# 5.2 Convergence of Stieltjes Series

If any eigenvector \{k) of H had a strictly negative eigenvalue Xlk\ then the quadratic form G of (1.13) would be negative, which is impossible. Hence all the eigenvalues of H are nonnegative, det H is the product of these eigenval­ues, and so D(m, 0 for all m^0, n>0.

This argument completes the proof except for the special case in which it might happen that D(m, n) = 0. This is associated with the existence of a nontrivial solution of the homogeneous linear equations

and therefore

From (1.13), we discover a nontrivial polynomialpn(u) for which

This can only occur if

except at the zeros of p( u), contradicting

the hypothesis that <j>(u) takes on infinitely many values, which is a requirement for/(2) to be a Stieltjes series.

./-o

Hence from (1.15),

Corollary 1. All [L/M] Pade approximants to Stieltjes series exist and are nondegenerate, if L^M— 1.

The proof follows immediately from the basic representation (1.1.8). The sequences of Pade approximants to Stieltjes series of the next section are

charactenzed by

with the natural condition that J> — \.

Corollary 2. For any given Stieltjes series, there exists a regular C-fraction with the same formal expansion denoted by

with a't >0 for all i>\. This justifies the nomenclature (4.5.5) for the S-fraction, and demonstrates the identification with Stieltjes series.

Exercise 2. Check the statements of Corollary 2. Exercise 3. Use the definitions (1.2), (1.12) to prove that

Proof. In Chapter 4, we saw from (4.2.13) and (4.4.5) the possibility of expressing a formal power series as a regular C-fraction. Theorem 5.1.2 enables the signs of the elements of the fraction to be determined. We find that a2M+ 2>0, a2M+[ <0 and

Elementary equivalence transformations of the form (4.1.6) may be used to show that the fraction has the representation (1.18) with positive def­inite elements {a't, i=l, 2,...}. For example, note that a\ =c0 >0, a'2 = — c,/c0 >0, etc.

As general historical references for Stieltjes series (and not just for this section) we cite Stieltjes [1889, 1894], Tchebycheff [1858] and Van Vleck [1903]. An excellent review is given by Perron [1957], and material related to continued fractions is treated by Wall [1948].

Exercises 1. Use the definitions (1.3), (1.12), (1.1.1), and (1.4.8) to prove that

By permuting the variables and summing, deduce that

Hence, deduce that

on the fact that all the poles of the

Theorem 5.2.1. If /(z) is a Stieltjes series, then the poles of the \

Pade approximants (with J> — 1) to f(z) are simple poles which lie on the negative real axis and have positive residues:

Remark. The proof uses the determinantal inequalities of Theorem 5.1.2 for the coefficients^, and does not require the fundamental representation of {/} given by (1.2) and based on the existence of 4>(u).

and that D(m,n)>0 [Bessis, 1979].

The convergence properties of Pade approximants of Stieltjes series hinge

■75s — 1) to Stieltjes series lie on the negative real axis and have positive residues. As a special exception (1.4) shows that if /(z) is a Stieltjes series. except that <f>(u) has precisely m points of increase, then /(z) is a rational function with poles on the negative real axis with positive residues. In thi> case, all the [M+J/M] Pade approximants to/(z) with — 1 and M>m are identical to /(z). For the case of genuine Stieltjes series, we have the following important theorem.Proof. For — 1, we define x) = 1, and then for M= 1,2,3,...,

(2.1)

Apart from a sign, these quantities are compact expressions for the de­nominators of a diagonal sequence of Pade approximants (see Section 1.3). They are related by

Notice that

i is a real-valued function of the real variable x. When we

consider sequences

M=0,1,2,...} with J fixed, we often omit the

superscript (J). We tirst show that the functions

then

and

if

form a Sturm sequence, ihis means that if, for some

have opposite signs. We apply Sylvester's determinant

given by (2.1), (2.2). Using subscripts to denote the

deleted rows and columns, Sylvester's identity is

By symmetry,

We identify

and therefore

Also, (2.3) is valid if we identify A0(jc)= 1 and take j= 1. We exclude the possibility that

by using the Frobenius (* * *) identity, which would imply that A7(jc)=0 foe all j if any two successive A's vanish at the same value of x. Therefore (2.31 is a Sturm sequence.

To locate the roots of AM(x), which are the poles of the [M+J/M] Pade approximant, we consider the Sturm sequence (23) and recall that / >0. The first few members of the sequence are

From the representation (1), we have the general properties that

We define Xjj. to be the A:th zero of Aj(x), as is shown in Figure 1 and according to the following interlacing scheme:

at

From (2.3),

and from (2.4)

therefore

at

Figure 1. The sign of the first four Pade denominators of Stieltjes series for x<0.

therefore

lhe complete argument tor the interlacing ot the zeros ot A -(jc) lollows by induction in an obvious way. In short, A^(.x) = 0 has roots x = xJk, for k = 1,2,..., j which lie on the negative real axis and are distinct, because each zero of A -_,(jc) lies in an interval (xjk , ,, xjk). Hence the poles of the Padé approximants are distinct and lie on the negative real axis. To prove that the residues are positive, we use the identity (3.5.18),

Writing P[M+J/M\x) = TM{x) for short, it follows from (2.6) that

and at the particular root

Because Am+1(a:) is a polynomial of degree M+ 1, A'm+1(a:) has precisely one zero in each interval (xM+[ k, xM+[ k+[). Therefore, since AM+1(0)>0,

and so

Thus the residues of the poles of the Pade approximants are positive, and the theorem is proved.

From (2.3) and (2.5)

We now proceed to consider properties of the diagonal and first subdiag- onal sequence of Pade approximants to Stieltjes series for x>0.Theorem 5.2.2. Let    z)j be a Stieltjes series. For z real and

positive, its Pade approximants obey the following inequalities'.

Remark. The proof uses properties based on the determinantal inequali­ties for the coefficients fj which are used to construct the Pade approxi­mants. Like the other theorems of this section, it does not assume the representation (1.2) which characterizes the basic function.

Proof. From (3.5.18), the (%) identity is

From the proof of Theorem 5.2.1, the denominator functions           (and

are positive for x>0. By virtue of (2.2), (2.7) and (2.8) follow from (2.13) with / = — 1 and /=0 respectively. From (3.5.20), the (**) identity is

Hence (2.9) and (2.10) follow from (2.14) with / = — 1 and/ = 0 respectively. From (3.5.17), the (* *) identity becomes

are positive; see Exercise 1. We define

for

(2.1 H follows from (2.15). The coefficients of every power of x in

and note that each r > 0, which implies that

Hence

and these also hold when differentiated once.

Proof. The proof follows that of the theorem in every respect.

The essential result of Theorem 5.2.2 which is carried further is that the [M~\/M] Pade approximants, for fixed ,x>0, form a strictly increasing sequence. This leads to a convergence theorem after some preliminary theorems.

These inequalities [(2.11), for example] can be extended into the complex plane. There they become nesting inclusion regions for the higher-order approximants. They can be derived by exploiting Theorem 5.5.9 and allied results. The inclusion regions are convex, and lens-shaped. The two vertices of the boundaries are the [M/M] and [M— 1 /M) approximants. The new lens-shaped region touches the boundary of the old in only two places, as shown in Figure 2. A more thorough discussion is given in EPA and by Henrici and Pfluger [1966], Gargantini and Henrici [1967], and Baker [1969].

Theorem 5.2.3. The sequence of [M—\/M] Pade approximants to a Stieltjes series is uniformly bounded as M—»co in the domain LT>( A). L'D(A) is a bounded region of the complex z-plane which is at least at a distance A from the cut — oo <z=£0 along the negative real axis.

Corollary. With the hypotheses of the theorem,

Proof. From Theorem 5.2.1, we may write

Figure 2. Nesting inclusion regions for the values of a staircase sequence of Padè approxi­mants of a Stieltjes series.

where /3, >0, y, >0 for /=1,2,...,M. Each of the Padé approximants ma> be bounded by

and therefore

and

and thus [M— \/M](z) is uniformly bounded for

because L'i'( A ) is bounded we may assume that

Figure 3. The bounded domain °D(A).

then

An elementary calculation shows that the minimum is achieved for y = — (Rez)/|z|2, and also that

(2.20)

Hence the theorem is proved.

Corollary. All paradiagonal sequences [M+J/M] of Pade approxi­mants (with — 1) to Stieltjes series are uniformly bounded in the domain

6D(A).

Proof. For J>0, the polynomial /(z) = 2/=0/.( —z)' is uniformly bounded on ^(A) and may be treated separately. Then ( — z)~y_1[/(z) — /(z)] ma> be expressed in the form (2.16). The rest of the proof is straightforward.

We have now established that the sequence of [M— 1 /M] Pade ap­proximants to a Stieltjes series is strictly increasing and bounded at an> given point on the positive real axis, and so it is convergent. In fact, we are building up to a much stronger result than pointwise convergence, and thi> requires the concept of equicontinuity of a sequence [Courant and Hilbert. 1953, Chapter 2].

for all

Definition. A sequence of functions fm(z), m = 0,1,..., defined on a domain ^ is equicontinuous if, given any e>0, there exists 5>0, depending only on e, such that

for any pairs of points

The significant part of the latter definition is that 8 = 8(e) does not depend on m, z, or z2. Thus equicontinuity embraces the properties of uniform continuity of each member of the sequence with independence of which member of the sequence is selected.

Theorem 5.2.4. The sequence of [M—l/M] Pade approximants to a Stieltjes series is equicontinuous on ^(A).

Proof. From Theorem 5.2.1, we may write

Therefore

From (2.16), we find that        =/,. By inspection of (2.20), we deduce

that

Hence the sequence of

Corollary. All paradiagonal sequences [M+J/M] of Pade approxi­mants (with — 1) to Stieltjes series are equicontinuous on 6D(A).

Proof. Completely parallel to that of the theorem.

The property of equicontinuity established is exploited by Arzela's theo­rem:

Theorem 5.2.5. For any set of functions which are uniformly bounded and equicontinuous on 6D, there exists a subsequence which converges uniformly to a continuous function defined on "D.

Proof Take a countable, dense set of points in 6D. This is easily done by using a countable set of rationals which is dense on ( — 00,00) and forming the countable set {zjk =rj +irk}. The subset of this contained in 6D is the point set required.

provided

mants is equicontinuous on ^(A).

Let Pj be the set of the first J points, so that PJ = {zj, i= 1,2,..., /} is a subset of 6D.

for

Let the given equicontinuous functions form a sequence 5.

Because S is a sequence of functions which is uniformly bounded on 6D, we may define 5, to be a subsequence of S convergent at z,, S2 to be a subsequence of 5, convergent at z2, etc. Then, by construction, 5, is an infinite subsequence of S which converges at z,, z2,..., zr Given any e>0, define 8 =8(e) (2.20), so that

whenever

8 is independent of n by the equicontinu­

ity hypothesis.

Choose J sufficiently large so that the «^-neighborhoods of all the points Pj cover 6D. This condition means that

and this choice is possible because the zy are dense in 6D. Then, for the chosen e and any given zG6!), zy exists with j^J and jzf — zj<S. Becausethe sequence Sj converges at z , N exists such that

for all

By equicontinuity

for all

By choosing the subsequence of functions to be the Jth element of Sj. J =1,2,..., we find a subsequence of the given sequence which satisfies Cauchy's condition for uniform convergence, and prove Arzela's theorem.

We have proved in Theorem 5.2.3 that the sequence of [M+J/M] Pade approximants to a Stieltjes are uniformly bounded on ^(A), and Theorem 5.2.4 establishes that the sequence is equicontinuous on <5D( A). Thus Arzela's theorem asserts that a subsequence of [M+J/M] Pade approximants to a Stieltjes series converges uniformly to a continuous limit function f{J){z) on ^(A).

Next we need a familiar theorem on uniformly convergent sequences of analytic functions:

Weierstrass's theorem [Titchmarsh, 1939, p. 95]. Let each member of a sequence of functions g{(z), g2(z), g3(z),... be analytic in a domain "D, and converge to a limit function g(z) in any domain 6D2 in the interior of 6D,. Then g(z) is analytic in °D2.

We apply Weierstrass's theorem directly to assert that f(n(z) is analytic in 6D(2A). But A was chosen (see Theorem 5.2.3) as an arbitrary small positive number, and can be replaced by \ A without further implications. Thus we may deduce that for arbitrary positive A, fiJ)(z) is analytic in <5D(A).

{with /Ss-1)

Stieltjes series

converges uniformly on 6D(A), as shown in

Figure 2, to a real symmetric function f^J\z), analytic on L'D( A).

Following Theorem 5.2.3, we noted that the entire sequence converges pointwise on the positive real axis, and so this pointwise limit is the real function f-J)(x). Since fiJ){x) is analytic on an interval of the real axis, the analytic continuation of f<-J\x) to ^(A), the domain of analyticity, is unique. Thus we have proved

Theorem 5.2.6. The sequence of

Having established convergence of the paradiagonal sequences (with — 1), the obvious questions are what the limit function f{J\z) is and how convergence is achieved. We answer the second question first by showing that the power-series coefficients of (—z)j of the expansions of the [M+J/M] Padé approximants approach the coefficients fj from below.

Theorem 5.2.7. Let f(z) be a Stieltjes series given by the formal power series

Its [L/M] Padé approximant has the power-series expansion

Then, for all i and

Proof. Each [L/M] Pade approximant of f(z) with L>M— 1 exists, so that (2.22) is a well-defined series with a nontrivial circle of convergence for fixed L and M. If /<L+M, the Pade equations and (1.2) require that

Otherwise, for i>L + M, consider

Provided

the last term vanishes by virtue of the Padé

equations. Now consider the expansion of (2.6),

are positive. The zeros of

occur at negative values. Thus we may write

and similarly

This expansion, and therefore the expansion of (2.25), is a power series in ( —z) with positive coefficients. Hence every bracket of (2.24) is positive, and

proving the theorem.

As a general reference, we cite EPA, Chapter 15, and we also refer to Baker [1970], Common [1968], Wynn [1968], and Brezinski [1977, p. 82].

Exercise Use the proof of Theorem 5.2.1 to show that the coefficients of each power of Jt in the polynomials A'^(x) and A^^(x) are positive.

and