ROCKY MOUNTAIN
JOURNAL OF MATHEMATICS
Volume 35, Number 1, 2005




KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA

               THOMAS CRAVEN AND GEORGE CSORDAS


            ABSTRACT. The paper answers an old question of Pólya
          involving Descartes’ Rule of Signs and a related conjecture of
          Karlin involving the signs of Wronskians of entire functions
          and their derivatives. Counterexamples are given along with
          classes of functions for which the conjecture is valid.


  0. Introduction. The purpose of this paper is to answer an old
unsolved question of Pólya (c. 1934) and to resolve a related conjecture
of Karlin (c. 1967). In Section 1 we state Pólya’s question, Karlin’s
conjecture, provide some background information and recall the deﬁ-
nitions and terminology that will be used in the sequel. For a general
class of polynomials closed under diﬀerentiation, we prove in Section 2
that Descartes’ Rule of Signs is equivalent to the sign regularity of
certain Hankel determinants, Theorem 2.3. The counterexamples we
give to Karlin’s conjecture, Section 3, also provide a negative answer to
Pólya’s question. (While this manuscript was in preparation, Dr. Dimi-
tar Dimitrov has kindly informed the authors that he has also obtained
a counterexample to Karlin’s conjecture.) In Section 4 we investigate
some classes of entire functions for which Pólya’s question has an aﬃr-
mative answer, Theorem 4.6 and Corollary 4.8, and for which Karlin’s
conjecture is valid, Theorem 4.5.

  1. Background information and definitions. In 1934, in connec-
tion with his investigation of the distribution of zeros of polynomials
and the total positivity of certain matrices, Schoenberg [24, p. 562]
cited the following question of Pólya.

  Pólya’s question. Let f (x) be a polynomial of degree n with
only real and simple zeros, x1 < x2 < · · · < xn . Let Z(xn ,∞) (f )
denote the number of real zeros of f in the interval (xn , ∞), and let
   2000 AMS Mathematics Subject Classiﬁcation. Primary 26C10, 30D15, Sec-
ondary 15A48.
   Key words and phrases. Hankel determinant, totally positive sequences, Turán
inequalities, Descartes’ Rule of Signs.

                             Copyright c
                                       2005 Rocky Mountain Mathematics Consortium


                                       61
62                     T. CRAVEN AND G. CSORDAS



V ({a0 , a1 , . . . , an }) denote the number of sign changes in the sequence
of real numbers {a0 , a1 , . . . , an } (where, as usual, the zero terms of the
sequence are deleted when counting the sign changes). Is it     true  that, for
                                                                   n
any sequence of real numbers {a0 , a1 , . . . , an } such that j=0 aj f (j) =
0,
                             
                              n                       
                                          (j)
(1.1)            Z(xn ,∞)          aj f         (x)       ≤ V ({aj }nj=0 )?
                             j=0




  In other words, Pólya’s question is whether or not Descartes’ Rule
of Signs holds for the sequence f (x), f  (x), . . . , f (n) (x) on the interval
(xn , ∞). We recall, see, for example, [23, Part V, pp. 87 90] that a
sequence of functions f0 (x), f1 (x), . . . , fn (x) satisﬁes Descartes’ Rule
of Signs on an interval (a,      nb), if for any sequence of real numbers
{a0 , a1 , . . . , an } such that j=0 aj fj = 0,
                             
                              n                  
(1.2)               Z(a,b)         aj fj (x)          ≤ V ({aj }nj=0 ),
                             j=0


where Z(a,b) (f ) denotes the number of real zeros of f in the interval
(a, b). The question of Pólya is natural in view of the plethora of
examples of sequences of entire and meromorphic functions which
satisfy (1.1), cf. [23, Part V, #36, #41, #77, #84 and #85]. While
from the vast literature dealing with Descartes’ Rule of Signs and its
generalizations we cite here only [9], [12, Chapter 6], [14], [16, pp.
191 193], [19, pp. 53 87], [21, pp. 40 50] and [22, pp. 22 32], we
speciﬁcally mention that it is known that (1.1) holds for sequences of
orthogonal polynomials [17, 20, 24]. Until the 1950’s various versions
of Descartes’ Rule of Signs (as for example Sylvester’s theorem [22,
Satz IV] were deemed interesting but isolated results. In the subsequent
decades, these “isolated” theorems played a pivotal role in the theory of
variation diminishing transformations, see, for example, [12, Chapters
5 6 and 9], in the theory of Chebyshev systems [14], [4, pp. 91 100]
and the theory of total positivity and combinatorics [2, 5, 12].
  Pólya’s question can also be formulated with the aid of certain de-
terminants. Indeed, Pólya and Szegö, see [23, Part V, #87 and
#90] have provided a criterion, cf. Lemma 2.2 below, expressed in
         KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                                63


terms of Wronskian determinants, which is both necessary and suf-
ﬁcient for the validity of Descartes’ Rule of Signs. We recall that
the Wronskian (determinant) of the suﬃciently smooth functions
f0 (x), f1 (x), . . . , fn (x) is deﬁned as
                                                  (j)
        W (f0 (x), f1 (x), . . . , fn (x)) := det(fi (x))ni,j=0
                                                                        (n)    
                                              f0 (x) f0 (x) · · ·
                                                                       f0 (x) 
                                              f (x) f  (x) · · ·      f1 (x) 
(1.3)                                                                    (n)
                                              1       1
                                           = .           ..               ..   .
                                              ..                               
                                                          .                .   
                                              f (x) f  (x) · · ·      f (x) 
                                                                         (n)
                                                n      n                 n


  In order to motivate and state Karlin’s conjecture referred to in the
title of the paper, we will need to introduce the following deﬁnitions.

  Definition 1.1. A TP-sequence (totally positive  ∞sequence)    is a
sequence {αk }∞k=0 , αk ∈  R, for  which   φ(x) =    k=0 αk x k
                                                                is  an
entire function and all the minors, i.e., the determinants of square
submatrices, of the matrix
                       ⎛                           ⎞
                         α0 0      0    0 0 ...
                       ⎜ α1 α0 0        0 0 ... ⎟
                       ⎜                           ⎟
(1.4)            M := ⎜ α2 α1 α0 0 0 . . . ⎟
                       ⎝                           ⎠
                         α3 α2 α1 α0 0 . . .
                                  ...
are nonnegative.

  Remark 1.2. (A remark on sign regularity.) We recall that a matrix
An = (ai,j )ni=1,j=1 is a Hankel matrix, if the general term of An
is of the form ai,j = αi+j−2 . By rearranging the columns of the
matrix M deﬁned by (1.4), we can form the Hankel (sub)matrices
                n
An = (αi+j−2 )i=1,j=1 , which, since {αk }∞
                                          k=0 is a TP-sequence, enjoy
the following sign regularity:
(1.5)
                                                                           
                                        α0             α1   ...      αn−1 
                                                                           
       n(n−1)/2               n(n−1)/2  α1             α2   ...       αn 
  (−1)          det An = (−1)                                               ≥ 0.
                                                            ...            
                                                                           
                                         αn−1           αn   ...      α2n−2
64                      T. CRAVEN AND G. CSORDAS



                                                      
   Definition 1.3. A real entire function φ(x) := ∞                   k
                                                         k=0 (γk /k!)x is
said to be of type I in the Laguerre-Pólya class, denoted φ(x) ∈ L-PI,
if φ(x) or φ(−x) can be expressed in the form

                                           ∞           
                                                     x
(1.6)                   φ(x) = cxn ebx            1+      ,
                                                     xk
                                           k=1


where c, b, xk ∈ R, xk > 0, b ≥ 0, n is a nonnegative integer and
  ∞                                         +
   k=1 1/xk < ∞. We will write φ ∈ L-P , if φ ∈ L-PI and if γk ≥ 0 for
all k = 0, 1, 2 . . . . For reasons of convenience we will allow 0 ∈ L-P + .

  With the foregoing terminology, Karlin’s conjecture [12, p. 390], see
also [8, p. 258] regarding a misprint in [12, p. 390], is as follows.

 Karlin’s Conjecture. Let φ ∈ L-P + . Then for all n ≥ 1

(1.7)   (−1)n(n+1)/2 W (f (x), f  (x), . . . , f (n) (x)) ≥ 0,       for all x ≥ 0,

where the Wronskian determinant is deﬁned by (1.3).

  What is the raison d’être for this conjecture? In the ﬁrst place, the
results and numerous examples of Karlin and Szegö’s 157 page paper
[15] tend to suggest that (1.7) is true. Also, by a celebrated result of
Schoenberg [25], see also [1] or [12, p. 412], if

                            ∞
                                             ∞
                                              
                              γk       k
                  φ(x) :=              x :=         αk xk ∈ L-P + ,
                                  k!
                            k=0               k=0


then {αk }∞k=0 is a totally positive sequence and consequently this
sequence possesses the sign regularity condition given by (1.5). One
can use this observation to construct nontrivial functions φ ∈ L-P + for
which Karlin’s conjecture is valid. While there are many other known
special cases when (1.7) is true, here we will conﬁne ourselves to the
remark that in 1989 the authors have investigated certain polynomial
invariants and used them to prove that (1.7) is true for 3×3 Wronskian
determinants [8, Theorem 2.13].
          KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                         65


  2. The equivalence of Descartes’ Rule of Signs and the sign
regularity of Hankel determinants.

 Notation. Let Cp be a set of polynomials with nonnegative coeﬃcients
which is closed under diﬀerentiation.
  Note that the set of all polynomials in L-P + forms such a set Cp .

  With the notation above, we consider the following two statements:
  (Dp ) (Descartes’ Rule of Signs). For every polynomial
                                                     m        f ∈ Cp ,
every sequence {aj }m
                    j=0 of real numbers, such that    j=0 aj f
                                                               (j)
                                                                   = 0,
and for every m ≤ degree f ,
                     m           
(2.1)            Z+       aj f (x) ≤ V ({aj }m
                              (j)
                                              j=0 ),
                          j=0

where Z+ (f ) denotes the number of positive real zeros of f , count-
ing multiplicities, and where V ({aj }m     j=0 ) denotes the number of sign
changes in the sequence {a0 , a1 , . . . , am }.
 (Hp ) (Sign Regularity of Hankel Determinants). For every
polynomial f ∈ Cp , f = 0, and for each positive integer m ≤ degree f ,

(2.2) (−1)m(m+1)/2 W (f (x), f  (x), . . . , f (m) (x)) > 0,   for all x > 0.

  We shall need the following two lemmas in order to prove that the
statements (Dp ) and (Hp ) are equivalent.

 Lemma 2.1 [12, Theorem 3.2, p. 59] or [10]. Let A = (aij ) be an
m × n matrix with 2 ≤ m ≤ n. Assume that
  (1) all m × m minors with consecutive columns have the same sign,
and
 (2) all (m − 1) × (m − 1) minors of the submatrix B consisting of A
with the last row deleted have the same (nonzero) sign.
Then all m × m minors of A have the same sign.

 Lemma 2.2 [23, Part V, # 87 and #90]. For any suﬃciently
smooth function f , Descartes’ Rule of Signs for f (x), f  (x), . . . , f (m) (x)
66                       T. CRAVEN AND G. CSORDAS



is equivalent to the statement that for any integers 0 ≤ i1 < i2 < · · · <
ik ≤ m, all Wronskian determinants W (f (i1 ) , f (i2 ) , . . . , f (ik ) ) of order
k are nonzero on (0, ∞) and have the same sign.

  Theorem 2.3. The statements (Dp ) and (Hp ) are equivalent.

  Proof. Lemma 2.2 gives us the following equivalent version of Dp :
  (D1) For every f ∈ Cp , each k = 1, . . . , m = degree f + 1 and any
integers 0 ≤ i1 < i2 < · · · < ik ≤ n = degree f , all Wronskian
determinants W (f (i1 ) , f (i2 ) , . . . , f (ik ) ) of order k are nonzero on (0, ∞)
and have the same sign.
  (D1) =⇒ (Hp ).        Assume that the polynomial f ∈ Cp has degree
n. Then the Wronskian W (f (n−k+1) (x), . . . , f (n) (x)) = (−1)k(k−1)/2 ×
(f (n) (x))k has sign (−1)k(k−1)/2 because f (n) (x) is a positive constant.
Since all kth order Wronskians have the same sign by (D1), we have
(−1)k(k−1)/2 W (f (x), f  (x), . . . , f (k−1) (x)) > 0 for all x > 0.
  (Hp ) =⇒ (D1).             Assume that the polynomial f ∈ Cp has degree
n. We proceed by induction on the order k of the Wronskian. If
k = 1, then we use the fact that f ∈ Cp implies that f (i1 ) ∈ Cp . Hence
W (f (i1 ) (x)) = f (i1 ) (x) > 0 on (0, ∞) implies that Z+ (f ) = 0 = V (a0 ).
Now assume that (D1) holds for all integers less than k. Then (Hp )
together with the induction hypothesis implies that all Wronskians of
order k − 1 have the appropriate sign (−1)(k−1)(k−2)/2 on the interval
(0, ∞). By (Hp ), we also have (−1)k(k−1)/2 W (f (l) , . . . , f (k+l−1) ) >
0 on (0, ∞) for each l = 0, 1, . . . , n − k + 1. Thus we can now apply
Lemma 2.1 to conclude the proof that (D1) holds for Wronskians of
order k.

  We state and prove another equivalent condition in terms of determi-
nants which plays an important role in the study of Descartes systems,
Haar systems and generalized convexity [12, p. 25], [14] as well as in
the theory of special functions dealing with determinants of Turán type
[15, Chapter 3]. Theorem 2.4 can be used in conjunction with our ex-
amples in the next two sections to provide further results and examples
in these contexts.
         KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                    67


  Theorem 2.4. The statements (Dp ) and (Hp ) are equivalent to the
following: For any f ∈ Cp , any integers 0 ≤ i0 < i1 < · · · < ik ≤ deg f
and any real numbers 0 < t0 < t1 < · · · < tk ,
                       (i0 )                                      
                       f (t0 ) f (i1 ) (t0 ) · · · f (ik ) (t0 ) 
                       (i0 )     (i1 )               (ik )        
                       f (t1 ) f (t1 ) · · · f              (t1 ) 
        (−1)k(k+1)/2      ..        ..                  ..        > 0.
                                                                   
                            .         .                   .       
                       (i0 )                                      
                        f (tk ) f (i1 ) (tk ) · · · f (ik ) (tk )


  Proof. The equivalence is nearly proved in [14, Theorem 4.4]. What
is left for us to prove is only that the signs of the determinants, known
to be dependent only on the size k, are in fact (−1)k(k+1)/2 . Assume
inductively that this is true for determinants of smaller size. Note
that the claimed sign is the sign of the permutation which reverses the
numbers 0 through k; that is, the sign associated with the secondary
diagonal of the matrix when one expands the determinant. Since f (i0 )
has the largest degree of the polynomials in the matrix, when we expand
the determinant along the ﬁrst column, the dominant term involving
the arbitrarily large number tk is the number (−1)k f (i0 ) (tk ) multiplied
by a determinant of sign (−1)k(k−1)/2 , using the induction hypothesis.
Therefore the sign of the larger determinant is as claimed.

  3. Counterexamples. In this section we exhibit some examples,
involving polynomials as well as transcendental entire functions having
an inﬁnite number of real negative zeros, which show that Karlin’s
conjecture, see (1.7), in the absence of additional assumptions, fails in
general. The polynomials in Example 3.1 below show that (Hp ), cf.
(2.2), does not always hold, and whence, by virtue of Theorem 2.3, we
have a negative answer to Pólya’s question (1.1) as well.

  Example 3.1. Let f (x) := x3 (x + 1)5 . Then the 5 × 5 Wronskian
of f ,
W (f (x), f  (x), f  (x), f  (x), f (4) (x))
= 414720(x+1)5 (175616x15 + 1317120x14 + 4751040x13 + 10885840x12
  + 17702475x11 + 21649908x10 + 20599540x9 + 15556680x8
  + 9410130x7 + 4548200x6 + 1727268x5 + 496920x4
  + 101015x3 + 12540x2 + 600x − 24),
68                       T. CRAVEN AND G. CSORDAS



is negative for small values of x. Such examples are rather delicate; for
example, x4 (x + 1)5 is not a counterexample in the 5 × 5 Wronskian
case.
  One can avoid the root at 0; (50x + 1)3 (x + 1)5 and (x + 1)3 (x + 50)5
also yield counterexamples. In fact, it can be shown that the 5 × 5
Wronskians of the polynomials of the form (x + a)3 (x + b)5 , where
0 < a < b and b/a ≥ 42, produce counterexamples. Moreover, a small
perturbation of the zeros will give examples with simple zeros.

  Example 3.2. A speciﬁc counterexample to (Dp ), cf. (2.1), can be
obtained as in the solution of [23, V #87]. Let r > 0 be the positive
zero of the Wronskian in Example 3.1. Then, evaluated at this point
r, the ﬁve rows are linearly dependent, say with

 c0 f (j) (r) + c1 f (j+1) (r) + c2 f (j+2) (r) + c3 f (j+3) (r) + c4 f (j+4) (r) = 0,

for j = 0, 1, 2, 3, 4. Thus the function

            c0 f (x) + c1 f  (x) + c2 f  (x) + c3 f  (x) + c4 f (4) (x)

has a positive root r of multiplicity 5, while there are at most 4 sign
changes in the sequence {c0 , c1 , c2 , c3 , c4 }, contradicting Descartes’ Rule
of Signs.

  Example 3.3. Counterexamples, to (2.1) and (2.2), of size 4 × 4 also
exist, though they are not quite as nice as the 5 × 5 case. The lowest
degree example we have found is f (x) = x2 (x + 1)11 , for which the 4 × 4
Wronskian at x = 0 is −1584.

  We next consider examples involving transcendental entire functions
in L-P + . To this end, set g(x) = ebx h(x), where h(x) is a suﬃciently
smooth function and b ∈ R. Then

(3.1) (−1)m(m−1)/2 W (g, g  , . . . , g (m−1) )
                                = (−1)m(m−1)/2 embx W (h, h , . . . , h(m−1) ),

as one easily sees by factoring out the exponential factors and then
doing row and column operations until the Wronskian in derivatives of
          KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                               69


h is obtained: begin with the second row and subtract b times the ﬁrst
row; then subtract b times the ﬁrst column from the second, obtaining
the proper 2 × 2 matrix in the upper left corner. Proceed with the next
row and column. Hence, it follows from Example 3.1 and equation
(3.1) that g(x) = ebx x3 (x + 1)5 , where b > 0, is a transcendental entire
function in L-P + for which (2.2) fails.
  In the next example we construct counterexamples involving func-
tions in L-P + which possess an inﬁnite number of zeros. The intuitive
idea behind the construction is that if we multiply the polynomial f (x)
in Example 3.3 by a function in g(x) ∈ L-P + all of whose zeros are
located “far” away from the origin, then the qualitative behavior of the
4 × 4 Wronskian of the product φ(x) = f (x)g(x) near the origin diﬀers
“little” from the 4 × 4 Wronskian of f (x). A precise formulation is
given in the next example.

  Example 3.4. (Functions in L-P + with an inﬁnite number of zeros.)
Set f (x) = x2 (x+1)11 as in Example 3.3 and
                                           let g(x) = ∞k=1 (1+x/αk ).
                                             ∞
Assume that the zeros αk > 0 satisfy         k=1 1/αk <  ε ≤ 0.2 and
consider the function φ(x) = f (x)g(x). We shall see that the 4 × 4
Wronskian is again negative at x = 0 and is positive at x = 1. To
estimate the derivatives, we note that
                               ∞
                                                        ∞
                                                         
                  log g(x) =          log(1 + x/αk ) ≤         x/αk
                               k=1                       k=1

for 0 ≤ x < 2 < α1 , so that 1 < g(1) ≤ eε < 2 and
                            ∞
                                            
                                1           x
                  g  (x) =            1+       < 2ε
                               αk          αl
                                k=1      l=k

at x = 1 and is less than ε at x = 0. Similarly, one sees that g (k) (0) < εk
and g (k) (1) < 2εk for k ≥ 2. When the 4 × 4 Wronskian is evaluated
at zero, we obtain

   − 48(5g (4) (0) − 20g  (0)g  (0) − 12g  (0)2
             + 54g  (0)2 g  (0) − 66g  (0) − 27g  (0)4 + 66g  (0)2 + 33)
           ≤ −48(33 − 20g  (0)g  (0) − 12g  (0)2 − 66g  (0) − 27g  (0)4 )
           ≤ −48(33 − 59ε4 − 66ε2 ) < 0,
70                    T. CRAVEN AND G. CSORDAS



since ε ≤ 0.61. The corresponding computation at x = 1 has 91
terms. Dropping all but one positive term, the one with g(1)4 > 1,
and estimating the negative terms using g (k) (1) < 2εk , yields

  68719476736(19001961 − 49152ε12 − 700416ε10 − 1824768ε9
              − 3843584ε8 − 29601792ε7 − 46972928ε6 − 68073984ε5
              − 299323392ε4 − 276908544ε3 − 406657152ε2 )
            > 0,
since ε ≤ 0.2.
  Of course, this same technique can be used with any polynomial
counterexample, though the precise determination of ε can be diﬃcult.

  Remark 3.5. (Polynomials of arbitrarily high degree.)We recall that
the Jensen polynomials, gn (x), associated with φ(x) = ∞              k
                                                              k=0 γk x /k! ∈
    +
L-P , are deﬁned by
                        n  
                              n
(3.2)          gn (x) =           γk xk , n = 0, 1, 2 . . . .
                              k
                          k=0

It is known, see for example [19, p. 40], that gn (x) ∈ L-P + , n =
0, 1, 2 . . . , and that the sequence {gn (x/n)} converges uniformly to
φ(x) on compact subsets of C. Now consider the Jensen polynomials
associated with the function φ(x) deﬁned in Example 3.4. Then
elementary continuity considerations show that, for all n suﬃciently
large, the 4 × 4 Wronskian of gn (x/n) changes sign on the interval [0, 1]
and thus provide additional counterexamples to (2.1) and (2.2).

   Example 3.6. Among the very nicest functions in the L-P + class
are those which arise
                    ∞from an     old theorem of Laguerre, cf. [7]. These
                               k
are the functions     k=0 γk x   /k! for which the coeﬃcients γk can be
interpolated by a polynomial with only real nonpositive zeros. These
also fail 
          the Wronskian condition in general. As an example, consider
            ∞    3        5 k
f (x) =     k=0 k (k + a) x /k!. The 5 × 5 Wronskian at x = 0 is a
polynomial in a which is negative for a ≥ 163.

 4. Positive results. It was noted in Section 1 that there are
many known classes of functions which satisfy Descartes’ Rule of Signs
           KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                          71


(1.2) and whose Wronskian determinants enjoy the sign regularity
condition of (2.2). Nevertheless, today the problem of characterization
of functions in L-P + which satisfy (1.2) or (2.2) remains open. Here
our primary goal is to construct a class of functions in L-P + which
satisfy both (2.1) and (2.2), without the degree restriction when the
function has inﬁnitely many zeros.
  To begin with, we remark that Karlin’s conjecture (1.7) is valid for
functions in L-P + which have only one distinct positive zero, as the
following example shows.

  Example 4.1. Let f (x) = (x + a)r , where r is an arbitrary positive
integer and a > 0. Factoring common factors out of the rows, we obtain

  W (f (x), f  (x), . . . , f (m−1) (x))
                           m−2
 = (x+a)mr−m(m−1)                 (r − k)m−1−k
                            k=0
                                                                            
     1        r             r(r−1)           ···     r(r−1) · · · (r−m+2) 
                                                                            
     1       r−1          (r−1)(r−2)         ···   (r−1)(r−2) · · · (r−m+1) 
    ·                                                                       ,
                                             ···                            
                                                                            
       1   r−m+1          (r−m+1)(r−m)        ···   (r−m+1) · · · (r−2m+3)

                                                                          m−1
in which the ﬁnal determinant has the value (−1)m(m−1)/2 k=1 k!
by [18, p. 106] (in which d = −1). Moreover by (3.1), the function
g(x) := ebx f (x), where we assume that b ≥ 0, also satisﬁes the sign
regularity condition (2.2).

  We introduce next the following class of entire functions in L-P + ,
(4.1)
              ∞                                                    
                 γk k            ∞
                               +            k        +
C0 := f (x) =         x ∈ L-P         γk+p x ∈ L-P for all p ≥ 0 ,
                   k!
                    k=0                     k=0


and proceed to derive several properties of this class of functions. In
the ﬁrst place,
             the   class C0 is closed under diﬀerentiation. To see this,
                ∞       k
let f (x) :=    k=0 γk x /k! ∈ C0 .  Then since L-P + is closed under
                                       ∞        k           +
diﬀerentiation [21, p. 119], f (x) =∞ k=0 γk+1 x k/k! ∈ L-P
                                                          +
                                                               . But then
by the deﬁnition of the class C0 , k=0 γk+1+p x ∈ L-P for all p ≥ 0
72                     T. CRAVEN AND G. CSORDAS



and so, a fortiori, f  (x) ∈ C0 . (Caveat. In general, the class      +
                                                               L-P is knot
closed under shift of indices in    the sense that if f (x) = ∞  k=0 αk x ∈
                                       ∞
L-P + , then the entire function k=0 αk+1 xk need not belong to L-P + ,
as the following simple
                       example shows.         The function ex ∈ L-P + , but
the entire function ∞     k=0 1/(k  + 1)!x k
                                             ∈
                                             / L-P +
                                                     ). Moreover,
                                                        ∞        C0 is closed
                                                                   k
under multiplication by x; that is, if f (x) :=               γ
                                                          k=0 k  x   /k! ∈ C0 ,
then xf (x) ∈ C0 . In order to verify this, it suﬃces to check that for
p = 0, 1, . . . ,
                             ∞
                             
                 Gp (x) :=       (k + p)γk+p−1 xk ∈ L-P + .
                             k=1
                                    
Since f (x) ∈ C0 , we have Fp (x) := ∞          k      +
                                      k=0 γk+p x ∈ L-P . But L-P is
                                                                   +
                                                  1−p    p+1
closed under diﬀerentiation and hence Gp (x) = x d/dx(x Fp (x)) ∈
L-P + . We remark also that if f (x) := xg(x) with f (x) ∈ C0 , then g
need not belong to C0 as f (x) = x(x2 /6 + x + 1) shows.
  To facilitate the exposition of the properties of functions in class C0 ,
it will be convenient to introduce the following deﬁnition.

  Definition 4.2. Let {γk }∞
                           k=0 be a sequence of real numbers. The
inequalities
(4.2)       γk2 ≥ αγk−1 γk+1 , where α ≥ 1,       k = 1, 2, 3, . . . ,
will be called Turán inequalities and the constant α will be referred to
as the Turán constant associated with the sequence {γk }∞ k=0 .



Proposition      4.3 (The order of a function inC0 .) Let f (x) =
  ∞       j                                           ∞         j
      γ
  j=0 j x   /j! ∈ C0 and for p ≥ 0, let Fp (x) =      j=0 γj+p x . Then
                   ∞
the sequence {γk }k=0 satisﬁes the Turán inequalities (4.2) with Turán
constant α ≥ 2. Moreover, both f (x) and Fp (x) are entire functions of
order zero.

  Proof. We ﬁrst suppose that γ0 > 0. Since f ∈ C0 and Fp ∈ L-P + , it
follows from Remark 3.5 that the Jensen polynomials associated with
Fp (x), p = 0, 1, 2, . . . ,
                            2  
                                2
(4.3)       g2,p (x) =             k!γk+p xk = γp + 2γp+1 x + 2γp+2 x2
                                k
                       k=0
          KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                           73


                                         2
have only real negative zeros. Hence, γp+1    ≥ 2γp γp+2 , p = 0, 1, 2, . . . .
Then a calculation shows that 0 ≤ γk ≤ γ0 /2k(k−1)/2 (γ1 /γ0 )k , see
[1, Lemma 4.2], where α2 should be replaced by α. But then these
estimates, together with the well-known formula which expresses the
order of an entire function in terms of its Taylor coeﬃcients, [3, p. 9,
formula (2.2.3)], imply that the order of f (x), as well as that of Fp (x),
is zero. If γ0 = γ1 = · · · = γm−1 = 0, but γm = 0, then we consider
the Jensen polynomials associated with f (p+m) (x). Since the rest of
the argument is, mutatis mutandis, the same as before, the proof of the
proposition is complete.

  The statement
              of the knext theorem            requires some additional notation.
Let f (x) = ∞   k=0 α k x   ∈  L-P     +
                                          , α0  = 0, and let A = (αi−j )∞
                                                                         i=1,j=1 ,
where αi−j = 0 if i − j < 0, denote the lower triangular matrix formed
from the αk ’s. Since f (x) ∈ L-P + , it is known [1] or [12, p. 412]
that the sequence {αk }∞   k=0 is a totally positive sequence and hence, cf.
Deﬁnition 1.1, all the minors of the matrix A are nonnegative. For a
positive integer n, let 1 ≤ i1 < i2 < · · · < in and 1 ≤ j1 < j2 < · · · < jn
be positive integers and let
                                                                   
(4.4)                 A i1 , i2 , . . . , in | j1 , j2 , . . . , jn

denote the n × n minor obtained from A by deleting all the rows and
columns except those labeled i1 , i2 , . . . , in and j1 , j2 , . . . , jn , respec-
tively. Using an induction argument, in conjunction with the Cauchy-
Binet formula [12, p. 1], Karlin proved a general result, involving cer-
tain meromorphic functions, which gives a necessary and suﬃcient con-
dition for the minors (4.4) to be strictly positive. In the sequel, we will
make use of the following special case of the result of Karlin.

  Theorem 4.4 [12, Theorem 10.1, p. 428]. Let
                                    ω
                                    
                          f (x) =         αk xk ∈ L-P + ,
                                    k=0

where 1 ≤ ω ≤ ∞ and αk > 0 for all k. Then, with the notation above,
the minor (4.4) is positive if and only if

(4.5)             ik − ω ≤ jk ≤ ik ,       k = 1, 2, . . . , n.
74                      T. CRAVEN AND G. CSORDAS



  Preliminaries aside, we next show that functions in C0 satisfy the sign
regularity condition (2.2), subject to certain restrictions to be made
precise below.

                              ω
  Theorem 4.5. Let f (x) = k=0 (γk /k!)xk ∈ C0 , where 1 ≤ ω ≤ ∞.
Let p be a nonnegative integer and let t0 ∈ R, t0 > 0. For each non-
negative integer m, either the Wronskian W (f (p) , f (p+1) , . . . , f (p+m) )
is identically zero or

(4.6)     (−1)m(m+1)/2 W (f (p) (t0 ), f (p+1) (t0 ), . . . , f (p+m) (t0 )) > 0.


                                                          
  Proof. First suppose that f (0) = 0. Since f (x) ∈ C0 , ∞           k
                                                             k=0 γk+p x ∈
    +                                  ∞
L-P for all p ≥ 0 and hence {γk+p }k=0 is a totally positive sequence.
Set αk = γk+p , k = 0, 1, 2, . . . , and let A = (αi−j )∞  i=1,j=1 , where
αi−j = 0 if i − j < 0, denote the lower triangular matrix formed from
the αk ’s. We will consider two cases. First suppose that ω = ∞. For
r = 1, 2, . . . , m, set

             Ar = (αi+j−2 )ri=1,j=1 = (f (p+i+j−2) (0))ri=1,j=1 .

By Remark 1.2 we already know that (−1)r(r−1)/2 det Ar ≥ 0. We
now claim that by Theorem 4.4, we have (−1)r(r−1)/2 det Ar > 0.
To see this, consider the lower triangular matrix A and note that for
r = 1, 2, . . . , m, the minor
                                                             
(4.7)             A r, r + 1, . . . , 2r − 1 | 1, 2, . . . , r > 0,

since the inequalities (4.5) are satisﬁed. Thus, after rearranging the
columns of the submatrix, whose minor is positive by (4.7), we obtain
that for all p = 0, 1, 2, . . . and r = 1, 2, . . . , m,

(4.8)          (−1)r(r−1)/2 det(f (p+i+j−2) (0))ri=1,j=1 > 0.

But then it follows from a theorem of Karlin and Loewner [13, Theorem
3], that for all p = 0, 1, 2, . . . , r = 1, 2, . . . , m and x > 0,

                (−1)r(r−1)/2 det(f (p+i+j−2) (x))ri=1,j=1 > 0
          KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                               75


and so in particular (4.6) holds. In the second case, when ω is ﬁnite, so
that f is a polynomial of degree ω, the argument is analogous to the ﬁrst
case except that now the validity of (4.6) is subject to the additional
constraint that m + p ≤ ω to avoid zero determinants caused by the
higher order derivatives being identically zero.
   To handle the case in which f (0) = 0, we proceed by induc-
ω on the multiplicity
tion
                      k
                                    of the zero at the origin. Set f (x) =
   k=0 (γk /k!)x          ∈ C0 with γ0 = 0 and assume that the theorem
holds for functions with           a zero k−1of lower
                                                      multiplicity          at the ori-
                                                                     k         +
gin.
        f  ∈    C 0   implies  x       γ
                                    k=1 k x     =      k=1 k  γ   x    ∈  L-P   , hence
                              k                       k−1              +
   k=0   (εγk+1    +   γk )x   = (x +  ε)       γ
                                             k=1 k  x       ∈    L-P     , and therefore
                                      k                                              
g(x) =          k=0 (εγk+1 + γk )x /k! ∈ C0 . Since g(x) = f (x) + εf (x)
has a zero of lower multiplicity at the origin, the Wronskians of g
have appropriate signs by the induction hypothesis.     m               Expanding the
Wronskian of g gives W (g, g  , . . . , g (m) ) =                      k
                                                            k=0 k , where D0 =
                                                                   D   ε
W (f, f  , . . . , f (m) ), D1 = d/dxW (f (x), f  (x), . . . , f (m) (x)), and in gen-
eral, Dk is the (m + 1) × (m + 1) determinant in which the column
headed by f (m+1−k) is deleted among columns headed by f, . . . , f (m+1) .
If the Wronskian of g is identically zero for all suﬃciently small ε,
then so is the Wronskian of f . Otherwise, for suﬃciently small ε,
(−1)m(m+1)/2 W (g(x), g  (x), . . . , g (m) (x)) > 0 for every x > 0, whence
(−1)m(m+1)/2 W (f (x), f  (x), . . . , f (m) (x)) ≥ 0. Assume the Wronskian
of f equals zero at some positive x0 . Then x0 must be a multiple root
of the Wronskian, so D0 = D1 = 0. We now have the ﬁrst m + 1
of m + 2 column vectors being linearly dependent, and also the linear
dependence of all columns but the one headed by f (m) , from D1 = 0.
It easily follows that all sets of m + 1 column vectors are linearly de-
pendent, that is, that all Dk are zero at x0 , which contradicts the sum
being nonzero. Therefore the Wronskian of f , when not identically
zero, is never zero for any positive value of x. In the case of derivatives
f (p) , p ≥ 1, the conclusion is immediate from the induction hypothesis
because the zero of f (p) at the origin has lower multiplicity.

  We next show that functions in C0 satisfy Descartes’ Rule of Signs.

                                        m         (k)
                                                           
  Theorem 4.6. Let f ∈ C0 . Then Z+         k=0 ak f (x) ≤ V ({ak })
                                                        m
for any sequence {ak }m
                      k=0 of real numbers such that     j=0 aj f
                                                                 (j)
                                                                     = 0.
76                      T. CRAVEN AND G. CSORDAS



   Proof. Let f ∈ C0 and ﬁrst suppose that f is a polynomial of degree
n. Then by Theorem 4.5 the sign regularity condition (4.6) holds
when m ≤ n. Since C0 is closed under diﬀerentiation, Theorem 2.3
implies that f satisﬁes Descartes’ Rule of Signs. If m > n, then the
assertion clearly remains valid. Next suppose that f is a transcendental
entire function in C0 . By the Pólya-Szegö criterion, see Lemma 2.2,
a suﬃcient condition for the validity of Descartes’ Rule of Signs for
f (x), f  (x), . . . , f (m) (x) is that for any integers 0 ≤ i1 < i2 < · · · <
ik ≤ m, all Wronskian determinants W (f (i1 ) , f (i2 ) , . . . , f (ik ) ) of order
k be nonzero on (0, ∞) and have the same sign. Since f (j) (x) > 0
for x > 0, an easy induction argument in conjunction with Lemma
2.1 shows that the suﬃcient condition above remains valid if it is only
assumed that the indices 0 ≤ i1 < i2 < · · · < ik ≤ m are consecutive
integers; i2 = i1 + 1, i3 = i2 + 1, . . . , ik = ik−1 + 1, cf. [24, p. 555]. Since
by Theorem 4.5 the Wronskian determinants are nonzero on (0, ∞), it
follows that Descartes’ Rule of Signs holds in this case as well.

  If f (x) ∈ L-P + , then it is clear that for a > 0, b ≥ 0, f (ax + b) is
also in L-P + . While the invariance of C0 under the dilation x → ax,
a > 0, is evident, Example 4.7 below shows that the class C0 is not
closed under the translation x → x + b for b ≥ 0.

                                            ∞         k
   Example 4.7. Let f0 (x) =                  k=0 γk x /k!, where γk > 0 and
                                           ∞
suppose that the sequence {γk }k=0 satisﬁes the Turán inequalities
γk2 ≥ αγk−1 γk+1 , k = 1, 2, 3,     . . . , with kTurán constant α ≥ 4. For
p = 0, 1, 2, . . . , set Fp (x) = ∞     k=0 γk+p x . Then the method used to
prove Proposition 4.3 shows that f0 (x) and Fp (x) are entire functions
of order zero. Moreover, a result of Hutchinson [11, p. 327], see also
[7, Section 4], implies that Fp ∈ L-P + . Thus f0 ∈ C0 . We can also
use Hutchinson’s result   nto show        that, for any positive integer n, the
                                     k    k2
polynomial pn (x) =           k=0 x    /(2   k!) is in C0 . But it is easy to see
                                              2              k  k2
that, for t0 > 0 suﬃciently large,               k=0 (x + t0 ) /2   / L-P + and
                                                                    ∈
consequently p2 (x + t0 ) ∈    / C0 .

  We note that both Descartes’ Rule of Signs and the sign regularity
of Hankel determinants, when they hold for a given function f (x), will
also hold for f (x + a) for any a ≥ 0. This observation, in conjunction
with Theorem 4.6 and (3.1), yields the following immediate corollary.
         KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                               77


  Corollary 4.8. Let
                                                                     
                 dn bx          
     C1 =                       
                    e f (x + a)  f ∈ C0 , a, b ≥ 0, n = 0, 1, 2, . . . .
                dxn
                     m         (k)
                                        
If f ∈ C1 , then Z+     k=0 ak f     (x) ≤ V ({ak }m     k=0 ) for any sequence
                                       m
{ak }m
     k=0 of  real numbers such  that     j=0 a j f (j)
                                                       
                                                       =   0.

                                  ∞
  Corollary 4.9. Let f (x) = k=0 γk xk /k! ∈ C0 be a transcenden-
tal entire function and let gn∗ (t) be the nth Appell polynomial associ-
ated with f (x); that is, the set {gn∗ (t)}∞

                                                                xt
                                           n=0 is generated by e f (x) =
   ∞    ∗     n
   n=0 gn (t)x /n!. Then, for t ≥ 0, n = 1, 2, . . . ,

                           ∗                                        
                           g0 (t)     g1∗ (t)   ···     ∗
                                                       gn−1     (t) 
                           ∗                                        
                           g1 (t)     g2∗ (t)   ···       ∗
                                                         gn (t) 
            (−1)n(n−1)/2     ..         ..                 ..      >0
                                                                     
                               .          .                  .      
                           ∗                                        
                            gn−1 (t)   gn∗ (t)   ···    ∗
                                                       g2n−2     (t)

or the determinant is identically zero.

  Proof. Deﬁne the matrix
                                  ⎛                                           ⎞
                                            1         0          ...       0
                       n
                i − 1 i−j         ⎜         t         1          ...       0⎟
      B=              t          =⎜
                                  ⎝         ..         ..                  .. ⎟ .
                j−1        i,j=1             .    n−1. n−2                .⎠
                                         tn−1       1 t          ...       1

Then we obtain
 ∗                                       
 g0 (t)    g1∗ (t) · · · gn−1∗
                                     (t) 
 ∗           ∗                 ∗         
 g1 (t)    g2 (t) · · ·      gn (t) 
                                         
    ..         ..                ..      
     .          .                 .      
 ∗          ∗               ∗            
  gn−1 (t) gn (t) · · · g2n−2 (t)
        ⎛ ⎛                                 ⎞ ⎞                                     
                  γ0  γ1 · · · γn−1                  γ0       γ1      ···     γn−1 
                                                                                    
        ⎜ ⎜ 1     γ   γ   2 · · ·       γ n ⎟ T⎟  γ1          γ2      ···      γn 
        ⎜  ⎜
= det ⎝B ⎝ ..           ..               .. ⎠B ⎠ =  ..
                                            ⎟  ⎟                ..               ..  ,
                   .     .                .          .          .                . 
                                                                                    
              γn−1 γn · · · γ2n−2                     γn−1     γn      ···     γ2n−2
78                    T. CRAVEN AND G. CSORDAS



where we note, in particular, that the result is independent of t. Since
f (x) is a transcendental entire function in C0 , it follows from the proof
of Theorem 4.5, see inequality (4.8), that
(−1)n(n−1)/2 det(f (i+j−2) (0))ni=1,j=1 = (−1)n(n−1)/2 det(γi+j−2 )ni=1,j=1
                                        ≥ 0,
proving the corollary.

  How special is the Turán constant α in Example 4.7? Hutchinson
[11] shows that α must be at least 2 and also indicates that 4 is not
necessary. For example, Hutchinson’s comments [11, p. 331] lead one
to note that 1 + 13.5x + 54x2 /2! + 54x3 /3! ∈ C0 , though the Turán
inequalities involve ratios 4 and 3.375. We examine this question more
carefully, in the next example, in the case of transcendental entire
functions.

                                  
  Example 4.10. Let f (x) = ∞             k    k2
                                     k=0 x /a . We show f (x) ∈ L-P
                                                                          +

for a ≥ 1.85. This will be accomplished by showing that the Jensen
polynomials
                 n  
                       n k! k
        gn (x) =               x
                       k ak 2
(4.9)            k=0
                      nx nx (n−1)x                nx (n−1)x           x
               =1+        +              + ··· +               · · · 2n−1
                       a     a     a3              a    a3          a
have only real zeros for each n = 1, 2, . . . . We use a reﬁnement of a
technique in [23, V, #176]. Since, a/n < a3 /(n − 1) < · · · , we can
apply the argument of [23, I, #117] to conclude that for any given x,
the absolute values of the terms of gn (x) monotonically increase from
the ﬁrst term 1 to the maximum term and then monotonically decrease
thereafter (allowing, of course, for the case that the initial or ﬁnal term
is the maximal one). Fix n ≥ 2 and x > 0. Assume that the kth term
is the maximal one for some k, 1 ≤ k ≤ n − 1, and consider
(4.10) (−1)k gn (−x)
                n! xk           n! xk−1             n! xk+1
          ≥             2 −                 2 −
              (n−k)! ak     (n−k+1)! a(k−1)     (n−k−1)! a(k+1)2
                n(n−1) · · · (n−k−1)xk+2   n(n−1) · · · (n−k−2)xk+3
            +                   2        −                          .
                         a(k+2)                     a(k+3)2
           KARLIN’S CONJECTURE AND A QUESTION OF PÓLYA                     79


From (4.9), we can see that the kth term is dominant precisely when
a2k−1 /(n − k + 1) < x < a2k+1 /(n − k), see [23, I, #117]. Accordingly,
we evaluate the right-hand side of (4.10) at x = a2k /(n − k), obtaining

                         2
                 a2k     ak −9 n(n − 1) · · · (n − k + 2)
     (−1)k gn −        ≥                                  G(n, k),
                n−k               (n − k)k−1

where
(4.11)
                                        
                 n−k+1             n−k+1
 G(n, k) := a9            − a8 1 +
                    n−k             n−k
                   (n−k+1)(n−k−1) (n−k+1)(n−k−1)(n−k−2)
              + a5                 −                    .
                       (n − k)2            (n − k)3


We claim that G(n, k) is positive for a ≥ 1.85. If we can establish that,
then we shall have

                                                      
                            a2                      a4
     gn (0) > 0,    gn −             < 0,   gn −             > 0, . . . ,
                           n−1                     n−2
     (−1)n−1 gn (−a2(n−1) ) > 0,      (−1)n gn (x) > 0 as x → −∞,


demonstrating that gn (x) has n negative real zeros.
  To show that G(n, k) is positive, we ﬁrst note that it increases with
k. Indeed, we have

                                                                     
         ∂               a8                   a5                4
            G(n, k) =          (a − 1.5) +              a3 −
         ∂k           (n − k)2             2(n − k)2           n−k
                                  2
                        2((n − k) + n − k − 3)
                      +                         .
                                (n − k)4

For a > 1.6 this is always positive, with the only diﬃculty being
when k = n − 1 makes the last term negative; in this case one has
a9 − a8 − 2a5 − 2 whose largest real root is less than 1.6. Therefore we
80                       T. CRAVEN AND G. CSORDAS



need only deal with k = 1 in (4.11). Since n ≥ 2, we have
(4.12)
             n        2n−1 8 n(n−2) 5 n(n−2)(n−3)
G(n, 1) =        a9 −       a +           a −
           n−1         n−1       (n−1)2             (n−1)3
                               1         2      1−a5                    
         = a9 − 2a8 + a5 − 1+       −          +         + 2 − a 8
                                                                   − a 9
                              n−1      (n−1)2     n−1
                                 1
         ≥ a9 − 2a8 + a5 − 1 +       (−2 + (1 − a5 ) + 2 − a8 + a9 )
                               n−1
         > a9 − 2a8 + a5 − 1 if a ≥ 1.33,

since the coeﬃcient of 1/(n − 1) is positive for a ≥ 1.33. The last
polynomial in (4.12), a9 − 2a8 + a5 − 1, has only one real root, that
being just under 1.85.
  Will smaller values of a suﬃce? A computation shows that if a = 1.7,
then g9 (x) has only 7 real zeros. The exact threshold value of a is
unknown.

  Example 4.11. From Example 4.10, we know that if a ≥ 1.85, then
the function                              k
               ∞                ∞
                     xk       1  1      x
                           =
                   a(k+p)2   ap2   ak2 a2p
                      k=0                  k=0
            +
is in L-P       for all p ≥ 0. Therefore
                         ∞
                           xk
                                  ∈ C0      for a ≥ 1.85
                           a 2 k!
                            k
                         k=0

and, in particular, a2 is somewhat less than the Turán constant α = 4
of Example 4.7. This provides new examples of functions for which
(Dp ) and (Hp ) both hold.


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82                      T. CRAVEN AND G. CSORDAS



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Department of Mathematics, University of Hawaii, Honolulu, HI 96822
E-mail address: tom@math.hawaii.edu

Department of Mathematics, University of Hawaii, Honolulu, HI 96822
E-mail address: george@math.hawaii.edu
