k

»k,m := μm .

(’1)k’m k’m

m

Then the operator Mμ applied to the Bernstein polynomial Bk,m gives the value Mμ (Bk,m ) =

»k,m . Moreover, we have

k

»k,m = μ0

m=0

for all m ∈ N0 .

Proof Elementary calculus gives

k’m

k’m

k

Bk,m (t) = (’1) j t j+m .

m j

j=0

Hence, application of Mμ to Bk,m leads to the representation

k’m

k’m

k

Mμ (Bk,m ) = (’1) j μ j+m

m j

j=0

k

= μm

(’1)k’m k’m

m

= »k,m .

Finally, since Mμ is a linear operator we have

k k k

»k,m = Mμ (Bk,m ) = Mμ = Mμ (1) = μ0 .

Bk,m

m=0 m=0 m=0

Obviously, if {μ j } is completely monotone then all the »k,m are nonnegative. The next

lemma shows how the coef¬cients {μ j } can be recovered by the operator Mμ from the »k,m

if the latter are nonnegative.

Lemma 7.9 Suppose that for μ = {μ j } j∈N0 we have »k,m ≥ 0 for k ∈ N0 and 0 ¤ m ¤ k.

Then

μn = lim Mμ (Bk (t n )), n ∈ N0 .

k’∞

Proof First, we prove the statement for n ≥ 1. Note that

n’1

kt ’ i

k ’i

i=0

90 Completely monotone functions

converges uniformly on [0, 1] to t n for k ’ ∞ because every factor converges uniformly

to t and the number of factors is ¬nite. Furthermore, we ¬nd for k ≥ n

t n = t n [(1 ’ t) + t]k’n

k’n

k ’ n m+n

= (1 ’ t)k’n’m

t

m

m=0

k

m(m ’ 1) · · · (m ’ n + 1)

= Bk,m (t).

k(k ’ 1) · · · (k ’ n + 1)

m=n

This means that

k

m(m ’ 1) · · · (m ’ n + 1)

μn = Mμ (t ) = »k,m .

n

k(k ’ 1) · · · (k ’ n + 1)

m=n

However, we can conclude from the representation of Mμ (Bk,m ) that

k

m n

Mμ (Bk (t )) = »k,m ,

n

k

m=0

so that

k

m(m ’ 1) · · · (m ’ n + 1) m n

μn ’ Mμ (Bk (t n )) = ’ »k,m

k(k ’ 1) · · · (k ’ n + 1) k

m=n

n’1

m n

’ »k,m

k

m=0

From the remark at the beginning of the proof and from k |»k,m | = k »k,m = μ0

m=0 m=0

we know that we can bound the ¬rst summand by μ0 for a given > 0 if k is suf¬ciently

large. Hence for n ≥ 1 we have

n n

|μn ’ Mμ (Bk (t n ))| ¤ μ0 + μ0 < 2 μ0 ,

k

for suf¬ciently large k. This shows convergence in the case n ∈ N. For n = 0 note that

Mμ (Bk (1)) = Mμ (1) = μ0 .

The proof implies that the condition »k,m ≥ 0 can be replaced by the condition

k

m=0 |»k,m | ¤ L for all k ∈ N0 with a uniform constant L > 0 without altering the re-

sult. The same is true for the next proposition if the nonnegative measure therein is replaced

by a signed measure.

This next result shows that every sequence {μ j } with »k,m ≥ 0 is actually a moment

sequence, meaning that it can be represented as a sequence of the moments of a certain

measure. This is in particular the case if the sequence is completely monotone.

7.2 The Bernstein“Hausdorff“Widder characterization 91

Proposition 7.10 Suppose that the sequence μ = {μ j } j∈N0 satis¬es »k,m ≥ 0 for k ∈ N0

and 0 ¤ m ¤ k. Then there exists a ¬nite nonnegative Borel measure ± on [0, 1] such that

1

μn = n ∈ N0 .

t n d±,

0

Proof Let us denote the Dirac measure centered at t ∈ [0, 1] by t. To be more precise,

this measure is de¬ned on the Borel sets B([0, 1]) by

if t ∈ A,

1

t (A) :=

0 otherwise,

for A ∈ B([0, 1]). This allows us to de¬ne for k ∈ N the discrete nonnegative measure

k

±k = »k,m m/k ,

m=0

which obviously satis¬es

k

1

m n