16.5. THE MÜNTZ THEOREMS 453

It follows that, since pn+1 = m

dn =∏ j<n+1

∣∣m− p j∣∣

∏ j<n+1 (m+ p j +1)√(2m+1)

=1√

(2m+1)∏

j<n+1

∣∣m− p j∣∣

(m+ p j +1)

The idea is to let n→ ∞ and see whether the distance between the best approximation andx→ xm converges to 0. That is, to determine whether dn → 0. I want m an integer to bearbitrary and for the sake of convenience, I want

∣∣m− p j∣∣= p j−m for all j large enough.

Therefore, one needs to have limk→∞ pk = ∞. From now on, this is assumed. It is desiredto have

−∞ = limn→∞

ln(

dn√

2m+1)= lim

n→∞∑

j<n+1ln

(1−

(1−

∣∣m− p j∣∣

(m+ p j +1)

))

Note that 0 <|m−p j|

(m+p j+1)= r j < 1 and so it is easily shown that

ln(1− (1− r j)) ∈ (−2(1− r j) ,−(1− r j)) .

It follows that dn→ 0 if and only if

∑j

1−p j−m

m+ p j +1= ∑

j

2m+1m+ p j +1

diverges. But by the limit comparison test from calculus, this happens if and only if ∑ j1p j=

∞. This proves most of the following theorem.

Theorem 16.5.1 Let g ∈C ([0,1]) . Then there exists a sequence hk consisting of a linearcombination of functions fp j such that |g−hn| → 0. Here we define Vn ≡ span( fp1 , ..., fpn)

where no pk is an integer, all are larger than −1/2, and limn→∞ pn = ∞ and ∑k1pk

= ∞.

Proof: Let ε > 0 be given. By the Weierstrass approximation theorem, there is apolynomial p(x) such that |g− p| ≤ ∥g− pn∥ < ε where the second norm is the supre-mum norm of the Weierstrass theorem. Thus p is a linear combination of functions fmfor m an integer. p = ∑

Lk=1 ck fk. Let hnk ∈ Vnk such that

∣∣hnk − fk∣∣ |ck| < ε

L . Let n >

max{nk : k ≤ L} . Consider h ∈ Vn defined by h ≡ ∑Lk=1 ckhnk . Then |h−g| ≤ |h− p|+

|p−g| ≤ ∑Lk=1

∣∣ckhnk − ck fk∣∣+ ε < L ε

L + ε = 2ε . It follows that letting εk → 0, thereexists hk consisting of a finite linear combination of functions of the form fp j such that|hk−g| → 0. ■

This is the first Müntz theorem. The second one involves approximation in the usualnorm for continuous function ∥ f∥ ≡ max{| f (x)| : x ∈ [0,1]}. It also depends on linearalgebra techniques.

Theorem 16.5.2 Let 12 < pk, none of the pk are zero, and limk→∞ pk = ∞ and ∑k

1pk

= ∞.

Let Vn ≡ span(1, fp1 , ..., fpn) . Then if g ∈C ([0,1]) and ε > 0, there exists h ∈Vn for somen such that ∥g−h∥ ≤ ε .

Proof: This follows in the same way as above if I can show that for all m a nonnegativeinteger, there is a function h∈Vn with ∥h− fm∥< ε . Since 1 is included, there is nothing to