# Operator Monotone Functions which Are Defined Implicitly and Operator Inequalities

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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 oe-parameter family of operator mootoe fuctios. 2. THE CONSTRUCTION OF NEW OPERATOR MONOTONE FUNCTIONS Let us defie a o-egative 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

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 ). 3-3 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 oe-parameter 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.

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 oe-parameter family of o-egative fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be o-egative 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 oe-parameter family of o-egative operator mootoe fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be o-egative 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 oe-parameter family of o-egative operator mootoe fuctios o [0, ), ad J a arbitrary iterval. Let f(t), h(t) be o-egative 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 oe-parameter 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 oe-parameter families of o-egative 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 o-egative strictly icreasig fuctios o a ifiite iterval J, ad let [. r : r>0] be the oe-parameter family of o-egative 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 well-defied 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:

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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