/2

< ε . This happens if and only if for all n large enough, the

distance in L2 (0,1) between the function, x→ xm and span(xp1 ,xp2 , · · · ,xpn) is less thanε. However, from Lemma 19.4.2 this distance equals

1√2m+1

n

∏k=1

|m− pk|m+ pk +1

=1√

2m+1

n

∏k=1

1−(

1− |m− pk|m+ pk +1

)Thus S is dense if and only if

∞

∏k=1

(1−(

1− |m− pk|m+ pk +1

))= 0

which, by Lemma 19.4.3, happens if and only if∞

∑k=1

(1− |m− pk|

m+ pk +1

)=+∞

But this sum equals∞

∑k=1

(m+ pk +1−|m− pk|

m+ pk +1

)which has the same convergence properties as ∑

1pk

by the limit comparison test. Thisproves the theorem.

The following is Müntz’s second theorem.

Theorem 19.4.5 Let S be finite linear combinations of {1,xp1 ,xp2 , · · ·} where p j ≥ 1 andlimn→∞ pn = ∞. Then S is dense in C ([0,1]) if and only if ∑

∞k=1

1pk

= ∞.

Proof: If S is dense in C ([0,1]) then S must also be dense in L2 (0,1) and so by Theorem19.4.4 ∑

∞k=1

1pk

= ∞.Suppose then that ∑

∞k=1

1pk

= ∞ so that by Theorem 19.4.4, S is dense in L2 (0,1) . Thetheorem will be proved if it is shown that for all m a nonnegative integer,

max{|xm− f (x)| : x ∈ [0,1]}< ε

for some f ∈ S. This is true if m = 0 because 1 ∈ S. Suppose then that m > 0. Let S′ denotefinite linear combinations of the functions{

xp1−1,xp2−1, · · ·}.