φ is nonnegative and nonincreasing. As a nonincreasing function, φ possesses limiting values

on the right and on the left for every r ∈ (0, ∞), which satisfy φ(r +) ¤ φ(r ) ¤ φ(r ’). If

we set n = 2, we see that φ is midpoint convex, meaning that φ(r + h) ¤ [φ(r ) + φ(r +

2h)]/2 for all r, h > 0. This gives in particular 2φ(r ’ h) ¤ φ(r ) + φ(r ’ 2h) and 2φ(r ) ¤

φ(r + h) + φ(r ’ h) and leads to φ(r ’) ¤ φ(r ) and 2φ(r ) ¤ φ(r +) + φ(r ’). But this

shows that φ is continuous at r > 0. Finally, a continuous and midpoint convex function is

convex.

After this preparatory step we are able to prove that the condition stated in the lemma

is equivalent to the requirement that φ is completely monotone, if it holds for all n ∈ N0 .

This seems to be quite natural, since n represents the discretization of the nth-order

derivative.

7.1 De¬nition and ¬rst characterization 87

Theorem 7.4 For φ : (0, ∞) ’ R the following statements are equivalent:

(1) φ is completely monotone on (0, ∞);

(2) φ satis¬es (’1)n n φ(r ) ≥ 0 for all r, h > 0 and n ∈ N0 .

h

Proof If φ is completely monotone on (0, ∞) then we have

h φ(r ) = h {φ(r + h) ’ φ(r )} = h hφ

n’1 n’1

n

(ξ1 )

with ξ1 ∈ (r, r + h). Iterating this statement gives for every r, h > 0 and every n ∈ N0 a

ξn ∈ (r, r + nh) such that n φ(r ) = h n φ (n) (ξn ). Hence, (1) implies that (’1)n n φ(r ) ≥ 0.

h h

Conversely, if (2) is satis¬ed we have to show that φ ∈ C ∞ (0, ∞) and that the alternation

condition is satis¬ed.

From Lemma 7.3 we know that φ is nonnegative and continuous on (0, ∞). This means

that φ satis¬es (’1)n φ (n) ≥ 0 for n = 0. We also know from Lemma 7.3 that φ is nonin-

creasing and convex. Hence, it possesses left- and right-hand derivatives satisfying

φ’ (r ) ¤ φ+ (r ) ¤ φ’ (y)

whenever 0 < r < y < ∞. If we can show that g := ’φ+ satis¬es the alternation condition

in (2), it follows, again from Lemma 7.3, that g is continuous on (0, ∞), meaning that φ is

in C 1 (0, ∞) and that (’1)n φ (n) ≥ 0 for n = 1 is satis¬ed. Moreover, since g satis¬es the

alternation condition in (2), we can carry out everything for g instead of φ. Iterating this

idea ¬nally proves that φ is completely monotone on (0, ∞).

To see that g satis¬es the alternation condition, we start by showing that (’1)k k φ is a

h

nonincreasing function. First of all we have

n’1 n’1

i +1

i i

h/n φ r + h = φ r+ h ’φ r + h

n n n

i=0 i=0

= φ(r + h) ’ φ(r )

= h φ(r ).

1

h φ(r )

k

Thus we can expand as follows:

n’1 n’1

h

h φ(r ) = ··· h/n φ r + (i 1 + · · · + i k ) .

k k

n

i 1 =0 i k =0

1

If we apply (’1)k to both sides of this equality we get

h/n

n’1 n’1

h

h φ(r ) = ··· h/n φ r + (i 1 + · · · + i k ) ¤ 0.

k+1

k 1 k

(’1)k

(’1) h/n

n

i 1 =0 i k =0

The last inequality holds because φ satis¬es the alternation condition in (2). Hence we know

that (’1)k k φ(r ) ≥ (’1)k k φ(r + h/n) for any n ∈ N. Iterating this process gives us

h h

h m

h φ(r ) ≥ (’1)k hφ r+ ≥ · · · ≥ (’1)k hφ r+

(’1)k k k k

h

n n

88 Completely monotone functions

for any n, m ∈ N. As φ is continuous this shows that (’1)k hφ

k

is nonincreasing. But this

implies that

h φ(r ) ’ (’1)k h φ(r + δ)

k k

(’1)k

¤0

’δ

for any δ > 0 and thus that (’1)k h φ+ (r ) ¤ 0, which ¬nishes the proof.

k

For later use we state the following obvious extension for completely monotone functions

on [0, ∞).

Corollary 7.5 If φ is completely monotone on [0, ∞) then we have

h φ(r ) ≥0

(’1)

for all h > 0, all r ≥ 0, and all ∈ N0 .

7.2 The Bernstein“Hausdorff“Widder characterization

Bochner™s characterization of positive semi-de¬nite functions demonstrated how powerful

an integral representation can be. Hence, it is now our goal to represent completely monotone

functions in such a way. This can be achieved in different ways. Here, we choose an approach

over completely monotone sequences. In the style of Theorem 7.4, a sequence {μ j } j∈N0 is

called completely monotone if (’1)k k μm ≥ 0 for all k, m ∈ N0 .

The Bernstein polynomials will play an important role in this context.

De¬nition 7.6 For k ∈ N0 we de¬ne the Bernstein polynomials by

km

Bk,m (t) = t (1 ’ t)k’m , 0 ¤ m ¤ k.

m

Associated with these Bernstein polynomials is the Bernstein operator, de¬ned by

k

m

Bk ( f )(t) = f Bk,m (t)

k

m=0

for any f : [0, 1] ’ R.

We will also need an operator that is de¬ned by a sequence of numbers {μ j } and acts on

polynomials.

De¬nition 7.7 For a sequence μ = {μ j } j∈N0 of real numbers, a linear operator Mμ :

π(R) ’ R is de¬ned by

n n

= ajμj.

ajt j

Mμ

j=0 j=0

The operator Mμ is de¬ned on polynomials given in the monomial basis. We need to

know how it acts on the Bernstein polynomials.

7.2 The Bernstein“Hausdorff“Widder characterization 89

Lemma 7.8 For k ∈ N0 , 0 ¤ m ¤ k, and a sequence {μ j } j∈N0 of real numbers we set