Operator Monotone Functions which Are Defined Implicitly and Operator Inequalities

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1 Joural of Fuctioal Aalysis 175, (2000) doi: jfa , available olie at o Operator Mootoe Fuctios which Are Defied Implicitly ad Operator Iequalities Mitsuru Uchiyama Departmet of Mathematics, Fukuoka Uiversity of Educatio, Muakata, Fukuoka, Japa uchiyamafukuokaedu.ac.jp Received November 5, 1999; accepted March 21, 2000 dedicated to professor t. furuta The fuctio t : (0<:<1) is operator mootoe o 0t<. This is kow as the Lo werheiz iequality. However, ot too may examples of cocrete operator mootoe fuctios are kow so far. We will systematically seek operator mootoe fuctios which are defied implicitly. This ivestigatio is ew, ad our method seems to be powerful. We will actually fid a family of operator mootoe fuctios which icludes t : (0<:<1). Moreover, by costructig oeparameter families of operator mootoe fuctios, we will get may operator iequalities; especially, we will exted the Furuta iequality ad the expoetial iequality of Ado Academic Press Key Words: operator mootoe fuctio; Pick fuctio; Lo wer theory; Lo wer Heiz iequality; Furuta iequality. 1. INTRODUCTION Throughout this paper, A ad B stad for bouded selfadjoit operators o a Hilbert space ad sp(x) for the spectrum of a operator X. A real valued fuctio f(t) is called a operator mootoe fuctio o a iterval I i R 1 if, for A, B with sp(a), sp(b)/i, AB implies f(a) f(b). Clearly a composite fuctio of operator mootoe fuctios is operator mootoe too, provided it is well defied. A holomorphic fuctio which maps the ope upper half plae 6 + ito itself is called a Pick fuctio. By Lo wer's theorem [13], f(t) is a operator mootoe fuctio o a ope iterval (a, b) if ad oly if f(t) has a aalytic cotiuatio f(z) to 6 + _ (a, b) so that f(z) is a Pick fuctio; therefore f(t) is aalytically exteded to the ope lower half plae by reflectio. Thus if f(t)0 ad Copyright 2000 by Academic Press All rights of reproductio i ay form reserved.
2 OPERATOR MONOTONE IMPLICIT FUNCTION 331 g(t)0 are operator mootoe, the so is f(t) + g(t) * for 0+, *1, ++*1. If f(t) is operator mootoe o (a, b) ad if f(t) is cotiuous o [a, b), the f(t) is operator mootoe o [a, b). It is kow that t : (0<:1), log(1+t), ad t t+* (*>0) are operator mootoe o [0, ). Thus, AB0 implies A : B : for 0<:<1, (1) which is called the Lo werheiz iequality [12, 13]. But AB0 does ot geerally imply A 2 B 2. We have show that if A, B0 ad (A+tB ) 2 A 2 for every t>0 ad =1, 2,..., the AB=BA [16].See[1,3,5,9,11,14] for details about operator mootoe fuctios. ChaKwog [4] posed the followig questio: Does AB0 imply (BA 2 B) 12 B 2? Furuta [7, 8] aswered it affirmatively as follows: AB0 implies { (Br2 A p B r2 ) 1q (B r2 B p B r2 ) 1q, (A r2 A p A r2 ) 1q (A r2 B p A r2 ) 1q, (2) where r, p0 ad q1 with (1+r) qp+r. This is called the Furuta iequality. I this iequality, the case p1 is a deformatio of the Lo wer Heiz iequality; further, the case (1+r) q>p+r follows from the case (1+r) q= p+r by the Lo werheiz iequality agai. So the essetially importat part of (2) is the case p>1 ad (1+r) q= p+r. Oe obtais the secod iequality of (2) from the first oe by takig iverses. Taahashi [15] showed that the expoetial coditio (1+r) qp+r is the best possible coditio for (2). Ado [2] obtaied the related iequality: for t>0, AB implies { (e(t2)b e ta e (t2)b ) 12 e tb, e ta (e (t2)a e tb e (t2)a ) 12. This was improved by use of the iequality itself ad (2), by Fujii ad Kamei [6] as follows: for p0, rs0, AB implies { (e(r2)b e pa e (r2)b ) s(r+ p) e sb, e sa (e (r2)a e pb e (r2)a ) s(r+ p). (3) It is evidet that the essetially importat part of this iequality is the case s=r. Recetly, by makig use of oly (2), we [18] got a simple proof of (3). Now we give a simple example that motivated us to ivestigate operator mootoe fuctios which are defied implicitly, A, B0 ad A 2 B 2 imply (A+1) 2 (B+1) 2,
3 332 MITSURU UCHIYAMA because AB follows from A 2 B 2. But we ca easily costruct 2_2 matrices A, B such that (A+1) 2 (B+1) 2,butA 2 3 B 2 ; for example, A= \ , B= \ The above results mea that,(t)=(t 12 +1) 2 is operator mootoe o [0, ), but (t)=(t 12 &1) 2 is ot o [1, ). We may say that, ad are implicitly defied by,(t 2 )=(t+1) 2 (t0) ad ((t+1) 2 )=t 2 (t0). Oe aim of this paper is to seek operator mootoe fuctios which are defied implicitly; this ivestigatio is ew, ad we will actually fid a family of operator mootoe fuctios which icludes t : (0<:<1). This meas that we ca get ot merely a extesio of (1) but also aother proof of (1). The other aim is to exted simultaeously (2) ad (3), by makig use of a oeparameter family of operator mootoe fuctios. 2. THE CONSTRUCTION OF NEW OPERATOR MONOTONE FUNCTIONS Let us defie a oegative icreasig fuctio u(t) o[&a 1, ) by k u(t)= ` i=1 (t+a i ) # i (a 1 <a 2 <}}}<a k,1# 1,0<# i ). (4) Theorem 2.1. Let the fuctio s=u(t) be defied by (4). The the iverse fuctio u &1 (s) is operator mootoe o [0, ). Proof. Sice u &1 (s) is cotiuous o [0, ), we have to show that u &1 (s) is operator mootoe o (0, ). We may assume that a 1 =0; for, settig v(t)=u(t&a 1 ) we have u &1 (s)=v &1 (s)&a 1 ; hece the operator mootoicity of u &1 (s) follows from that of v &1 (s). Set D=C"(&, 0], ad restrict the argumet by &?<arg z<? for z # D. For#>0 defie the sigle valued holomorphic fuctio z # o D by z # =exp #(log z +i arg z), which is the pricipal brach of the aalytic fuctio exp(# log z). Usig this we defie a holomorphic fuctio u(z) o D by k u(z)= ` i=1 (z+a i ) # i, 0=a1 <a 2 <}}}<a k
4 OPERATOR MONOTONE IMPLICIT FUNCTION 333 which is a extesio of u(t). Sice u$(z)= { k ` i=1 (z+a i ) # i=\ : k j=1 # j z+a j+, it is ecessary ad sufficiet for u$(z)=0 i D that k j=1 # j(z+a j )=0. Sice # j >0 ad a j 0, the roots of k j=1 # j(z+a j )=0 are all i (&, 0). Therefore, u$(z) does ot vaish i D. Let us cosider the fuctio w=u(z) as a mappig from the zplae to the wplae. We deote D i the zplae by D z ad D i the wplae by D w. Take a t 0 >0 ad set s 0 =u(t 0 ). Sice u$(t 0 ){0, by the iverse mappig theorem, there is a uivalet holomorphic fuctio g 0 (w) from a disk 2(s 0 ) with ceter s 0 oto a ope set icludig t 0 such that u(g 0 (w))=w for w # 2(s 0 ). We show that for a arbitrary poit w 0 i D w ad for a arbitrary path C i D w from s 0 to w 0, the fuctio elemet (g 0, 2(s 0 )) admits a aalytic cotiuatio (g i, 2(`i)) 0i with `0=s 0 alog C, which satisfies the coditio (C) { g i(w) is uivalet from 2(`i) ito D z, u(g i (w))=w for w # 2(`i). For ` # C let us deote the subpath of C from s 0 to ` by C`, ad let E be the set of poits ` i C such that (g 0, 2(s 0 )) admits a aalytic cotiuatio satisfyig (C) alog C`. Sice E icludes s 0 ad is a relatively ope subset of C, ife is closed i C, the w 0 # E. Thus we eed to show the closedess of E; actually we show that if C`"[`] is icluded i E, sois`. Take a sequece [`] i C`"[`] which coverges to `, ad costruct a family [(g, 2(`))] so that [(g i, 2(`i))] 1i is the aalytic cotiuatio of (g 0, 2(s 0 )) alog satisfyig (C); C`"[`] may be covered by fiite C` umbers of 2(`i), but eve i this case we ca costruct a ifiite umber of 2(`i) as above. If a ifiite umber of the radii of disks 2(`) are larger tha a positive costat, the ` is i some 2(`) ad hece i E. Therefore, we assume that the sequece of radii of 2(`) coverges to 0. The sequece of z :=g (`) is bouded i D z, which is obvious from the form of the fuctio u(z) ad the boudedess of the sequece of `=u(g (`)). Hece it cotais a coverget subsequece [z i ], whose limit we deote by z 0. We prove, by cotradictio, that z 0 is i D z. Assume that z 0 =0; the from the defiitio of u(z), `i =u(z i ) 0; this implies `=0, which cotradicts C` /D w. Assume that arg z i A?; the, because # 1 1 ad a 1 =0, lim arg `i =lim arg u(z i )? or lim u(z i )=0; this implies that C` itersects (&, 0], which cotradicts C` /D w. Similarly assume that arg z i a &?; the C` itersects (&, 0], which cotradicts C` /D w. Therefore, z 0 is i D z. Thus u(z) is cotiuous at z 0. Hece u(z 0 )=lim u(z i )=lim `i =`. Sice u$(z 0 ){0, by the iverse mappig theorem, there is a disk 2(`) ad a holomorphic
5 334 MITSURU UCHIYAMA iverse fuctio g` from 2(`) ito D z such that g`(`)=z 0 ad w=u(g`(w)) for w # 2(`). Sice ` ` ad sice the radii of the disks 2(`) ted to 0, 2(`)$2(`) for >N. Therefore, by (C), we have g`(w)=g (w) for >N ad for w # 2(`). This implies z z 0 ; i fact, for >N z = g (`)= g` (`) which coverges to g`(`)=z 0. Let us joi (g`, 2(`)) to [(g i, 2(`i))] 1iN. The this ew family is a aalytic cotiuatio of (g 0, s 0 ) satisfyig (C). Hece ` # E. Thus we have show that a aalytic elemet (g 0, s 0 ) has a aalytic cotiuatio satisfyig (C) alog every path i D w. By the moodromy theorem, this aalytic cotiuatio is a sigle valued holomorphic fuctio. We deote it by g(w). The g(w) is a holomorphic fuctio from D w ito D z such that u(g(w))=w (w # D w ) ad g(s)=u &1 (s) (0<s<). We fially show that g(w) is a Pick fuctio. We deote the ope lower half plae by 6 &. Set 1= i=1 # i. Sice g(w) is cotiuous, there is a eighbourhood W of s 0 such that Here we ote that g(w)v :=[z:&?1<arg z<?1]. u(v & 6 + )/6 +, u(v & 6 & )/6 &, ad u((0, ))=(0, ). I fact, to see the first iclusio, take z # V & 6 + ; sice 0=a 1 <a i for i>1, z+a i # V & 6 +, ad hece 0<arg(> k i=1 (z+a i) # i )<?, which meas that u(v & 6 + )/6 + ; similarly we ca see the secod iclusio, ad the last equality is clear. From these iclusios, it follows that g(w & 6 + )6 +. I fact, take a arbitrary w # W& 6 + ; the g(w)#v. Assume g(w) 6 + ; the, by the above argumet, we have w=u(g(w)) 6 + ; this is a cotradictio. Because u((0, ))=(0, ) ad u(g(w))=w for w # D w it follows that g(6 + ) & (0, )=<. This ad the coectedess of g(6 + )id z, together with the iclusio <{ g(w & 6 + )/6 +, show that g(6 + )6 +. Hece g is a Pick fuctio. K For 0<:<1, a fuctio u(t)=t 1: satisfies (4). Hece the above theorem says u &1 (s)=s : is operator mootoe o [0, ): which is (1). I the above proof we used the coditio # 1 1. To see that we caot weake this coditio to i r i 1, we give a
6 OPERATOR MONOTONE IMPLICIT FUNCTION 335 Couterexample. Set u(t)=t 12 (t+1). The u$(t)= 1 2 t&12 (3t+1) ad u"(t)= 1 4 t&32 (3t&1). Therefore u"(t)<0 (0<t<13) hece (u &1 )" (s)>0 (0<s< 4 ). 33 Sice a operator mootoe fuctio o [0, ) is cocave, this implies that u &1 (s) is ot operator mootoe o [0, ). Theorem 2.2. Defie a fuctio v(t) by l v(t)= ` j=1 (t+b j ) * j (t&b1 ), b 1 <b 2 <}}}<b l, 0<* j. (5) The, for u(t) represeted by (4), if the coditios { a 1b 1, bj <t * j ai <t # i for every t # R (6) are satisfied, the fuctio, defied o [0, ) by,(u(t))=v(t) (&a 1 t), that is,,(s)=v(u &1 (s)) (0s) is a operator mootoe fuctio o [0, ). Proof. The otatio of the precedig proof is retaied. Set 1 i =# 1 +}}}+# i, 4 i = : b j <a i+1 * j, where a k+1 =. The the secod coditio of (6) is equivalet to 4 i 1 i (1ik). We have see that u &1 (s) admits a aalytic cotiuatio g(w) which is a Pick fuctio. Let us defie a holomorphic fuctio v(z) od z by l v(z)= ` j=1 (z+b j ) * j i the same way that we defied u(z) i the precedig proof. The v(g(w)) o D w is a aalytic cotiuatio of v(u &1 (s)). We eed to show that v(g(w)) is a Pick fuctio. Take w such that 0<arg w<?. Sice, for a i b j <a i+1, 0<arg(g(w)+b j )arg(g(w)+a i ),
7 336 MITSURU UCHIYAMA we have 0<arg v(g(w)) =: * j arg(g(w)+b j ) j =: i : i : * j arg(g(w)+b j ) a i b j <a i+1 : * j arg(g(w)+a i ) a i b j <a i+1 =: (4 i &4 i&1 ) arg(g(w)+a i ) (4 0 =0) i k&1 = : i=1 k&1 : i=1 k = : i=1 k = : i=1 4 i [arg(g(w)+a i )&arg(g(w)+a i+1 )]+4 k arg(g(w)+a k ) 1 i [arg(g(w)+a i )&arg(g(w)+a i+1 )]+1 k arg(g(w)+a k ) (1 i &1 i&1 ) arg(g(w)+a i ) # i arg(g(w)+a i )=arg u(g(w))=arg w<?. This completes the proof. K 3. THE FURTHER CONSTRUCTION OF OPERATOR MONOTONE FUNCTIONS This sectio is a cotiuatio of the precedig sectio. We start with a simple lemma. Lemma 3.1. Let f (=1, 2,...) be strictly icreasig cotiuous fuctios o [a, ) (a # R) with f (a)=0, f ()=, ad let f (t) f +1 (t) for t #[a, ). If f (t) coverges poitwise to a strictly icreasig cotiuous fuctio f(t), the f &1 (s) coverges uiformly to f &1 (s) o every bouded closed iterval [0, b] (0<b<). Furthermore, if a sequece [h ] of cotiuous fuctios o [0, ) satisfies h (t)h +1 (t) ad coverges to a cotiuous fuctio h(t), the h ( f &1 (s)) coverges uiformly to h( f &1 (s)) o [0, b] as well.
8 OPERATOR MONOTONE IMPLICIT FUNCTION 337 Proof. Sice f &1 (s) f &1 &1 +1 (s) f (s), it is easy to see that f &1 (s) coverges poitwise to f &1 (s). Therefore, by Dii's theorem the sequece coverges uiformly o [0, b]. By makig use of Dii's theorem agai, [h (t)] coverges uiformly to h(t) o [a, f &1 1 (b)], ad it is equicotiuous there. Sice a f &1 (s) f &1 1 (b) for 0sb, i virtue of h ( f &1 (s))&h( f &1 (s)) =h ( f &1 (s))&h ( f &1 (s))+h ( f &1 (s))&h( f &1 (s)), we obtai the uiform covergece of h ( f &1 (s)) o [0, b]. K Theorem 3.2. Let u(t), v(t) be the fuctios defied by (4), (5). Suppose that coditio (6) is satisfied. The, if 0;:, the fuctio, o [0, ) defied by,(u(t) e :t )=v(t) e ;t (&a 1 t<) is operator mootoe o [0, ). Proof. We assume ;>0; the proof below is modified whe ;=0. The two fuctios u~ (t)=u(t) :+ \t+ ad v~ (t)=v(t) ;+ \t+ satisfy (6) of Theorem 2.2 for sufficietly large. Thus the fuctio, defied o [0, ) by, (u~ (t))=v~ (t) (t&a 1 ) is operator mootoe. I geeral, if (t) is operator mootoe o [0, ), so is c 1 (c 2 t)(c 1, c 2 >0). The fuctio, defied by, (u (t))=v (t), where u (t)=u(t) \1+: t, v + (t)=v(t) \1+; t + satisfies, (s)= + \;, \\ : & s + +, so that it is operator mootoe o [0, ). By Lemma 3.1,, (s)=v (u &1 (s)) coverges uiformly to,(s) o every fiite closed iterval as. Hece the limit fuctio, is operator mootoe o [0, ). K Usig the above theorem we costruct a oeparameter family of operator mootoe fuctios.
9 338 MITSURU UCHIYAMA Corollary 3.3. Let u(t), v(t) be the fuctios give by (4), (5). Suppose that coditio (6) is satisfied ad that 0;:, 0c1. The, for each r>0 the fuctio, r (s) o [0, ) defied by is operator mootoe., r (u(t) e :t (v(t) e ;t ) r )=(v(t) e ;t ) c+r (&a 1 t<) Proof. Let us represet u(t) v(t) r ad v(t) c+r as (4) ad (5), respectively. The it is easy to see that their expoets satisfy (6). Sice (c+r) ;:+;r, all coditios i the theorem are satisfied. Hece operator mootoicity of, r follows from it. K It is ot difficult to derive the ext corollary from Lemma 3.1 ad Theorem 3.2. Corollary 3.4. Suppose that two ifiite products u~ (t):=` i=1 (t+a i ) # i (a i <a i+1,1# 1,0<# i ) ad v~(t) :=` j=1 (t+b j ) * j, (bj <b j+1,0<* j ) are both coverget o &a 1 t<. If coditio (6) is satisfied ad if 0;:, the the fuctio, defied by,(u~ (t) e :t )=v~(t) e ;t (&a 1 t<) is operator mootoe o [0, ). Moreover, if 0c1 ad r>0, the the fuctio, r (s) o [0, ) defied by is operator mootoe., r (u~ (t) e :t (v~(t) e ;t ) r )=(v~(t) e ;t ) c+r (&a 1 t<) We remark that each family [, r ] of operator motoe fuctios costructed above satisfies the followig relatio,, r (h(t) f(t) r )= f(t) c+r (r>0), where h(t) ad f(t) are appropriate icreasig fuctios.
10 OPERATOR MONOTONE IMPLICIT FUNCTION AN ESSENTIAL INEQUALITY AND AN EXTENSION OF THE FURUTA INEQUALITY The aim of this sectio is to give a essetial iequality which leads us to extesios of (2) ad (3). We eed some tools from the theory of operator iequalities. We adopt the otio of a coectio (or mea) that was itroduced by Kubo ad Ado [10]: the coectio _ correspodig to a operator mootoe fuctio,(t)0 o [0, ) is defied by A_B=A 12,(A &12 BA &12 ) A 12 if A is ivertible, ad A_B=lim t +0 (A+t) _B if A is ot ivertible. I this paper we eed the followig property: AC ad BD imply A_BC_D. From ow o, we assume that a fuctio meas a cotiuous fuctio, I, J represet itervals (maybe ubouded) i the real lie, ad J i the iterior of J. To simplify future proofs, we make a prelimiary remark. Remark. Suppose that sp(a)[a, b]j, ad that f is a fuctio o the iterval J. The for each =>0 there is a affie fuctio p = (t)=ct+d such that c>0, p = (a)=a+=, p = (b)=b&= ad p = (t) coverges uiformly to t o [a, b] as= 0. We have & f( p = (A))&f(A)& 0 (= 0), ad sp( p = (A))[a+=, b&=]. Therefore, to show somethig about f(a) uder a coditio sp(a)j we will ofte assume that sp(a) is i the iterior of J. Lemma 4.1. Let,(t)0 be a operator mootoe fuctio o [0, ). Let k(t) be a oegative ad strictly icreasig fuctio o a iterval I[0, ). Suppose,(k(t) t)=t 2 (t # I). The sp(a), sp(b)i, AB O {,(B12 k(a) B 12 )B 2, A 2,(A 12 k(b) A 12 ).
11 340 MITSURU UCHIYAMA Proof. Let us assume that sp(a), sp(b) are i the iterior of I, soa ad B are ivertible. By makig use of the coectio _ correspodig to,, we have B &12,(B 12 k(a) B 12 ) B &12 =B &1 _k(a)a &1 _k(a) =A &1,(Ak(A))=AB. Here we used B &1 A &1 ad the property of the coectio metioed above. Thus we obtai the first iequality,(b 12 k(a) B 12 )B 2. For geeral A, B, sice p = (A)p = (B) for p = (t) as i the Remark, we ca apply the result that we have just show to p = (A) adp = (B). By lettig = 0, we obtai the first iequality. We ca similarly obtai the secod iequality. K Lemma 4.2. Let [, r : r>0] be a oeparameter family of oegative fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be oegative strictly icreasig fuctios o J. If, for a fixed real umber c : 0c1, the coditio is satisfied, the, r (h(t) f(t) r )= f(t) c+r (t # J, r>0) (7), c+2r (s, &1 r (s))=s 2 (s= f(t) c+r ). Proof. Sice h(t) f(t) r =, &1 r (s), by (7) with 2r+c i place of r,, c+2r (s, &1 r (s))=, c+2r ( f(t) c+r h(t) f(t) r ) This completes the proof. =, c+2r (h(t) f(t) c+2r )= f(t) 2c+2r =s 2. K We call the followig iequality the essetial iequality. Theorem 4.3. Let [, r : r>0] be a oeparameter family of oegative operator mootoe fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be oegative strictly icreasig fuctios o J. If coditio (7) is satisfied for a fixed c: 0c1, the sp(a), sp(b)j i, f(a) f(b)= O {, r( f(b) r2 h(a) f(b) r2 ) f(b) c+r, f(a) c+r, r ( f(a) r2 h(b) f(a) r2 ). (8) Proof. We will oly prove the first iequality of (8). Sice sp(a), sp(b) are i the iterior of J, f(a) adf(b) are ivertible, because f(t) is strictly icreasig. We first obtai (8) i the case 0<r1. By makig use of the coectio _ correspodig to, r, we have
12 OPERATOR MONOTONE IMPLICIT FUNCTION 341 f(b) &r2, r ( f(b) r2 h(a) f(b) r2 ) f(b) &r2 = f(b) &r _h(a) f(a) &r _h(a)= f(a) &r f(a) c+r = f(a) c f(b) c. Thus (8) follows. We ext assume (8) holds for all r such that 0<r. Take ay fixed r such that <r+1. Because (r&c)2, we have, (r&c)2 ( f(b) (r&c)4 h(a) f(b) (r&c)4 ) f(b) (r+c)2. Here we simply deote the left had side by H ad the right had side by K; clearly HK. Set I :=[f(t) (r+c)2 : t # J]. The I[0, ) ad sp(k)i. To see sp(h)i, take a, b i J such that aa, Bb. Sice h(a)h(a)h(b), h(a) f(a) (r&c)2 f(b) (r&c)4 h(a) f(b) (r&c)4 h(b) f(b) (r&c)2. I cojuctio with (7), this shows sp(h)i. It follows from Lemma 4.2 that, r (s, &1 (r&c)2 (s))=s2 for s # I. Thus we ca apply Lemma 4.1 to get which meas, r (K 12, &1 (r&c)2 (H) K 12 )K 2,, r ( f(b) r2 h(a) f(b) r2 ) f(b) c+r. K I the above proof, the strict coditio sp(a), sp(b)j i was ecessary just to say that f(a) ad f(b) are ivertible. Eve if we replace A ad B by p = (A) ad p = (B), respectively, f( p = (A)) f( p = (B)) does ot ecessarily hold, so that we caot weake the coditio to sp(a), sp(b)j. I additio to the coditios of the above theorem, let us assume that f(t) is operator mootoe. The we get Theorem 4.4. Let [, r : r>0] be a oeparameter family of oegative operator mootoe fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be oegative strictly icreasig fuctios o J. If f(t) is operator mootoe, ad if coditio (7) is satisfied for a fixed c : 0c1, the sp(a), sp(b)j, AB= O {, r( f(b) r2 h(a) f(b) r2 ) f(b) c+r, f(a) c+r, r ( f(a) r2 h(b) f(a) r2 ). (9)
13 342 MITSURU UCHIYAMA Proof. Sice p = (A)p = (B), ad sice sp( p = (A)), sp( p = (B))J i,wemay assume that sp(a), sp(b)j i. The operator mootoicity of f(t) esures that f(a) f(b). Hece (9) follows from (8). K We explai why the above theorem icludes the Furuta Iequality. Let p1, ad put f(t)=t, h(t)=t p (0t<). Defie a oeparameter family of operator mootoe fuctios [, r : r>0] by The, r (t)=t (1+r)( p+r) (0t<)., r (h(t) f(t) r )=t 1+r = f(t) 1+r. Thus (7) with c=1 is satisfied. Therefore, from Theorem 4.4 it follows that AB0O (B r2 A p B r2 ) (1+r)( p+r) B 1+r. If q(1+r)p+r, take * such that 1 1+r =* q p+r. The 0<*1, hece by the Lo werheiz iequality (1) we have This is just the Furuta iequality. (B r2 A p B r2 ) 1q B ( p+r)q. Remark. I the above theorems, we assumed that coditio (7) is satisfied for all r>0. However, it is evidet from the above proof that if we assume that (7) is satisfied for r i a iterval (0, :), the (8) ad (9) hold for r #(0,:). Equatios (8) ad (9) are abstract iequalities; however we ca get cocrete iequalities by usig oeparameter families of oegative operator mootoe fuctios o [0, ) i Corollary 3.3. Corollary 4.5. Uder the coditios of Corollary 3.3, suppose A, B&a 1. The v(a) e ;A v(b) e ;B O, r ((v(b) e ;B ) r2 u(a) e aa (v(b) e ;B ) r2 )(v(b) e ;B ) c+r.
14 OPERATOR MONOTONE IMPLICIT FUNCTION 343 Proof. Set J=[&a 1, ), h(t)=u(t) e :t ad f(t)=v(t) e ;t. The the operator mootoe fuctio, r i Corollary 3.3 satisfies (7). Thus, if sp(a), sp(b)j i, we ca apply (8). For geeral A, B, take a arbitrary =>0. Sice A is bouded ad f(t) is strictly icreasig, there is $>0 so that $ f(a+=)&f(a). Moreover, for this $ there is =$>0 so that 0 f(b+=$)&f(b)$. Thus we obtai f(a+=) f(b+=$). Sice sp(a+=), sp(b+=$)j i, we ca apply (8), the let = 0. K Corollary 4.6. Let u(t), v(t) be the fuctios give by (4), (5). Let us assume that a 1 b 1 ad * j <1. For fixed :, c such that 0:, 0c1, defie the fuctio, r (s) o [0, ) by The, r (u(t) e :t v(t) r )=v(t) c+r (r>0). AB&a 1 O, r (v(b) r2 u(a) e aa v(b) r2 )v(b) c+r. Proof. The operator mootoicity of v(t) o[&a 1, ) is clear, ad that of, r (s) o[0,) follows from Corollary 3.3 with ;=0. Thus this corollary follows from Theorem 4.4. K 5. EXTENSIONS OF THE EXPONENTIAL TYPE OPERATOR INEQUALITY OF ANDO I this sectio, we treat oly a ifiite iterval with the right ed poit, so we deote it by J. Recall the iequality (3): for p0, rs>0 AB O (e (r2) B e pa e (r2) B ) s(r+ p) e sb. I this sectio we will obtai a extesio. We cosider (7) uder the coditio c=0, ad deote the fuctio by. r istead of, r. I additio to the coditios of Theorem 4.3 we assume that log f(t) is operator mootoe. The we have
15 344 MITSURU UCHIYAMA Theorem 5.1. Let f(t) ad h(t) be oegative strictly icreasig fuctios o a ifiite iterval J, ad let [. r : r>0] be the oeparameter family of oegative operator mootoe fuctios o [0, ) satisfyig. r (h(t) f(t) r )= f(t) r (t # J ; r>0). (10) If log f(t) is a operator mootoe fuctio i the iterior of J, the sp(a), sp(b)j, AB= O {. r( f(b) r2 h(a) f(b) r2 ) f(b) r f(a) r. r ( f(a) r2 h(b) f(a) r2 ). (11) Proof. We remark that f(t)>0 o the iterior of J, so that log f(t) is welldefied there. We may assume that sp(a), sp(b)(j ) i. Suppose AB. The, by assumptio, log f(a)log f(b). Take ' #(J ) i so that B', ad ote that for every =>0, sp(a+=)(j ) i. We claim that there is a>0 such that f(a+=) a f(b) a. (12) Sice a operator mootoe fuctio o J is cocave, for $ :== dt} d log f(t)>0, &A&+= we have log f(t+=)log f(t)+$ ('t&a&) ad hece log f(a+=)log f(a)+$log f(b)+$. Now, we ote that for every bouded selfadjoit operator X such that X' we have 0< f(') f(x) f(&x&), ad hece " f(x)* &I &log f(x) 0 (* +0). * " Therefore, from the above it follows that f(a+=) * &I f(b)* &I * * for sufficietly small *>0. Thus we have derived (12). Sice. ar (h(t) f (t) ar )= f (t) ar (t # J,0<r), by settig.~ r =. ar, f (t)= f(t) a we have.~ r (h(t) f (t) r )= f (t) r (t # J,0<r).
16 OPERATOR MONOTONE IMPLICIT FUNCTION 345 Therefore, coditio (7) with c=0 is satisfied. Sice f (A+=)= f(a+=) a f(b) a = f (B), ad sice sp(a+=), sp(b)(j ) i, by Theorem 4.3 we have This implies Sice r is arbitrary, for every r.~ r (( f (B) r2 h(a+=) f (B) r2 ) f (B) r.. ar ( f(b) ar2 h(a+=) f(b) ar2 ) f(b) ar.. r ( f(b) r2 h(a+=) f(b) r2 ) f(b) r, ad hece, by lettig = 0, we get (11). Now we explai why this theorem is a extesio of (3). For p, r>0, put. r (s)=s r( p+r) for s0, f(t)=e t ad h(t)=e pt for t # J :=(&, ). The (10) ad all the other coditios of Theorem 5.1 are satisfied. Thus AB implies (e (r2)b e pa e (r2)b ) r(r+p) e rb. By the Lo werheiz theorem, we get (3). r( p+r) Sice. r (s)=s ( p, r>0) is operator mootoe o [0, ) ad satisfies. r ( f(t) p f(t) r )= f(t) r for every fuctio f(t), we ca obtai Corollary 5.2. Let 0 f(t) be a strictly icreasig fuctio o a ifiite iterval J, ad let sp(a), sp(b)j. If log f(t) is a operator mootoe fuctio i the iterior of J, the for r>0, p>0 AB O {( f(b)r2 f(a) p f(b) r2 ) r( p+r) f(b) r f(a) r ( f(a) r2 f(b) p f(a) r2 ) r( p+r). By usig this we ca get a cocrete iequality: let us recall the fuctio u(t) defied by (4) i Sectio 2; sice log(u(t) e :t ) is operator mootoe o the iterior of J :=[&a 1, ), we obtai K Corollary 5.3. If :, p, r>0, the AB&a 1 O {[(u(b) e:b ) r2 (u(a) e :A ) p (u(b) e :B ) r2 ] r( p+r) (u(b) e :B ) r, (u(a) e :A ) r [(u(a) e :A ) r2 (u(b) e :B ) p (u(a) e :A ) r2 ] r( p+r). By applyig this iequality to u(t)=1, we get (3) agai. We ed this paper with a slightly complicated iequality:
17 346 MITSURU UCHIYAMA Corollary 5.4. Let u(t), v(t) be the fuctios defied by (4), (5), ad let a 1 b 1. For fixed :, ;0, defie. r (s) (r>0) o [0, ) by. r (u(t) v(t) r e (:+;r)t )=v(t) r e ;rt (t&a 1 ). The, for each r>0,. r (s) is operator mootoe ad AB&a 1 O {. r((v(b) e ;B ) r2 (u(a) e :A )(v(b) e ;B ) r2 )(v(b) e ;B ) r, (v(a) e ;A ) r. r ((v(a) e ;A ) r2 (u(b) e :B )(v(a) e ;A ) r2 ). ACKNOWLEDGMENTS The author expresses his thaks to Professor T. Ado for readig the origial mauscript ad for givig may commets. He also thaks Professors S. Takahasi, Y. Nakamura, H. Kosaki, T. Hara, M. Hayashi, ad M. Hasumi. Fially, he is grateful to the referees ad Professor D. Saraso for their kid suggestios. REFERENCES 1. T. Ado, ``Topics o Operator Iequalities,'' Lecture ote, Sapporo, T. Ado, O some operator iequalities, Math. A. 279 (1987), R. Bhatia, ``Matrix Aalysis,'' SprigerVerlag, New York, N. Cha ad M. Kwog, Hermitia matrix iequalities ad a cojecture, Amer. Math. Mothly 92 (1985), W. Dooghue, ``Mootoe Matrix Fuctios ad Aalytic Cotiuatio,'' Spriger Verlag, BerliNew York, M. Fujii ad E. Kamei, Furuta's iequality ad a geeralizatio of Ado's theorem, Proc. Amer. Math. Soc. 115 (1992), T. Furuta, AB0 assures (B r A p B r ) 1q B ( p+2r)q for r0, p0, q1 with (1+2r) q p+2r, Proc. Amer. Math. Soc. 101 (1987), T. Furuta, A elemetary proof of a order preservig iequality, Proc. Japa Acad. A 65 (1989), R. Hor ad C. Johso, ``Topics i Matrix Aalysis,'' Cambridge Uiv. Press, Cambridge, UK, F. Kubo ad T. Ado, Meas of positive liear operators, Math. A. 246 (1980), F. Hase ad G. K. Pederse, Jese's iequality for operators ad Lo wer's theorem, Math. A. 258 (1982), E. Heiz, Beitra ge zur Sto rugstheorie der Spektralzerlegug, Math. A. 123 (1951), K. Lo wer, U ber mootoe Matrixfuktioe, Math. Z. 38 (1934), M. Roseblum ad J. Rovyak, ``Hardy Classes ad Operator Theory,'' Oxford Uiv. Press, Oxford, 1985.
18 OPERATOR MONOTONE IMPLICIT FUNCTION K. Taahashi, Best possibility of the Furuta iequality, Proc. Amer. Math. Soc. 124 (1996), M. Uchiyama, Commutativity of selfadjoit opeators, Pacific J. Math. 161 (1993), M. Uchiyama, Further extesio of HeizKatoFuruta iequality, Proc. Amer. Math. Soc. 127 (1999), M. Uchiyama, Some expoetial operator iequalities, Math. Iequal. Appl. 2 (1999),
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