A NOTE ON RATIONAL OPERATOR MONOTONE FUNCTIONS. Masaru Nagisa. Received May 19, 2014 ; revised April 10, (Ax, x) 0 for all x C n.

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1 Scientiae Mathematicae Japonicae Online, e-014, A NOTE ON RATIONAL OPERATOR MONOTONE FUNCTIONS Masaru Nagisa Received May 19, 014 ; revised April 10, 014 Abstract. Let f be oeprator monotone for some open interval I of R. It is known that f has the analytic continuation on H + I H, where H + (resp. H ) is the upper (resp. the lower) half plane of C. In this note, we determine the form of rational operator monotone functions by using elementary argument, and prove the operator monotonicity of some meromorphic functions. 1 Introduction. We denote the set of all n n matrices over C by M n and set H n = {A M n A = A} and H + n = {A H n A 0}, where A 0 means that A is non-negative, that is, the value of inner product (Ax, x) 0 for all x C n. Let I be an open interval of the set R of real numbers. We also denote by H n (I) the set of A H n with its spectra Sp(A) I. A real continuous function f defined on the open interval I is said to be operator monotone if A B implies f(a) f(b) for any n N and A, B H n (I). In this note, we assume that an operator monotone function is not a constant function. Let f be a real-valued continuous function on the interval I. We call f a Pick function if f has an analytic continuation on the upper half plane H + = {z C Imz > 0} into itself. It also has an analytic continuation to the lower half plane H, obtained by reflections across I. We denote by P(I) the set of all Pick functions on I. It is well known that f P(I) is equivalent to that f is operator monotone on I ([1], [4], [5]). We characterize the rational Pick function (rational operator monotone function) by an elementary method in Section and give some examples using this characterization in Section 3. Rational operator monotone functions. Let I be an open interval and f(t) = at + b (a, b, c, d R, ad bc 0). It is well known that f is operator monotone on ct + d (, d c ) or ( d, + ) (see [1], [5]). So the following rational function is also operator c monotone on I: a i b 0 + a 0 t, where b 0 R, a 0, a 1,..., a n 0 and α 1, α,..., α n R \ I. 000 Mathematics Subject Classification. 47A63. Key words and phrases. Monotone matrix function, Operator monotone function, Pick function, Rational operator monotone function.

2 146 Let g P(I) be rational. Then there exists polynomials p(t) and q(t) with real coefficients such that g(t) = p(t) (t I), q(t) where common devisors of p(t) and q(t) are only scalars and a coefficient of the highest degree term of q(t) is 1. The polynomial q(t) with real coefficients is represented as products of the following factors: t a, t + at + b (a, b R). Since g has the analytic continuation to the upper half plane H + and the lower half plane H, g(z) = p(z) (z H + I H + ) q(z) and g has no poles on H + I. So we may assume that g(z) has the following form: g(z) = p(z) (z c 1 ) n(1) (z c ) n() (z c k ) n(k), where c 1, c,..., c k R I c and each n(i) (i = 1,,..., k) is a positive integer with n(1) + n() + + n(k) = deg q(z). By the partially fractional decomposition of g(z), g(z) = r(z) + n(i) k j=1 (z c i ) j, where r(z) is the remainder of p(z) by q(z) and {b i.j } R. Lemma.1. In above setting, g P(I) satisfies the following conditions: (1) There exist r 0, r 1 R such that r 1 0 and r(z) = r 0 + r 1 z. () n(i) = 1 and b i,1 0 for all i = 1,,..., k. Proof. (1) We set where d = deg r(z). Put r(z) = r 0 + r 1 z + + r d z d, θ = { 3π d if d and r d > 0 π d if d 1 and r d < 0. For a sufficiently large R > 0 and z = Re θ 1 H +, we may assume that Then we have d 1 r d R d = r d z d > r i z i + i=0 n(i) k j=1 Img(z) = Im( r d R d d r i z i + i=0 d 1 r d R d + r i z i + i=0 n(i) k j=1 (z c i ) j. n(i) k j=1 (z c i ) j ) (z c i ) j < 0.

3 147 This contradicts to g(z) H +. So we have that r(z) = r 0 + r 1 z and r 1 0. () In a suitable neighborhood of c i in H + I H, g P(I) has the form g(z) = b i,1 + + b i,n(i) + h(z), z c i (z c i ) n(i) where h(z) is holomorphic on the neighborhood of c i. Put π if n(i) and b i,n(i) > 0 n(i) θ = 3π if n(i) 1 and b i,n(i) < 0. n(i) For a sufficiently small r > 0, z = c i + re θ 1 H + and we may assume that Then we have b i,n(i) b i,n(i) = r n(i) (z c i ) > n(i) n(i) 1 j=1 (z c i ) j + h(z). Img(z) = Im( b n(i) 1 i,n(i) 1 + r n(i) (z c j=1 i ) j + h(z)) b n(i) 1 i,n(i) + + h(z) < 0. r n(i) (z c j=1 i ) j This contradicts to g(z) H +. So we have that n(i) = 1 and b i,1 0 for all i = 1,,..., k. We can now prove the following theorem: Theorem.. The following are equivalent: (1) f P(I) is rational. () There exist b 0 R, non-negative numbers a 0, a 1,..., a n and real numbers α 1, α,..., α n / I such that a i f(t) = b 0 + a 0 t. (3) There exist a 0, c 0, b 0 R, α 1, α,..., α n / I and β 1, β,..., β n 1 R satisfying that Proof. (1) () This is proved by Lemma.1. () (3) We assume f(t) = b 0 + a 0 t c(t β 1)(t β ) (t β n 1 ) (t α 1 )(t α ) (t α n ) and α 1 < β 1 < α < β < < β n 1 < α n. f(t) = b 0 + a 0 t a i,

4 148 where b 0 R, a 0, a 1,..., a n 0, α 1, α,..., α n g(t) as follows: / I and α 1 < α < < α n. We define a i g(t) = (t α 1 ) (t α n ), that is, Since we have g(t) = a i (t α 1 ) ( 1 )(+1 ) (t α n ). g(α i ) = (α i α 1 ) (α i α i 1 )(α i α i+1 ) (α i α n ), sign g(α i ) = ( 1) n i (i = 1,,..., n). By the fact deg g(t) = n 1 and the continuity of g, there exist a positive number c and β 1, β,..., β n 1 such that and α 1 < β 1 < α < β < < β n 1 < α n. (3) () Set g(t) = c(t β 1 )(t β ) (t β n 1 ) g(t) = c(t β 1)(t β ) (t β n 1 ) (t α 1 )(t α ) (t α n ), where c > 0, α 1 < β 1 < α < β < < β n 1 < α n. Then g(t) has the following form: g(t) = b i, for some b i R \ {0}. It suffices to show that b i > 0 for i = 1,,..., n 1. When we choose t such that β i 1 < t < α i and α i t is sufficiently small, we have sign g(t) = sign b i. Because α 1 < < β i 1 < t < α i < < α n, So we have b i > 0. sign g(t) = ( 1) (n 1) (i 1)+n (i 1) = 1. For a rational function f(t), we can choose polynomials p(t) and q(t) such that f(t) = p(t) q(t) and common devisors of p(t) and q(t) are only scalars. Then we call f of order n if n = max{deg p(t), deg q(t)}.

5 149 Corollary.3. The followings are equivalent: (1) f P(I) is rational of order n. () f has one of the following forms: or where a > 0, α i / I and Proof. (1) () When f(t) has the form (a) f(t) = a(t β )(t β 3 ) (t β n+1 ), (t α 1 )(t α ) (t α n ) (b) f(t) = a(t β 1)(t β ) (t β n ) (t α 1 )(t α ) (t α n 1 ), (c) f(t) = a(t β 1)(t β ) (t β n ) (t α 1 )(t α ) (t α n ) (d) f(t) = a(t β )(t β 3 ) (t β n ) (t α 1 )(t α ) (t α n ), β 1 < α 1 < β < α < < α n < β n+1. n 1 a i f(t) = b 0 + a 0 t ( ), where a 1, a,..., a n 1 > 0. Since f is rational of order n, we have a 0 > 0. We set that is, Then we have g(t) = (b 0 + a 0 t)(t α 1 )(t α ) (t α n 1 ) a i (t α 1 ) ( 1 )(+1 ) (t α n 1 ), n 1 So f has the form (b). When f(t) has the form f(t) = g(t) (t α 1 )(t α ) (t α n 1 ). sign( lim t g(t)) = 1, sign g(α n 1 ) = 1, sign g(α n ) = 1,, sign g(α 1 ) = ( 1) n 1, sign( lim t g(t)) = ( 1)n. f(t) = b 0 + a 0 t a i ( ), where a 1, a,..., a n > 0. Since f is rational of order n, we have a 0 = 0. We set g(t) = b 0 (t α 1 )(t α ) (t α n ) a i (t α 1 ) ( 1 )(+1 ) (t α n ),

6 150 that is, f(t) = g(t) (t α 1 )(t α ) (t α n ). Using the same argument as above, f has the form (d) if b = 0, the form (a) if b > 0 and the form (c) if b < 0. () (1) When f has the form (a),(b),(c) or (d), f is rational of order n. When f has the form (d), f P(I) by Theorem.. When f has the form (a), f is represented as the following form: f(t) = b i + a, where a > 0 and some b i R (i = 1,,..., n). Since we get b i < 0 from the fact lim f(t) = lim a(t β )(t β 3 ) (t β n+1 ) =, t α i +0 t α i +0 (t α 1 )(t α ) (t α n ) lim t α i +0 b i + a =. So f P(I). By the similar reason, f P(I) if f has the form (b) or (c). 3 Examples. The following Example 3.1 has been announced by M. Uchiyama in many Conferences (cf. [7], [8]). Example 3.1. Let {p n (x)} be the orthogonal polynomials on a closed interval [a, b] whose leading coefficient is positive. It is well known that the zeros {c 1, c,..., c n } of p n (x) satisfies that a = c 0 < c 1 < c < c n < c n+1 = b, and each interval (c i, c i+1 ) (i = 0, 1,..., n) contains exactly one zeros of p n+1 (x) ([6]). So p n+1 (x)/p n (x) has the form (b) in Corollary.3. This means that p n+1 (x)/p n (x) is operator monotone on any interval which does not contain any zeros of p n (x). Example 3.. Let 0 = a 0 < a 1 < a < < a n 1 < a n = π. Then f(x) = cos(x a 1) cos(x a 3 ) cos(x a n 1 ) cos(x a 0 ) cos(x a ) cos(x a n ) (m + 1)π is operator monotone on any interval I contained in R \ { 0, 1,..., n 1}. + a i m Z, i = In particular, tan x is operator monotone on any interval contained in R \ {mπ π m Z} (when n = 1, a 0 = 0, a 1 = π ).

7 151 Proof. The function cos x is represented by the infinite product as follows: where Remarking the fact we have that = f m (x) = cos x = lim m f m(x), m 1 k= m (1 f m (x) = ( 1)m 4m ((m 1)!) ((m 1)!) x (k + 1)π ). m 1 k= m g m (x) = f m(x a 1 )f m (x a 3 ) f m (x a n 1 ) f m (x a 0 )f m (x a ) f m (x a n ) m 1 k= m (x k + 1 π), (x ( (k+1)π + a 1 ))(x ( (k+1)π + a 3 )) (x ( (k+1)π + a n 1 )) (x ( (k+1)π + a 0 ))(x ( (k+1)π + a )) (x ( (k+1)π + a n )) belongs to P(I) by Corollary.3. Since f(x) is operator monotone on I. f(x) = lim m g m(x), Example 3.3. Let a 0 < a 1 < a < < a n 1 < a and k(1), k(),..., k(n) Z. Then f(x) = Γ(x a 0 k(1))γ(x a k()) Γ(x a n k(n)) Γ(x a 1 k(1))γ(x a 3 k()) Γ(x a n 1 k(n)) is operator monotone on any interval I contained in R\{a i 1 +k(i) m i = 1,,..., n, m = 0, 1,,...}, where Γ(x) is the Gamma function, i.e., Γ(x) = Proof. We use Gauss s Formula of Γ(x) as follows: 0 e t t x 1 dt (x > 0). Γ(x) = lim m g m(x), m where g m (x) = x m! x(x+1) (x+m) and the convergence is uniformly on any compact subset of R \ {0, 1,,...} ([3]). For a < b < a + 1, g m (x a) (x b)(x (b 1)) (x (b m)) = mb a g m (x b) (x a)(x (a 1)) (x (a m)) is operator monotone on any interval contained in R \ {a, a 1,..., a m} by Corollary.4. Then we have that h m (x) = g m(x a 0 k(1))g m (x a k()) g m (x a n k(n)) g m (x a 1 k(1))g m (x a 3 k()) g m (x a n 1 k(n)) also has the form (a) in Corollary.3, and is operator monotone on I. So is f(x), because f(x) = lim m h m(x).

8 15 Acknowledgement. This work was partially supported by Grant-in-Aid for Scientific Research (C)5400. References [1] R. Bhatia, Matrix Analysis, Graduate Texts in Math. 169, Springer-Verlag, [] R. Bhatia, Positive Definite Matrices, Princeton University Press, 007. [3] J. B. Conway. Functions of One Complex Variable, Second Edition, Graduate Texts in Math. 11, Springer-Verlag, [4] W. F. Donoghue, Jr., Monotone matrix functions and analytic continuation, Springer-Verlag, [5] F. Hiai, Matrix Analysis: Matrix monotone functions, matrix means, and majorization, Interdecip. Inform. Sci. 16 (010) [6] G. Szegö, Orthogonal polynomials, Amer. Math. Soc. Colloquium Publ. Vol. 3 Revised ed., [7] M. Uchiyama, Inverse functions of polynomials and orthogonal polynomials as operator monotone functions, Trans. Amer. Math. Soc. 355(003), [8] M. Uchiyama, Operator monotone functions, Jacobi operators and orthogonal polynomials, J. Math. Anal. Appl. 401(013), Communicated by Hiroyuki Osaka Department of Mathematics and Informatics, Graduate School of Science, Chiba University, Chiba, 63-85, Japan. nagisa@math.s.chiba-u.ac.jp